Post by: TrueTears on January 12, 2009, 11:55:27 pm
Two physics students are trying to determine the instantaneous speed of a bicycle 5m from the start of a 1000m sprint. They use a stop watch to measure the time taken for the bicycle to cover the first 10m. If the acceleration was constant and the measured time was 4s. What was the instantaneous speed of the bicycle at the 5 m mark?
Post by: mullums1 on January 13, 2009, 12:01:20 am
x= 10
t=4
u=0
a=?
Find "a" using x= u + .5 at^2
then after finding "a", find v using v= u + at
(sub in your calculated "a" value, t= 4 secs and U= 0)
Post by: TrueTears on January 13, 2009, 12:04:09 am
Post by: mullums1 on January 13, 2009, 12:05:42 am
yeah thanks XD i just didnt know wat that 1000m sprint meant, why did they even put that info? lol
Yeh, they will try to put you off with extra info. Just try to extract the relevation info and assosicate that with the right formula.
Post by: Flaming_Arrow on January 13, 2009, 12:07:31 am
Post by: vce08 on January 13, 2009, 12:07:49 am
You are starting Physics (my favourite subject in the VCE)
I'll be here happy to help you
Post by: Flaming_Arrow on January 13, 2009, 12:10:08 am
your variables are..
x= 10
t=4
u=0
a=?
Find "a" using x= u + .5 at^2
then after finding "a", find v using v= u + at
(sub in your calculated "a" value, t= 4 secs and U= 0)
after you find a you can't sub in t = 4 because that will mean he travelled 10m, we r trying to find speed when its 5m
Post by: TrueTears on January 13, 2009, 12:12:00 am
hey hey
You are starting Physics (my favourite subject in the VCE)
I'll be here happy to help you
haha thanks Table, gonna be fun XD
Post by: TrueTears on January 13, 2009, 02:56:10 pm
Post by: Flaming_Arrow on January 13, 2009, 02:58:02 pm
Post by: vce08 on January 13, 2009, 03:11:59 pm
Post by: danieltennis on January 13, 2009, 03:29:44 pm
Post by: /0 on January 13, 2009, 04:02:42 pm
An old light globe hangs by a wire from the roof of a train. What angle does it make with the vertical when the train is accelerating at
Dealing with accelerating reference frames can be tricky. I found the most intuitive way to go is to let the tension in the rope be
The force
Post by: TrueTears on January 13, 2009, 07:00:57 pm
also:
1. A railway cart of 500kg filled with coal weighing 250kg moves at a speed of 2
i did 2(250 + 500) = 500v + 250 v
v = 2ms^-1 . which is correct answer but how do u know that the coal and the cart travels at the same final speed?
Post by: Flaming_Arrow on January 13, 2009, 07:07:07 pm
Post by: TrueTears on January 13, 2009, 07:09:31 pm
momentum at the start is equal to the momentum at the end
yeah i know that but u got
Post by: TrueTears on January 13, 2009, 10:12:39 pm
2. Two ice skaters, Dean and Melita, are performing an ice dancing routine, in which Dean( with a mass of 70kg) glides smoothly at a velocity of 2 ms^-1 due east towards a stationary Melita(with a mass of 50kg), holds her around the waist and they move off together. During the whole move, no significant friction is applied by the ice. Where is the centre of mass of the system comprising Dean and Melita 3s before impact.
(the answer in the book is, between Dean and Melita, 2.5m from Dean). How do u get that?
3. A car of mass 1500kg travelling due west at a speed of 20 ms^-1 on an icy road collidies with a truck of mass 2000kg travelling in the opposite direction at the same speed, the vehicles lock together after impact.
a) Which vehicle experiences the greatest change in velocity?
b) which vehicle experiences the greatest change in momentum?
c) which vehicle experiences the greatest force?
4. A billard player claims that he can make a red stationary ball move north with a speed of 2.5ms^-1 by striking it with a white ball moving at a speed of 2ms^-1. When challenged he said that according to the Law of Conservation of Momentum the white ball would rebound with a speed of 0.5ms^-1, ie 0.5ms^-1 south. Explain, using calculations, why the player's claim is not correct - even though it is consistent with the Law of Conservation of Momentum.
Post by: /0 on January 13, 2009, 10:27:07 pm
momentum at the start is equal to the momentum at the end
yeah i know that but u gotwhere
is the final speed of the cart and
is the final speed of the coal. The momentum is the same, but the speeds could be different, how do u know here the speed of the coal and cart are the same?
While the coal is in the cart, it has the same speed as the cart. An instant after coal falls out, there are no forces acting on it apart from gravity, so it must have the same speed as the train.
also..
2. Two ice skaters, Dean and Melita, are performing an ice dancing routine, in which Dean( with a mass of 70kg) glides smoothly at a velocity of 2 ms^-1 due east towards a stationary Melita(with a mass of 50kg), holds her around the waist and they move off together. During the whole move, no significant friction is applied by the ice. Where is the centre of mass of the system comprising Dean and Melita 3s before impact.
(the answer in the book is, between Dean and Melita, 2.5m from Dean). How do u get that?
3. A car of mass 1500kg travelling due west at a speed of 20 ms^-1 on an icy road collidies with a truck of mass 2000kg travelling in the opposite direction at the same speed, the vehicles lock together after impact.
a) Which vehicle experiences the greatest change in velocity?
b) which vehicle experiences the greatest change in momentum?
c) which vehicle experiences the greatest force?
2. The formula for centre of mass is
At 3s before impact, the skaters are
Hence, the centre of mass is 2.5m from Dean.
3.
a) Intuitively, the car with less mass should experience a greater change in velocity. You can confirm this by solving
b) The change in momentum can be derived from part c). Since
c) By Newton's third law, both vehicles experience the same force.
4. It breaks the law of conservation of energy. You can derive through conservation of kinetic energy that
and
ANYWAY, Back to the problem
BEFORE:
AFTER:
But
Setting up the standard conservation of momentum equation:
But
So if the masses of the two balls are equal, then the speed of the ball being hit will be equal to the speed of the hitting ball beforehand. The hitting ball loses all its velocity in order to conserve momentum. This is the basis for Newton's pendulum. The reason why, in pool, you sometimes see the hitting ball continue, is because it does not directly hit the stationary ball.
Post by: TrueTears on January 14, 2009, 09:43:35 pm
Post by: TrueTears on January 14, 2009, 10:21:49 pm
Post by: vce08 on January 14, 2009, 10:29:42 pm
Then find out the angle at which when banked where the road will provide a force towards the centre of the circle which will equal this force.
( i think you would have to work it out by doing angle stuffs with the normal reaction force or something)
Post by: TrueTears on January 14, 2009, 10:38:45 pm
Find out the centripetal force required which = mv^2/ryeah so
Then find out the angle at which when banked where the road will provide a force towards the centre of the circle which will equal this force.
( i think you would have to work it out by doing angle stuffs with the normal reaction force or something)
where
so
so
but problem is u cant solve that lol out of range... so it must be wrong equation
so any help?
Post by: vce08 on January 14, 2009, 11:13:02 pm
Post by: /0 on January 15, 2009, 12:10:22 am
Also,
Post by: TrueTears on January 15, 2009, 02:47:02 pm
Post by: TrueTears on January 15, 2009, 03:11:22 pm
A space craft leaves earth to travel to the moon. How far from the centre of the is the spacecraft when it experiences a net force of zero?
(extra info
Earth mass :
Earth's radius of body:
Earth's radius of orbit around sun:
Moon's mass:
Moon's radius of body:
Moon's radius of orbit around earth:
Any help would be much appreciated, ive been stuck on these for a while x.x"" lol
Post by: vce08 on January 15, 2009, 09:20:30 pm
You would get equations like this
Fnet (from moon) = GM(moon)m(satellite)/r(moon)^2
Fnet (from earth = GM(earth)m(satellite)/r(earth)^2
where r(moon) is the distance from moon centre to the satellite
and r(earth) is the distance from the earth centre to the satellite
hence at the point where net force on satellite = 0
m(moon)/r(moon)^2 = m(earth)/r(earth)^2
and since r(moon) = 3.84 x 10^8 - r(earth)
m(moon)/ (3.84 x 10^8 - r(earth))^2 = m(earth)/r(earth)^2
sub in relevant numbers to get the answer
p.s. the answer i think i get is 3.46 x 10^8 m. dunno how correct that is
Post by: TrueTears on January 15, 2009, 09:33:08 pm
Post by: TrueTears on January 16, 2009, 02:55:21 pm
Post by: TrueTears on February 24, 2009, 11:22:09 am
a) Find h
b) calculate the percentage of energy lost during the collision
thanks
Post by: TrueTears on February 24, 2009, 11:28:49 am
A 100g glider on a linear air track is moving at 0.5ms^{-1} towards a stationary 200g glider. Magnets mounted on the glides prevent them from actually colliding, but they undergo an elastic interaction. Determine the velocity of each of the gliders after this interaction.
Post by: /0 on February 24, 2009, 04:38:05 pm
Post by: TrueTears on February 24, 2009, 04:52:34 pm
Elastic means energy is conserved:and momentum is conserved:
Elastic means kinetic energy is conserved.
Post by: TrueTears on February 24, 2009, 04:53:11 pm
Post by: kamil9876 on February 25, 2009, 03:39:53 pm
and also...
A 100g glider on a linear air track is moving at 0.5ms^{-1} towards a stationary 200g glider. Magnets mounted on the glides prevent them from actually colliding, but they undergo an elastic interaction. Determine the velocity of each of the gliders after this interaction.
conservation of momentum:
0.1*0.5=0.1v+0.2v' (v' is the velocity of the 0.2kg glider, whereas v is that of the 0.1kg one)
0.05=0.1v+0.2v'
5=10v+20v'
1=2v+4v'
conservation of total kinetic energy:
0.5*0.1(0.5^2)=0.5*0.1*(v^2)+0.5*0.2*(v'^2)
0.25=(v^2)+2(v'^2)
sub in the equation derived from the momentum thingy:
0.25=(v^2)+2((1-2v)/4)^2
Some quadratic will come out from this. (Note v and v' can be positive or negative because of direction). I think this is the right idea so far.
Post by: kamil9876 on February 25, 2009, 04:17:56 pm
In an elastic collision between 2 objects of massand
show that the speeds of the approach
is equal to the speed of the seperation
. The symbols
represent speeds not velocities.
I did this question twice and I got the result u2+u1=v1+v2. On the third attempt I got the result that the question gives since I realised that this question is worded very shittly(adverb of the adjective "shit"?). Anyways, what's happening is that initially both objects are travelling in the same direction, but once they collide, they move apart. I deduced this from the equations that were given for speed of approach and seperation. At first, I assumed that they move towards eachother and then move apart, but in that case the two equations given for speed of speration and approach would be different, and that's the shitty wording i was refering to earlier.
Take the initial direction to be positive. Hence the conservation of momentum implies:
The negative sign comes from the assumption that object 2 is the one that rebounds back while object 1 continues going in the positive direction. This is because the expression u2-u1 implies that u2>u1 since this is a speed(always positive) Hence object 2 must be comming from behind, catching up to object 1 in order to make the collision. Because object 2 is behind, it is the one that rebounds back.
Kinetic Energy:
momentum equation implies that:
By doing a similair thing to the kinetic energy equation, u can get the result:
By equating the two expressions for
Post by: TrueTears on February 26, 2009, 10:57:00 am
Post by: TrueTears on February 26, 2009, 09:57:27 pm
1. A block of wood of mass 1.5kg is suspended from a fixed point by two light strings so that it is able to swing through the arc of a circle. A bullet of mass 25g, travelling horizontally at 300ms^{-1} embeds in the block. The block swings and rises through a vertical height of h before coming to rest.
a) Find h
b) calculate the percentage of energy lost during the collision
thanks
Post by: TrueTears on February 26, 2009, 09:59:21 pm
Calculate the spring constant for this spring.
Post by: TrueTears on February 26, 2009, 10:03:18 pm
4. Predict the effect on the gravitational force between 2 objects of
a) Halving the distance between them
b) Doubling both masses
Post by: Mao on February 27, 2009, 12:28:07 am
2. you can work this out from the first piece of info alone:
3. (again, not entirely sure, but here's a crack at it, hopefully it's the right answer)
Ug = 1500 * 5 * g = 75000 J
Elost = 0.15 * 75000 = 11250 J = W = F * d
F = 11250 / 0.1 = 112,500 N
(I think...)
4.
halving R:
Doubling both m:
Post by: TrueTears on February 28, 2009, 01:07:48 pm
Post by: TrueTears on February 28, 2009, 01:11:16 pm
A 50kg passenger in a car is held firmly by her seatbelt during a collision. The seatbelt, in stopping her, puts an average force of 12000N on her for
So... Isn't
I =
But my book says its
Post by: TrueTears on February 28, 2009, 02:10:37 pm
Post by: TrueTears on February 28, 2009, 02:14:21 pm
If the distances of these moons from the planet centre are R, 4R and 9R respectively, calculate the ratio of their orbital speeds.
Post by: vce08 on February 28, 2009, 02:28:06 pm
Calculate how far an astronaut would need to be away above the Earth in order for his weight to be 0.01 his weight on the Earth's surface.
we need the value of g to be 0.098
since g = GM/r^2
and the value of M is already known then r can be found out very easily. (this is 5.98 x 10^24 kg)
and G = 6.67 x 10^-11
then the value of r required is
Pretty much 10 times the radius of the earth which can simply be seen by looking at the formula
Post by: Mao on February 28, 2009, 05:44:58 pm
Also this question:
A 50kg passenger in a car is held firmly by her seatbelt during a collision. The seatbelt, in stopping her, puts an average force of 12000N on her fors. Calculate the impulse of the force if the collision time was 0.01s.
So... Isn't
I == 120Ns.
But my book says its= 12Ns, Impulse doesn't change? But the time changed... So shouldn't the impulse also change?
the seatbelt only exerts the force for 0.001s, as the question states.
last one: Three moons around planet X have masses M, 4M and 9M
If the distances of these moons from the planet centre are R, 4R and 9R respectively, calculate the ratio of their orbital speeds.
Post by: TrueTears on February 28, 2009, 07:57:47 pm
and mao for that seatbelt question, how do you work out the impulse? If you don't know the force exerted for the 0.01s...?
Post by: Mao on February 28, 2009, 09:56:12 pm
thanks mao and table,
and mao for that seatbelt question, how do you work out the impulse? If you don't know the force exerted for the 0.01s...?
impulse is related to force, rather than measuring momentum [the overall motion of bodies], it is a measure of how a force contributes to motion.
In this case, the force exerted by the seatbelt is only applied for 0.001 seconds, hence the impulse of this force is 12Ns. It does not concern with what happens to the passenger for the duration of the collision, but only this force itself. The sum of impulse of all forces is the net change in momentum, an impulse is specific to a force, much like how work is specific to a force also. [An example of this is pushing someone down a hill, you do work on that person, gravity also does work on that person, both you and gravity contribute to the overall motion, but neither are concerned with what the other force is doing, or what the final motion will be]
Post by: TrueTears on March 01, 2009, 11:04:59 am
Post by: Mao on March 01, 2009, 06:19:56 pm
ah k thanks mao, so when doing impulse the time is how long the FORCE is applied and it has nothing to do with how long the COLLISION was, so the collision could have been 10 seconds, but the force was only applied for, then impulse would still be 12Ns?
correct, the impulse of the force would still be 12Ns.
Post by: TrueTears on March 01, 2009, 07:50:36 pm
Post by: Over9000 on March 05, 2009, 08:13:24 pm
Post by: MattPritchard on March 09, 2009, 05:36:44 pm
May sound like a stupid question but how did you get from the 2nd line to the 3rd?
Also what does the
Post by: TrueTears on March 09, 2009, 05:38:36 pm
in this case because its 1 on something, it will be inversely proportional to
Post by: Flaming_Arrow on March 09, 2009, 05:39:58 pm
May sound like a stupid question but how did you get from the 2nd line to the 3rd?
Also what does thesymbol mean?
Post by: Flaming_Arrow on March 09, 2009, 05:40:23 pm
"proportional to"
in this case because its 1 on something, it will be inversely proportional to
damit beaten :D :D
Post by: MattPritchard on March 09, 2009, 05:55:38 pm
Post by: TrueTears on March 17, 2009, 06:30:34 pm
A: the acceleration is always constant and in the same direction
B: the acceleration is at right angles to the direction of motion
C: The magnitude of the acceleration is constant
D: the acceleration is proportional to the square of the instantaneous speed
I said the answer was A, is that right?
Thanks :)
Post by: kamil9876 on March 17, 2009, 06:44:30 pm
A is wrong because the acceleration direction does change! remember that the acceleration is in the opposite direction of
Post by: TrueTears on March 17, 2009, 06:58:46 pm
and in projectile motion, acceleration is always down towards the earth ?
Post by: Damo17 on March 17, 2009, 07:02:07 pm
but in circular motion, doesn't acceleration always point towards the centre?
and in projectile motion, acceleration is always down towards the earth ?
I would say it changes direction from south at the top to east and then north and then west as it goes around.
Post by: kamil9876 on March 17, 2009, 07:03:18 pm
Yes you are right that it's always to the centre, but always to the centre is not the same direction!
Think about it in terms of vectors:
then
So the acceleration is not constant. Constant in magnitude yes, but not constant in direction since this vector changes over t, just like position changes over t.
Post by: kamil9876 on March 17, 2009, 07:04:49 pm
but in circular motion, doesn't acceleration always point towards the centre?
and in projectile motion, acceleration is always down towards the earth ?
I would say it changes direction from south at the top to east and then north and then west as it goes around.
Yep, 'always to the centre' does not mean 'always the same direction' so when it is at the top, the acceleration is south, whereas at the west the acceleration is east. So even tho 'always to the centre' is true doesnt mean 'always the same vector'(direction)
Post by: kaanonball on March 17, 2009, 07:19:06 pm
but in circular motion, doesn't acceleration always point towards the centre?
and in projectile motion, acceleration is always down towards the earth ?
in projectile motion the Vertical acceleration is due to gravity (9.8m/s/s downwards) i.e. as you said towards the center of the earth but their also exists a Horizontal acceleration which is always 0
Post by: TrueTears on March 17, 2009, 07:21:33 pm
thanks kamil and damo
yeah just didn't include that in my passage :)but in circular motion, doesn't acceleration always point towards the centre?
and in projectile motion, acceleration is always down towards the earth ?
in projectile motion the Vertical acceleration is due to gravity (9.8m/s/s downwards) i.e. as you said towards the center of the earth but their also exists a Horizontal acceleration which is always 0
Post by: TrueTears on March 24, 2009, 06:08:58 pm
Do you draw the forces which parts of the MACHINE exerts, or the forces that is exerted ON the machine?
for example, if you had a string and 2 people pull on the 2 ends in opposite directions. Then the string would be in tension and there would be forces acting towards the right and left (where the 2 people pulling it), but because of newtons 3rd law, the string is also trying to pull INwards, so what forces do I actually label on this diagram? (Because our teacher stressed that when something is in tension then the material in tension is pulling inwards and if it is in compression then its actually pushing outwards. However, the forces being EXERTED ON the material for tension is 'stretching' it longer and for compression it's 'pushing it inwards'.) So yeah, what forces do I put in here? Thanks
(http://img21.imageshack.us/img21/8764/trusses.jpg)
Post by: Mao on March 24, 2009, 08:19:26 pm
Post by: TrueTears on March 24, 2009, 08:21:28 pm
Post by: Mao on March 24, 2009, 08:27:15 pm
question says: "Draw arrows on the diagram to show which parts of the machine are in tension and which are in compression." How do you interpret that?
oh, as in show the forces acting on the cable, supporting frame, etc [so parts in tension have forces acting outwards, parts in compression have forces acting inwards]
Post by: TrueTears on March 24, 2009, 08:28:13 pm
cool thanks.question says: "Draw arrows on the diagram to show which parts of the machine are in tension and which are in compression." How do you interpret that?
oh, as in show the forces acting on the cable, supporting frame, etc [so parts in tension have forces acting outwards, parts in compression have forces acting inwards]
Post by: TrueTears on April 08, 2009, 07:50:50 pm
(http://img24.imageshack.us/img24/9311/structure.jpg)
Post by: TrueTears on April 08, 2009, 07:57:22 pm
thanks guys
(http://img24.imageshack.us/img24/9007/structure2.jpg)
Post by: Mao on April 09, 2009, 08:40:37 pm
The maximum torque by the weight is experienced when the arm is horizontal. Since the truck is required not to tip, we hence have rotational equilibrium:
Post by: dejan91 on April 10, 2009, 05:28:07 pm
This is assuming stress is loaded at a constant rate:
For a) He would feel a constant increase in length - No sudden changes.
For b) He would feel the line extending minimally for a brief time, then at a quickening rate (reaching a maximum in between the arrows of graph b.), and then once again extending at a depreciating rate.
You can confirm this by considering Young's Modulus,
From this,
For c) He would feel the line change length at an exponential rate.
Now, I'm not sure if you were meant to answer the question in terms of tension or change in length or something else, but I hope this helps :)
Post by: TrueTears on May 03, 2009, 06:53:59 pm
(http://img14.imageshack.us/img14/9551/physicssinewave.jpg)
1. When the student measures the input and output voltages of a simple amplifier on a two-channel oscilloscope, the graphs(above) are displayed. The peak - to - peak voltage of the output is 2V. What is the voltage gain of the amplifier, based on the information given? (indicate in your answer whether the amplifier is inverting or non-inverting.)
so
so
so gain is
2. What are the DC and AC(peak) components of the input voltage?
Erm, I don't really get what this question is asking... 'components' ? 'AC (peak)' ? peak to peak or just the peak?
Thanks in advance :)
Post by: TrueTears on May 03, 2009, 08:00:44 pm
Sketch the graph of DC input voltage against DC output voltage for this amplifier between
If anyone could do a rough sketch on paint or something and explain how they did it, that would be very much appreciated :)
Post by: dejan91 on May 03, 2009, 08:21:29 pm
(http://i40.tinypic.com/2v0cp3m.jpg)
Post by: dejan91 on May 03, 2009, 08:23:09 pm
Post by: TrueTears on May 03, 2009, 08:30:34 pm
(http://img261.imageshack.us/img261/5746/q445.jpg)
But I don't get why.
Post by: dejan91 on May 03, 2009, 08:36:54 pm
Inverting amlpifier, so basically exactly what I did except negative gradient.
Post by: TrueTears on May 03, 2009, 08:38:50 pm
Post by: dejan91 on May 03, 2009, 08:39:42 pm
I don't see any reason why it should be 1.5. I'm gonna going for the old 'the book is wrong' thing, unless you maybe obtained or read a wrong gain.
Post by: TrueTears on May 03, 2009, 08:42:38 pm
Post by: TrueTears on May 03, 2009, 10:09:02 pm
thanks Mao :P
Post by: TrueTears on May 08, 2009, 10:49:06 pm
The sketch shows an electro-optical system that allows sound to be transmitted over a distance via a fibre-optical cable, using light.
Explain what is happening at each device
Can anyone check if what I said is correct and if I missed anything please point it out :P
The input analog signal(sound) from the microphone is amplified at the first amp which acts as a transducer and converts the sound analog signal into an electric signal. At the LED, the electrical signal is converted into optical signal by the LED, ie modulation occurs. The light from the LED then acts as the carrier wave which contains all the information from the original analog signal. This optical signal then travels through the optical-fibre wire where it reaches the photodiode at Q. At Q, the optical signal is converted back into an electrical signal, ie demodulation occurs. It is then amplified in the amplifier and then exits through the loudspeaker as sound
Post by: Mao on May 08, 2009, 11:03:44 pm
2. amplifier amplifies the electric signal from the microphone (very small)
3. amplified signal is transmitted by electro-optic transducer (LED)
4. optic signal is received by opto-electric transducer (photodiode)
5. signal from the opto-electric transducer is amplified
6. amplified signal is transformed from electric wave to sound waves via loudspeakers
Post by: TrueTears on May 08, 2009, 11:12:12 pm
Post by: TrueTears on May 08, 2009, 11:30:34 pm
You alter the amplitude of the carrier wave, in this case the light from the LED, to match that of the electrical signal being fed into the LED. So then the LED can 'carry' the electric signal and transfer it through the optical fibre to the photodiode.
And for the demodulation does it happen like this?
At the photodiode the light, which is acting as the carrier wave, is removed and all we are left is the electric signal which is then amplified in the amplifier etc etc.
Thanks again!
EDIT: Oh and also are capacitors in the unit 3 course for physics?
Post by: kurrymuncher on May 08, 2009, 11:57:40 pm
Oh and also for the modulation that takes place at the LED does it happen like this?
You alter the amplitude of the carrier wave, in this case the light from the LED, to match that of the electrical signal being fed into the LED. So then the LED can 'carry' the electric signal and transfer it through the optical fibre to the photodiode.
yeah, thats how it works.
The modulated signal will be sent along the optical fibre cable. At the end, the modulated signal is detected by a photodiode or photoresistor. The photodiode or photoresistor converts the light back into a fluctuating current in the circuit and so on....
I dont think there are capacitors in unit 3. not sure though
Post by: TrueTears on May 09, 2009, 12:06:21 am
And yeah when it converts the light, does it just remove the carrier wave (light) so all we are left is just the original electrical signal? Hence the term "demodulation"
Post by: kurrymuncher on May 09, 2009, 12:18:21 am
I dont think that makes sense
also,did you know that a machine that can modulate and demodulate a signal is called a Modem. lol
Post by: TrueTears on May 09, 2009, 12:21:34 am
Yep. The carrier wave is removed and the photodiode changes the signal into a current. This photodiode will be connected to a circuit, therefore the current from the signal will cause a variation in Vout, and somehow they can detect the original signal.You are one mad dog
I dont think that makes sense
also,did you know that a machine that can modulate and demodulate a signal is called a Modem. lol
changes to TT female version <3
Post by: Mao on May 09, 2009, 10:08:35 am
LED is a diode, once reached the activation voltage, power output varies with current. That is, the electric signal (current variation) loses some energy (voltage) to release photons at the metal surface on the LED. The intensity of the light emitted is proportional to the power output hence current, i.e. the peaks in electric signal becomes peaks in light intensity. the carrier wave is amplitude-modulated.
At the photodiode (reverse biased), photons provide energy for the electrons to cross the depletion region. Hence the current is proportional to the number of photons (i.e. intensity of light). Peaks in light intensity becomes peaks in current, thus the signal is demodulated.
I'm not keen on using phrases such as 'carrier wave is removed' and things of that like.
Post by: TrueTears on May 09, 2009, 01:39:13 pm
Post by: TrueTears on May 09, 2009, 03:01:20 pm
We had to do a Practical SAC to work out the Young's Modulus of a fishing string. I have attached my data and everything that is required is in it.
The question I want to know is how do I work out the percentage error for each variable? Such as Force, New length, change in L, stress, strain. The ruler I used to measure the original length of string is just your normal standard ruler.
Our teacher has never taught us how to do it but he says you must put the percentage error for each variable, how do you work that out?
Many thanks!
Post by: Over9000 on May 09, 2009, 04:23:36 pm
U get original length of string is 150mm say for example
And ruler manufacturer states accurate
so then you have
So that's your error for that.
Then you have the weights we used
Say manufacturer states accurate
Where each was 100
so you have
Then u add both to have 5.33% error margin for those 2 factors
Post by: TrueTears on May 09, 2009, 06:05:11 pm
Post by: Over9000 on May 09, 2009, 07:09:28 pm
ahhh okay thanks over9000, how do we know the uncertainty for the certain equipments? It was never stated on them or anything :SAsk your teacher
Post by: TrueTears on May 11, 2009, 06:49:13 pm
This a linear optoisolator. The current travelling through the circuit with the LED is denoted by
I'm just wondering what does the graph of
Many thanks!
Post by: TrueTears on May 15, 2009, 04:18:11 pm
A few new questions:
1. A rower on the Yarra is able to reach a velocity of 10
a) With what velocity is the rower moving relative to the otter when rowing upstream?
b) What is the velocity of the rower relative to the riverbank?
c) What is the velocity of the otter relative to the rower?
d) What is the velocity of the rower relative to the riverbank when rowing downstream?
2. When Newton published his laws of motion he identified some assumptions about space and time that he assumed to be true and that underpin all his laws. Which one or more of the following would best describe his assumptions?
A. Time passes uniformly for observers in any frame of reference
B. The measurement of space is always relative to the observer's frame of reference
C. Time is not the same for all observers at very high speeds
D. Space is absolute therefore 1 m will be 1 m for all observers despite being in different reference frames
E. Time can only be measured by a clock
F. Distance can only be measured by a ruler
Many thanks!
Post by: kamil9876 on May 15, 2009, 05:12:34 pm
a.)The otter "floats" which implies that it is stationary relative to the water. Hence we will pick a reference frame where the water is stationary. Relative to the water the guy rows at 10m/s. Hence in our reference frame the otter has speed of 0, while the rower has 10m/s. Hence 10m/s.
b.) The answer could be 8m/s or 12m/s, depending on whether he is rowing in the direction to the flow of water or against it. I think we are to assume that he is going upstream since part a says so. Hence 8m/s. This is easily done with vector addition. In one second he would travel 10m in the river's frame however because the river would move 2m back in that one second the displacement in the frame of the riverbank is 10+(-2)=8m in one second.
c.) 10m/s. (refer to part a)
d.) 12m/s as explained in part c.
Post by: kamil9876 on May 15, 2009, 05:18:30 pm
A,B(depends on what u mean by "space"),D.
B - the measurement of distance isn't any different in different frames, however the measurement of space co-ordinates is (and I think Galilleo made this explicit before Issac, hence "gallilean relativity").
Imagine the riverbank scenario again, the guy may be on the boat and say his bag is at co-ordinates (1,0,0) in his frame. In the riverbanks frame say the co-ordinates of the bag are the same at t=0, (1,0,0). However because the bag is moving at 12m/s after one second the bag will be at (13,0,0) in the river bank's frame. However in the rower's frame it is still at (1,0,0)
Post by: /0 on May 24, 2009, 05:35:38 am
Post by: kamil9876 on May 24, 2009, 09:54:00 am
Post by: TrueTears on June 06, 2009, 01:40:58 pm
A 50V DC voltage is applied across the terminal P-Q what is the voltage across the terminals X-Y?
Thanks
Post by: arthurk on June 06, 2009, 01:56:34 pm
2000 and 3000 ohms in series then calculate the voltage out across the 3000 ohm resistor
not sure if thats right though
Post by: TrueTears on June 06, 2009, 02:05:41 pm
30V? seeing as the 1000 ohm resistor is in parallel thus is negligible when determinging voltage outyeap thanks just got it.
2000 and 3000 ohms in series then calculate the voltage out across the 3000 ohm resistor
not sure if thats right though
Post by: naved_s9994 on June 06, 2009, 02:20:37 pm
Post by: dejan91 on June 06, 2009, 03:14:18 pm
Post by: Damo17 on June 06, 2009, 03:41:03 pm
Post by: TrueTears on June 06, 2009, 03:42:08 pm
Also this one:
The yield strength of mild steel is
What does it mean by safety factor? Because sometimes in the questions I've done in the past, I divide the stress with the safety factor and sometimes multiply the force applied with the safety factor. What is the actual systematic way of understanding "safety factor"?
Post by: kurrymuncher on June 06, 2009, 04:03:02 pm
the safety factor is used to pretty much ensure that the object does not reach its maximum stress and then fail, they take a safety factor so that can safely apply a force without any danger of failing.
this of the top of my head, hopefully sum1 can confirm
Post by: Damo17 on June 06, 2009, 04:04:22 pm
Thanks. ^^^
Also this one:
The yield strength of mild steel is. what is the minimum diameter of mild steel rod that can support a 75kg man with a safety factor of 3 before breaking?
What does it mean by safety factor? Because sometimes in the questions I've done in the past, I multiply the stress with the safety factor and sometimes the force applied. What is the actual systematic way of understanding "safety factor"?
I'm not doing structures but I'll try and help as much as possible. Safety factors, usually from 4 to 10, are a way of accounting for any unknown weaknesses in the materials actually being used. For example, if a steel cable is capable of supporting a load of 500 kg, the manufacturer might consider a safety factor of 5 is appropriate, and so recommend that it be used for masses no greater than 100 kg.
So for your question:
so max weight the steel can hold is 75kg* 3=225kg
Then the max force it can handle is 225 *10=2250N
You are trying to find the diameter but first you need the Area.
Post by: Damo17 on June 06, 2009, 04:05:33 pm
If you are given the max stress of an object and a safety factor of 3. You then divide that max stress by 3, and that is the new stress of the object. You then use this new stress value with the weight of the 75kg man, to find the area etc etc.
Wants it before breaking therefore you use the max stress.
Post by: kurrymuncher on June 06, 2009, 04:07:11 pm
Post by: TrueTears on June 06, 2009, 04:11:16 pm
Safety factor:
A multiplier applied to the calculated maximum force or a number that the tensile strength is divided by in order to determine the safe working load.
Both will give same answer :)
Post by: Damo17 on June 06, 2009, 04:14:35 pm
Post by: methodsboy on June 06, 2009, 04:52:56 pm
Recommended max stress (given in question) = stress(use this for other calculations) / safety factor
ONLY USE when it states the "recommended max stress"
Post by: methodsboy on June 06, 2009, 04:54:17 pm
Thanks guys!!! I got itwhere safe working load = allowable stress =]
Safety factor:
A multiplier applied to the calculated maximum force or a number that the tensile strength is divided by in order to determine the safe working load.
Both will give same answer :)
Post by: dejan91 on June 06, 2009, 06:18:23 pm
Post by: TrueTears on June 06, 2009, 06:42:34 pm
Thanks guys!!! I got itOk there's another question but using what I said before it doesn't work...
Safety factor:
A multiplier applied to the calculated maximum force or a number that the tensile strength is divided by in order to determine the safe working load.
Both will give same answer :)
A sample of rope fractures under a tension at 50 000N. The builder wishes to place a safety factor of 2.5 on the rope. The cross section area is
Using
Quote from: TrueTears
A multiplier applied to the calculated maximum force
stress
But answer is
However if you do
Quote from: TrueTears
a number that the tensile strength is divided by in order to determine the safe working load.You get the right answer. Ie:
stress =
then
Why doesn't both ways work here?
Is there a bullet proof way to do these safety factor questions?
Why did both ways work for this question?
Quote from: TrueTears
The yield strength of mild steel isPa. what is the minimum diameter of mild steel rod that can support a 75kg man with a safety factor of 3 before breaking?
Post by: arthurk on June 06, 2009, 06:55:05 pm
Post by: TrueTears on June 06, 2009, 06:56:51 pm
well the second way is definitely correct seeing as the maximum stress coinciding with a safety factor must be below the stress at the fracture point which method 1 clearly did not provide.Yeah, so how do I know which method? What's the difference in the 2 problems which didn't allow both methods to work?
Post by: Damo17 on June 06, 2009, 07:05:22 pm
Post by: kamil9876 on June 06, 2009, 07:14:14 pm
Quite obviously, twice as much Maximum stress means its twice as safe, and half as much applied stress also means twice as safe. The more stress you apply the less safe it is.
In this situation you are after the denominator (Actual stress you plan to apply)
Post by: kamil9876 on June 06, 2009, 07:18:29 pm
Post by: TrueTears on June 06, 2009, 07:31:18 pm
Post by: TrueTears on June 06, 2009, 09:42:56 pm
Explain what would happen to the light globes (Lighter, brighter, short circuit, burnout ?) if only
Thank you all once again!
Post by: naved_s9994 on June 06, 2009, 10:11:29 pm
I.e, if resistance equal - then therefor brightness equal.. Switch A will NOT make a difference. The Flow of CURRENT does.
So it all in the end means , it transforms into a parralel circuit and it all depends on the resistance of each globe
Post by: Mao on June 07, 2009, 02:41:07 am
ULTIMATE stress: the point when the material breaks
MAXIMUM stress: ULTIMATE/safety factor
Post by: Mao on June 07, 2009, 02:48:47 am
The intensity of the other globes, however, depend on the type of source (fixed or variable current), and the type of globe (fixed current drawing or non-fixed).
Post by: TrueTears on June 07, 2009, 02:56:20 am
Post by: TrueTears on June 07, 2009, 02:56:59 am
(http://img190.imageshack.us/img190/6160/graphattachedtolaserdio.jpg)
(Soz for such a bad diagram, but all the sin curves are the same, they are simply just shifted up, the top right and bottom left diagram have exactly the same sin curve except the axis are different, one is brightness vs time the other is voltage vs time.)
The top left diagram is the signal that is being sent into W
At X the signal would be the top right diagram.
At Y what would the signal be like?
The answer is the bottom left signal. But I thought it should just be the original, ie, top left? Why is it the bottom left signal? Wouldn't that have a higher max voltage then the original?
Isn't the laser just using light as the carrier wave and transporting the original signal across the optical wires? So why does the entire original signal get shifted upwards (as indicated in the bottom left diagram)?
Post by: Mao on June 07, 2009, 03:04:38 am
Modulation/demodulation is not a process where the input signal is 'exactly copied' to the output device. Only the frequency and amplitude are preserved.
Post by: jules on June 07, 2009, 12:36:28 pm
Post by: Flaming_Arrow on June 07, 2009, 12:44:06 pm
Post by: TrueTears on June 07, 2009, 01:55:41 pm
Bias of the diode prevents current flow from smoothly alternating (due to switch-on voltage). Hence to preserve the signal, the demodulated signal at Y is the bottom left.Ah I see, thanks Mao! So what type of signal would you expect to see at Z? (This is not part of the actual question but I'm just wondering)
Modulation/demodulation is not a process where the input signal is 'exactly copied' to the output device. Only the frequency and amplitude are preserved.
Post by: dejan91 on June 07, 2009, 02:40:54 pm
Also, does it really matter if you put in a positive or negative gain? Apparently they accept both. Strictly speaking, if the amplifier is inverting, and you didn't have a graph to work with (just the gain formula, nor did they tell you it was an inverting amplifier) you should be able to plot a Vout graph just as well as if you had the vin graph. If you had the gain as positive, you would get the wrong graph.
Post by: NE2000 on June 07, 2009, 02:44:58 pm
Bias of the diode prevents current flow from smoothly alternating (due to switch-on voltage). Hence to preserve the signal, the demodulated signal at Y is the bottom left.Ah I see, thanks Mao! So what type of signal would you expect to see at Z? (This is not part of the actual question but I'm just wondering)
Modulation/demodulation is not a process where the input signal is 'exactly copied' to the output device. Only the frequency and amplitude are preserved.
Amp would include decoupling, which will result in it being back to being centred around zero I would imagine. I made that mistake on that question too, but if you think about a photodiode circuit, you can't produce negative voltage when the brightness is low, if the brightness is low then the current through the circuit is zero (or a bit more perhaps) and the voltage out will be around zero as well.
Post by: naved_s9994 on June 07, 2009, 09:29:36 pm
Post by: Damo17 on June 07, 2009, 09:35:40 pm
Are questions like (A) relative to (B) still on course?
no, relative motion is out.
Post by: naved_s9994 on June 07, 2009, 09:36:40 pm
Post by: TrueTears on June 15, 2009, 11:31:54 pm
Post by: Mao on June 16, 2009, 09:53:44 am
[note, a stationary charge experiences no force because it does not generate any magnetic field. only a moving charge can generate a magnetic field.]
Post by: mark_alec on June 16, 2009, 10:46:48 am
So if you were sitting on an electron moving through a magnetic field, what would you observe?An electric field with the correct direction and magnitude so that the force you experience would be the the same as the force observed from someone who is stationary with respect to the magnetic field.
Post by: TrueTears on June 16, 2009, 09:23:35 pm
Another one:
When current is connected to solenoid containing 2 iron rods side by side the 2 rods move apart, why does this happen?
Post by: Mao on June 16, 2009, 10:47:26 pm
Post by: TrueTears on June 17, 2009, 06:57:32 pm
Post by: methodsboy on June 17, 2009, 07:07:14 pm
Post by: TrueTears on June 17, 2009, 07:08:19 pm
Post by: TrueTears on June 17, 2009, 07:40:08 pm
(http://img191.imageshack.us/img191/2291/lenzslaw.jpg)
How do you work out the the direction of the change in magnetic flux? (In this case upwards)
And when it says "the induced magnetic field will then be down at this point" At what "point" is it down? How do you work out the direction of the current?
Thanks!
Post by: Mao on June 17, 2009, 07:57:48 pm
i.e. for the above case figure a
in that position, as it rotates flux to the right decreases
hence the direction of magnetic field generated by the induced current points to the right [since it is a coil, treat it as a solenoid].
by the right hand rule, current flows from A to B.
Post by: TrueTears on June 17, 2009, 08:07:31 pm
Also when it says in part (a) the magnetic flux is entering the loop from above and in part (b) it enters from below. What does "entering from above" and "entering from below" mean?
Post by: Mao on June 17, 2009, 08:12:01 pm
Ah cool thanks!
Also when it says in part (a) the magnetic flux is entering the loop from above and in part (b) it enters from below. What does "entering from above" and "entering from below" mean?
it means a weird explanations that you should pay no attention to :)
always consider it as a change in flux.
in part b, as the coil rotates there is a net increase in flux to the right, hence by lenz's law, the induced current will have a magnetic field opposing this, i.e. to the left.
by the right hand screw rule, current flows from C to D.
Post by: TrueTears on June 17, 2009, 08:41:30 pm
Post by: TrueTears on June 19, 2009, 01:29:42 am
Thanks.
Post by: Mao on June 19, 2009, 11:53:55 am
:)
Post by: TrueTears on June 19, 2009, 01:27:33 pm
Yeap, that's what I did and I got
Post by: cns1511 on June 19, 2009, 06:17:51 pm
-Mao
Post by: TrueTears on June 19, 2009, 06:22:43 pm
stupid shittycaranda XD
Post by: TrueTears on June 20, 2009, 02:10:26 pm
Thanks.
Post by: NE2000 on June 20, 2009, 04:10:22 pm
Just out of interest, when an electromagnet is formed normally it would be iron and you would put copper wires around it with a current flowing through it, some of the electron domains in the iron would line up with the magnetic field hence become an electromagnet. But why can't all metals become electromagnets? (Or not work as well as iron). Don't all metals have electrons?
Thanks.
I think it something to do with iron, nickel and cobalt having electrons that tend to align in the same direction. All metals have electrons, all electrons spin and all spinning electrons generate magnetic fields. In the case of many metals these magnetic fields cancel out very well so the metals don't have the ability to be turned into electromagnets. In the case of these three metals nearby on the periodic table the magnetic field seems to add together in these domains which allows them to be permanently or temporarily magnetised. It's something like that anyway.
Post by: TrueTears on June 20, 2009, 04:15:01 pm
Post by: TrueTears on June 20, 2009, 09:03:28 pm
Many thanks!
EDIT: Okay, I think I know what they do now, can anyone check if my understanding of these 2 devices are correct?
1. Split rings
Split rings are used in DC motors. They are connected to the coils so they can spin with them. Their purpose is to connect to the opposite terminal, so if one half of the split ring was connected to the -ve terminal and the other half was connected to the +ve then each time the coil is rotated into 90 degree position (ie, when it's vertical), the 2 halfs of the split ring would now be connected to its opposite terminals. As a result the current is now reversed so the net torque is still in its original direction (If it was clockwise, it would continue to be clockwise) hence the coil would still rotate in its original direction.
2. Slip rings
Slip rings are used in generators. They act like "terminals" of a battery. During a period of time when the coil is being rotated, one slip ring (let's call this P) would act as the +ve terminal and the other slip ring (let's call it Q) would act as the -ve terminal. Then as the coil continues to rotate, P would act as the -ve terminal and Q would act as the +ve terminal. What kind of terminal the 2 slip rings would act can easily be determined using Lenz Law and finding the direction of current flow (a diagram would be much easier to explain this). This means the current is reversed after a certain period of time hence it also explains why generators only produce AC current (due to the reversal of current).
Thanks :)
Post by: TrueTears on June 20, 2009, 09:17:27 pm
Okay, just another question which is related between flux and emf.
The graph attached shows that when flux is a maximum then emf = 0.
I understand this perfectly through a mathematical sense,
But what I don't understand is from a more intuitive angle. My basic understanding of emf is (for example) you have a metal rod and it falls through a magnetic field, the electrons are pushed to one end and the other end would be positively charged, hence we have a clear -ve and +ve end, therefore if we connected wires to this rod, a current would run through the wire from the +ve end of the rod to the -ve end. In a way this rod acts like a battery (with a +ve and -ve end with conventional current running from +ve to -ve end). Furthermore, a "large" emf value means that more electrons are pushed to that side of the rod and hence there is a larger difference between charge, a "smaller" emf value means that there is less electrons on that side of the rod hence the rod is not as "polar". 0 emf means that the rod does not have 2 distinct charged ends, the electrons are evenly distributed through the rod.
Now compare this with the generator diagram attached (top right). Why is it when flux is maximum the emf is 0? In other words, why is it that the electrons in the coil are evenly distributed when the flux is max? (max flux occurs when the coil is 'perpendicular' to the magnetic field). Furthermore, why is it when the flux is 0 the emf is a max? ie, when the coil is 'parallel' to the magnetic field, the electrons in the coil are very far from the +ve charges hence creating a "large" emf.
I understand all of that in a mathematical sense, yes when the flux is max the gradient is 0 so emf is 0, when the flux is 0, it has a max gradient hence emf is max. But how do I go about understanding this in a more "intuitive" sense? As in why emf is related to the gradient of a flux - time graph?
My understanding may be totally wrong, so any help would be appreciated.
Thanks everyone! :)
Post by: /0 on June 20, 2009, 09:37:40 pm
Let's say the wire is rotating clockwise.
Using "magnetic force on charges in the wire", when flux is max, the wire is making a 90 degree angle with the horizontal, so the top of the wire is instantaneously moving horizontally right, parallel to the magnetic field. At the same time, all other parts of the wire are also moving either instantaneously right or left (as in circular motion), parallel to the field, so the magnetic force is zero and the emf is zero.
I guess if you draw a diagram and let
EDIT: finally got it
(http://img221.imageshack.us/img221/7628/generator.jpg)
This is a cross-sectional viewing of the rectangular wire.
Let's say a positive charge is moving at a speed
By alternate angles,
so
So the force on the positive charge is
Now let L be the width of the wire (coming out of the page).
Since
So the emf is
And the flux in the top half of the wire is given by
(If we found the flux in both the top and bottom halves, we would need to consider the other positive charge on the opposite side of the wire)
But, since
And since
Using "Lenz's Law", look at diagram (a) and (b) on page 269. Just imagine that the wire's angle with the horizontal in diagram (b) is the same as the wire's angle with the horizontal in diagram (a). Therefore, the change in flux as we go from position (b) to position (a) should be 0, since they both have the same flux, so the average emf is 0. Now let the angle with the horizontal,
i.e.
Post by: TrueTears on June 20, 2009, 09:46:48 pm
what about when the square coil in the diagram is parallel to the magnetic field (which is going from left to right)
Applying your reasoning, instantaneously the side of the coil would be falling vertically down (consider the "right" side of the coil when it is parallel). Which means the electrons are falling down at that instant, use the Left hand slap rule, fingers points left to right, thumb points down in the direction of where electrons move, then your palm is facing into the page. But how do you know this supplies the most force hence separates the +ve and -ve charges the most?
Post by: /0 on June 20, 2009, 09:52:03 pm
Post by: TrueTears on June 20, 2009, 09:55:26 pm
Post by: kamil9876 on June 20, 2009, 09:57:33 pm
Post by: /0 on June 20, 2009, 10:03:19 pm
When the wire is horizontal and the right side is moving down, that's when its velocity perpendicular to the magnetic field is greatest, so the force on charges will be greatest. The left-hand palm should go into the page and the right-hand palm should go out of the page to separate charges and create the emf.
It has no horizontal velocity component at that point, so it's all vertical.
/0: you're lenz law explanation, is it any similair to http://en.wikipedia.org/wiki/Symmetric_derivative Because I used a similair idea on msn.
Yeah that's what I was trying to get at lol, is it mathematically valid?
Post by: TrueTears on June 20, 2009, 10:07:16 pm
and I understood how the Lenz Law mathematical approach worked before posting this question, it was just the "intuitive" part that confused me.
But yeah I understand now, thanks heaps.
Post by: kamil9876 on June 20, 2009, 10:12:02 pm
When the wire is horizontal and the right side is moving down, that's when its velocity perpendicular to the magnetic field is greatest, so the force on charges will be greatest. The left-hand palm should go into the page and the right-hand palm should go out of the page to separate charges and create the emf./0: you're lenz law explanation, is it any similair to http://en.wikipedia.org/wiki/Symmetric_derivative Because I used a similair idea on msn.
Yeah that's what I was trying to get at lol, is it mathematically valid?
I thought it's when the magnetic field is perpendicular to the direction in which it moves, not the velocity?
"B perpendicular to v" is the same as "v perpendicular to B"
i.e:
So the angle can be associated with either quantity.
It's analogous to torque:
So the angle can be associated with either quantity.
Post by: TrueTears on June 20, 2009, 10:40:07 pm
Interesting how fundamentals of emf for rods can still be applied to coils :)
Post by: /0 on June 20, 2009, 11:01:08 pm
Post by: kamil9876 on June 20, 2009, 11:01:29 pm
When the wire is horizontal and the right side is moving down, that's when its velocity perpendicular to the magnetic field is greatest, so the force on charges will be greatest. The left-hand palm should go into the page and the right-hand palm should go out of the page to separate charges and create the emf.
It has no horizontal velocity component at that point, so it's all vertical./0: you're lenz law explanation, is it any similair to http://en.wikipedia.org/wiki/Symmetric_derivative Because I used a similair idea on msn.
Yeah that's what I was trying to get at lol, is it mathematically valid?
Good question, here is what i came up with now:
lef f(x) be differentiable
this implies:
=
Our limit it is hence:
=============================================================================
This could also be done using mean value theorem(and that was my original plan as it looked similair):
However this approach is void of generality as it requires that f'(x) is continous.
Post by: TrueTears on June 20, 2009, 11:05:44 pm
Post by: TrueTears on June 20, 2009, 11:52:10 pm
Can you apply emf = Blv in this case? (Coz I thought Blv is only used in situations where you have a rod and it's falling through a magnetic field, in this case its a coil that's rotating so...)
Thanks :)
Post by: TrueTears on June 21, 2009, 12:03:54 am
Post by: /0 on June 21, 2009, 12:23:14 am
Why is it that when a coil spins faster in a generator, the emf is larger?
Can you apply emf = Blv in this case? (Coz I thought Blv is only used in situations where you have a rod and it's falling through a magnetic field, in this case its a coil that's rotating so...)
Thanks :)
If the wire is spinning faster, then the same
Also with
Post by: TrueTears on June 21, 2009, 12:25:15 am
How would you prove emf = Blv still works?Why is it that when a coil spins faster in a generator, the emf is larger?
Can you apply emf = Blv in this case? (Coz I thought Blv is only used in situations where you have a rod and it's falling through a magnetic field, in this case its a coil that's rotating so...)
Thanks :)
If the wire is spinning faster, then the samewill be covered in a shorter
, so the emf is greater.
Also withI have a hunch that the formula still works for rotating wires, but not 100% sure
Post by: TrueTears on June 21, 2009, 01:57:11 am
Can you apply emf = Blv in this case? (Coz I thought Blv is only used in situations where you have a rod and it's falling through a magnetic field, in this case its a coil that's rotating so...)
2. If a generator produces AC current then does this imply that the voltage is AC voltage?
3. Why does a DC motor contain DC current? The current is reversed each half turn by a split ring commutator, so doesn't that become AC current?
Is it because the terminals of the battery doesn't change (unlike generators), it is just the split rings which attach themselves alternatively to the 2 terminals of the battery hence the current is still DC but the it is just "manipulated" to reverse at each half turn?
4. Describe the function performed by a commutator in a DC generator.
I thought commutators are only used in DC motors???
Post by: james23 on June 21, 2009, 02:01:39 am
4. Describe the function performed by a commutator in a DC generator.
I thought commutators are only used in DC motors???
It makes the current generated unidirectional by reversing the output connections each half rotation when the current through the coil should change direction, hence making the output current DC.
Post by: TrueTears on June 21, 2009, 02:18:44 am
At the instant when the wire is horizontal, you can approximate it to be a rod falling through a magnetic field... well, at least, one side is falling up and the other is falling down.But that's not a formal proof? When a rod is falling don't you need it to be totally falling down? Not just parts of it?
Anyway... I usedin the derivation of
and it worked out fine.
Post by: kamil9876 on June 21, 2009, 02:25:36 am
Quote
Anyway... I used Blv in the derivation of emf = \frac{\Delta \Phi}{\Delta t} and it worked out fine.
Can I see? :D
Post by: /0 on June 21, 2009, 02:26:37 am
I think expression is
QuoteAnyway... I used Blv in the derivation of emf = \frac{\Delta \Phi}{\Delta t} and it worked out fine.
Can I see? :D
It's on the previous page
Post by: TrueTears on June 21, 2009, 02:29:35 am
Dw, I'll just remember when you increase the speed
Post by: kamil9876 on June 21, 2009, 02:37:40 am
Generators "naturally" produce AC current (if you focus on the actual coil, it will always be AC no matter if you have a split ring, slip ring or fuck all). However we can change this AC into DC by making it such that the ends of the coil switch every half turn. By switching, I mean that the ends of the coil swap with each other the output-wire/terminal thingy that they are touching. SO the switch of connectioncancels out the "natural" change in direction of current. The split ring commutator, just like in DC motors, does this switching.
Quote
It makes the current generated unidirectional by reversing the output connections each half rotation when the current through the coil should change direction, hence making the output current DC.
As pointed out by james23
In addition to that, it's a
Post by: /0 on June 21, 2009, 02:43:30 am
Post by: /0 on June 21, 2009, 02:45:08 am
Post by: /0 on June 21, 2009, 02:46:24 am
Post by: dcc on June 21, 2009, 03:08:49 am
Post by: dcc on June 21, 2009, 12:23:58 pm
Post by: TrueTears on June 22, 2009, 10:13:34 pm
emf = Blv , could you use this formula for a coil in a AC/DC generator (because what 'l' be? the length of the entire "bent" rod or the coils that's wrapped around it?) Or is it just for rods following through a uniform magnetic field?
Thanks!
Post by: Mao on June 22, 2009, 10:18:23 pm
in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.
Post by: TrueTears on June 22, 2009, 10:20:51 pm
it can be used for coils in a generatorOh okay thanks, so 'l' would just be the length of the side of the square/rectangle I'm dealing with? But doesn't that just give the emf on that single side of the entire square/rectangle? So does that mean I have to calculate emf = Blv for each of the sides (that are perpendicular/that have a perpendicular component to the mag field) and add them in order to get the 'total' emf?
in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.
Post by: TrueTears on June 22, 2009, 10:26:41 pm
I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?
Post by: Mao on June 22, 2009, 10:33:57 pm
it can be used for coils in a generatorOh okay thanks, so 'l' would just be the length of the side of the square/rectangle I'm dealing with? But doesn't that just give the emf on that single side of the entire square/rectangle? So does that mean I have to calculate emf = Blv for each of the sides (that are perpendicular/that have a perpendicular component to the mag field) and add them?
in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.
Ahh, ooops, I did not read your post properly, I was talking about F=IlB.
For your case, emf = Blv will not apply because you will not given v, since the coil is in rotational motion. (It is possible to derive the tangential speed, but that is not within the scope of VCE Physics).
You will be using
Slightly beyond VCE level, you may be using
Post by: Mao on June 22, 2009, 10:37:46 pm
In a regular DC motor, which of the following would decrease the speed of the rotating coil in the motor. The answer is these 3 "Decrease the number of turns of the rotating coil." "Decrease the area of the rotating coil" and "Decrease the current in the rotating coil"
I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?
The coil moves in rotational motion, hence force is not the only factor.
By reducing the area, you will be reducing either (or both) of length (l) and distance from axle (x). Reducing l reduces force, whereas reducing x directly decreases the torque.
[however, this is not a conventional way of achieving slower speeds]
Post by: TrueTears on June 22, 2009, 10:38:25 pm
Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formula
Post by: TrueTears on June 22, 2009, 10:40:55 pm
Ah yes yes yes, thanks again Mao, you lifesaver!In a regular DC motor, which of the following would decrease the speed of the rotating coil in the motor. The answer is these 3 "Decrease the number of turns of the rotating coil." "Decrease the area of the rotating coil" and "Decrease the current in the rotating coil"
I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?
The coil moves in rotational motion, hence force is not the only factor., hence the distance from axle at which the force act is also important.
By reducing the area, you will be reducing either (or both) of length (l) and distance from axle (x). Reducing l reduces force, whereas reducing x directly decreases the torque.
[however, this is not a conventional way of achieving slower speeds]
Post by: Mao on June 22, 2009, 10:41:34 pm
Ahhh, thanks I get it. So then how do you explain that increasing the speed of the rotating coil increases the average emf?
Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formulahence emf is larger?
yes
Post by: TrueTears on June 22, 2009, 10:44:04 pm
Awesome, thanks for the quick and helpful replies !Ahhh, thanks I get it. So then how do you explain that increasing the speed of the rotating coil increases the average emf?
Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formula
yes
Post by: TrueTears on June 23, 2009, 01:59:11 pm
Do I have to take in consideration the number of turns here? ie flux = nBA ?
Because book specifically mentions not to times BA by n, but teacher says I should include the n. So my question is when do I include n to work out the flux when do I don't include n?
Thanks!
Post by: Mao on June 23, 2009, 06:45:42 pm
you only include the number of turns when dealing with emf, that
Post by: TrueTears on June 23, 2009, 06:47:35 pm
noo,Aiight, thanks!
you only include the number of turns when dealing with emf, thatand
Post by: TrueTears on June 23, 2009, 10:21:22 pm
Ok I came up with a "proof" but I'm not sure if this is correct or not. Could someone please check and let me know if it's valid or not.
My proof is: "If you decrease the wavelength of light being shone through the 2 slits (this is basically just the young's experiment) then the pattern of bright/dark bands created on the plate/wall will get narrower(more tightly packed) than the wavelength it was before."
So looking at the diagram, say A is a fixed point on the wall. Now let L1 be original wavelength and L2 be the decreased wavelength. Let n1 be the integer in front of L1 and n2 be the integer in front of L2. The dotted line is the central maximum which produces a bright band in the middle of the wall. (This is the case for every young's slit experiment anyway).
Now case 1: Assume that both wavelength L1 and L2 produce a point at A which is a bright band (light band) due to constructive interference.
So because the path difference is the same for both wavelength, since S1 to A and S2 to A does not change no matter what the wavelength is (remember A is fixed on the wall). This means we can equate n1L1 = n2L2. In this case both n1 and n2 are WHOLE numbers.
So looking at this equation, as L1 decreases to L2, n1 must INCREASE to n2.
So that means for L2 (the decreased wavelength) the whole number in front of the L2 is larger than the whole number in front of L1. So that means for the same distance from the middle line to A, there are more bright/dark bands on the wall for L2 than there are for L1, hence the the pattern of bright/dark bands get narrower (more tightly packed) for L2 (lower wavelength)
Now case 2: Assume that wavelength L1 and L2 produce a point at A which is a dark band. Due to destructive interference.
This is basically the same as case 1 but just let n1' = n1+0.5 and n2' = n2+0.5 Both n1 and n2 are whole numbers, hence n1' and n2' are positive integers.
so we get n1'L1 = n2'L2 hence using the same argument in case 1, the result is the same.
Now case 3: Assume that wavelength L1 produces a light band but L2 produces a dark band. (Or vice versa ofcourse)
So let the positive integer (ignoring 0 because that's middle line) in front of L2 be n2' = n2 + 0.5. Obviously n1 is just a whole number. n2 is a whole number which makes n2' a positive integer.
Using same argument as case 1, since Path difference is the same for both wavelength we can equate n1L1 = n2'L2.
From the same reasoning as L1 decrease to L2, n1 must INCREASE to n2' hence there is a larger positive integer in front of L2, hence it is narrower(more tightly packed)
In conclusion, by showing that the top half of the central maximum is narrow then by symmetry the same can be shown for the bottom half, so by decreasing the wavelength we narrow the pattern produced on the wall.
---------------------------------------------------------------------------------------------------------------
That's the end of the proof, is this correct? Or was what I did totally wrong?
Thank you!!!
Post by: kamil9876 on June 23, 2009, 10:24:58 pm
Post by: TrueTears on June 23, 2009, 10:35:46 pm
Yes, or just learn theahhh ok, thanks for that kamil!!formula :P. Jk, nice stuff. Nice to see these sorts of proofs and enthusiasm for them in vce physics :)
Now I wonder what would happen if the points at A was a partial (as in not a bright or dark band...) hehehe, or is that not in the VCE course?
Post by: kamil9876 on June 23, 2009, 10:41:38 pm
You can assume WLOG that A is a bright one and that
Post by: TrueTears on June 23, 2009, 10:43:18 pm
:P:PAwesome thanks!
You can assume WLOG that A is a bright one and thathas a maximum at B where B is a point between A and the maximum for
just below A. Also, WLOG you can assume that
. That's how i would prove it if I would be as persistent as you in avoiding
Can anyone confirm is partials are in the VCE course?
Post by: kamil9876 on June 23, 2009, 10:51:22 pm
Post by: TrueTears on June 24, 2009, 12:14:35 am
Also, for the diffraction ratio
Or can't it ever equal to 1?
Thanks!
Post by: Mao on June 24, 2009, 10:04:35 am
Warning, the following materials are beyond the VCE course
Formally, intensity profile of a diffraction can be described mathematically by
The first minimum occurs at
When the two values are 'comparable' (ratio ~1), they will give a diffraction angle large enough to be observed.
When d is significantly larger than wavelength (e.g. ratio is 0.01 or something like that), the diffraction angle is too small to be noticed, hence no observable diffraction pattern.
When wavelength is significantly larger than d, the diffraction angle will be very large. The result is it looking more like diffusion than diffraction, i.e. no visible minima/maxima, just a blob of light. And hence, no observable diffraction pattern.
The key note here is the use of the word 'significantly'. There is no rule saying whether 2 degrees is observable or not, hence the exam will not be this ambiguous. (you won't be asked to distinguish whether 0.5 or 1.7 are good enough ratios) However, you need to be aware that when you get 0.001 as a ratio, you probably won't notice any diffraction. (and VCAA has never thrown a question where wavelength is significantly larger than d, because wavelength of light is to the order of 10^-7, it's kinda hard to be 'significantly smaller' than that)
Post by: TrueTears on June 24, 2009, 10:22:33 am
Post by: TrueTears on June 24, 2009, 03:27:18 pm
1st formula:
Is this formula only used for single or can it also be used for double slit or both?
Why is it only for nodal points on the wall/plate? Why can't you determine the
2nd formula:
Again is this formula used for single or double slit or both?
Also why is it only for nodal points on the wall/plate?
Post by: kamil9876 on June 24, 2009, 04:26:22 pm
Double slit:
Now that implies that
However for constructive interference
Equating the two expressions gives:
For the nodal version just use
You should find the above proof in your textbook and it's pretty quick to do in an exam once your understanding of it becomes natural
To be continued....
Post by: kamil9876 on June 24, 2009, 05:27:33 pm
Now let C be the midpoint of the slit, and let A be the top end of the split (refer to diagram). Let
in order for wave A to destruct with with C at E:
(it's like AC=d now basically)
Now
B and D are another two points, but such that the distance between them is w/2 as well. Note that we can pair up every point from top and bottom half of the slit in such a way that the distance between them is always w/2. Because B and D are parralel the same formula comes out:
Hence you will find a nodal line at at
However, if we let L be the horizontal distance from the screen to slit and we let y_n be the distance of the nth nodal line from central maximum we get:
Now combining (1) and (2) we get:
However
if you let w/2=d you get your first formula.
=============================================================
Edit:
Looking at formula (1):
Which falsely suggests that ONLY odd numbers can be put in front of the lambda. It is true that ALL odd numbers work, however it is incomplete as it can be shown that indeed even numbers work too. My proof was simply a bad path to take, albeit still correct, but incomplete:
Instead of pairing up points on the slit that are w/2 apart, let's pair up points that are w/4 apart, w/6 apart etc. Using similair arguments as above, it can be shown that:
Now k can be even or odd, hence the coefficient of the
And so if we use this isntead of equation (1) in our trigonometry we get:
Funilly enough, this equation and my previous are both true. The previous one is identical to this one for odd N, however the previous one doesn't find the even N's and so it can muck up the order if by n you mean "which order of the band" since it skips every second one.
edit: forgot a few w's. omg im phailing bad :/
Post by: kamil9876 on June 24, 2009, 05:45:05 pm
Quote
Is this formula only used for single or can it also be used for double slit or both?
Why is it only for nodal points on the wall/plate? Why can't you determine the \lambda of the light by using an antinodal point on the wall/plate?
Ok, the reason why it only works for nodal lines is this:
We saw that if we pair up A and C, B and D etc. and gaurantee that they all interfere destructively then we get overall destructive inteference. This can be mathematically described as follows:
Total Amplitude at E=A+B+C+D....
=(A+C)+(B+D)+....
=0+0...
=0 hence destruction
Hence if these pairs all destruct, then the overall affect is destruction.
Now the inverse of this statement, "If pairs construct, then overall effect is construction" is false and it can be demonstrated as follows:
Total Amplitude at E=A+B+C+D...
=(A+C)+(B+D)...
=2a+2b....
however even though the pairs construct, the pair sum can destruct, ie: b=-a:
this would give:
2a+2b...=0
or similairly, they don't construct as strongly. Hence this pairing technique only works for finding destruction hence the formula only applies to destruction and there isnt some antinodal version where u put in n instead of n-0.5.
Ensuring construction requires that the pairs themselves(or simply put, the total amplitudes) are constructive and so this is quite a complicated calculus problem and requires a good quantitative description of waves(Huygen's principle).
Post by: TrueTears on June 24, 2009, 07:08:58 pm
Post by: TrueTears on June 24, 2009, 08:03:08 pm
(http://img15.imageshack.us/img15/5511/lernadexperiment.jpg)
So I understand the threshold frequency, if the wave is below the threshold frequency for that metal then it wouldn't produce a photocurrent etc.
But what I don't get is how would increasing/decreasing the frequency affects the stopping/minimum voltage. So from the graph light intensity stays the same but frequency is changing [also do we assume the changing frequency is above the threshold frequency or else no current would flow hence no stopping voltage, right?], as the
What about the converse, if the
And what's the "intuitive" reason for the answers?
Thanks!
Post by: Mao on June 24, 2009, 08:35:06 pm
Post by: TrueTears on June 24, 2009, 08:43:55 pm
Post by: Mao on June 24, 2009, 09:00:28 pm
Over9000, your post has been deleted. It was highly offensive and I warn you to not do this again. Such overreaction will not be looked upon kindly next time.
Post by: TrueTears on June 24, 2009, 09:45:37 pm
The voltage in photoelectric cell can be retarding but it can also help accelerate the electron.
So when voltage (electric energy) is ADDED to the electron's kinetic energy that means the the speed of the electron increase right?
So on a photocurrent - voltage graph, assuming adding voltage to the kinetic energy is positive, then the positive side of the graph would just be constant horizontal line right? Because adding voltage to the kinetic energy doesn't increase the maximum amount of photocurrent it just increases its speed right? [only a change light intensity increases/decreases the max photocurrent yeah?]
But when voltage opposes the motion of the electron it converts the kinetic energy of the electron into electric energy so as this continues eventually the electron will have no kinetic energy hence no current, which is why the negative part of the photocurrent - voltage graph is decreasing until the current reaches 0 and at this point is the
Thankie :)
Post by: TrueTears on June 24, 2009, 10:05:23 pm
Current is how many electrons pass through a point in one second, it makes sense to me that if you increase the speed of electrons current should increase? Hence it shouldn't be a horizontal line? shouldn't it be a line with positive gradient?
Post by: Mao on June 24, 2009, 11:59:13 pm
For photoelectric cells, the current depends on the light intensity. Remember that each photon knocks out one electron.
Work is done by the photon on the electron to knock it off the metal surface. The amount of work required varies (depending on location and how tightly bound the electron is to the metal cations). In this case, the threshold frequency gives the minimum work required (to liberate the least tightly bound electrons).
When a greater frequency is used, more electrons can be liberated as there are more energy available to liberate more tightly bound electrons. However, the least bound electrons still require the minimum work, whilst the higher frequency light provides more photon energy, hence these electrons have the greatest 'residual' kinetic energy. The less bound electrons have more kinetic energy than tighter bound electrons (as tighter bound electrons require more energy to liberate).
When a negative voltage is applied, work is done against electrons to reduce their motion. For electrons with not enough kinetic energy, they get knocked back onto the metal surface, but the more energetic ones can still escape. Since some are knocked back, the current decreases. As you apply more reverse voltage, more electrons get knocked back (because fewer electrons have sufficient energy to escape), hence current decreases. Until the reverse voltage reaches and exceeds the max KE, at this point, not even the most energetic can escape, hence no current flows.
Greater velocity does not necessitate greater current. It's difficult to explain, but the general gist is because distance between electrons is not fixed.
Post by: TrueTears on June 25, 2009, 12:26:53 am
Post by: TrueTears on June 25, 2009, 06:10:11 pm
If a photon doesn't have mass, then how come it has momentum?
Is there a intuitive way of thinking about this? Or would I need beyond vce physics?
Post by: dcc on June 25, 2009, 06:20:53 pm
Another question.
If a photon doesn't have mass, then how come it has momentum?
Is there a intuitive way of thinking about this? Or would I need beyond vce physics?
Intuition and quantum mechanics rarely join forces.
Post by: kamil9876 on June 25, 2009, 06:25:58 pm
But i can say that p=mv is not momentum when v is near c, let alone equal to it:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
Notice on the bottom how it says:
and so p=E/c at the limit. and use
Actually, it even says on the bottom:
Quote
with the limiting case applying for the momentum of a photon.
Post by: kamil9876 on June 25, 2009, 06:26:46 pm
Another question.
If a photon doesn't have mass, then how come it has momentum?
Is there a intuitive way of thinking about this? Or would I need beyond vce physics?
Intuition and quantum mechanics rarely join forces.
Haha was just about to say
edit: quote fail
Post by: TrueTears on June 25, 2009, 06:28:26 pm
Post by: TrueTears on June 25, 2009, 07:23:01 pm
Is it still called constructive interference? And for the interference that gives maximum amplitude is that called maximum constructive interference?
Is this the same for destructive?
Thanks.
Post by: Mao on June 25, 2009, 11:28:50 pm
but yes, it'll still be constructive.
Post by: TrueTears on June 26, 2009, 12:29:45 am
Thanks for confirmation!
Post by: TrueTears on June 26, 2009, 01:41:09 am
What if we are given a question like, what is the ratio of
The ratio
We just simply divide the wavelength of the slit by the width of the slit to work out extent of diffraction?
Thank you!
Post by: kamil9876 on June 26, 2009, 01:23:34 pm
sub in n=1 and u get ur formula. The angle
Post by: TrueTears on June 26, 2009, 03:38:15 pm
So we don't even need to care about whether there's dark/bright bands. All we need is the wavelength and width of slit. right?
Thanks.
Post by: kamil9876 on June 26, 2009, 05:06:16 pm
Post by: TrueTears on June 26, 2009, 06:50:16 pm
"what is the ratio of
Post by: NE2000 on June 26, 2009, 07:45:06 pm
haha my relativistic dynamics are piss poor:
But i can say that p=mv is not momentum when v is near c, let alone equal to it:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
Notice on the bottom how it says:
and so p=E/c at the limit. and use
Actually, it even says on the bottom:Quote
with the limiting case applying for the momentum of a photon.
lol this bothered me for a bit as well so thanks for that...that and the idea that I have a de Broglie wavelength when moving :)
This may seem a bit superficial but sometimes I like drawing diagrams to confirm concepts and I was wondering how people picture a photon in their mind. I was sort of getting an idea of it and then Taylor's experiment for the diffraction of individual photons threw me off. So just wondering how people visualise it.
Post by: TrueTears on June 26, 2009, 07:48:17 pm
Post by: NE2000 on June 26, 2009, 07:53:55 pm
Ahh okay thanks kamil, so how would you answer the question before?
"what is the ratio ofif the point on the wall/plate is a dark band occurring at n = 2.5 and the central maximum is n = 0."
Yeah I'm pretty sure the current course just requires us to determine whether the diffraction will be significant or not by looking at the ratio. Maybe it's just my lack of knowledge at this point but it seems like there should be more information... :S
Post by: NE2000 on June 26, 2009, 07:55:58 pm
Yeah the Taylor's experiment with the diffracting photon, you will need to get into quantum mechanics to understand that; where wave/particle nature of light/matter is unified. For now I just think of photons as a discrete amount of energy, kind of like a ball of energy.
Yeah but how does a ball of energy diffract...
btw, do you use Jacaranda? It's just that from what I remember Jacaranda said something similar about the wave and particle theories on light and matter being "unified successfully" in quantum mechanics...leaving us in suspense (clearly a ploy to get more people to do physics at uni) :)
Post by: TrueTears on June 26, 2009, 07:58:11 pm
Yeap, that's right.Yeah the Taylor's experiment with the diffracting photon, you will need to get into quantum mechanics to understand that; where wave/particle nature of light/matter is unified. For now I just think of photons as a discrete amount of energy, kind of like a ball of energy.
Yeah but how does a ball of energy diffract...
btw, do you use Jacaranda? It's just that from what I remember Jacaranda said something similar about the wave and particle theories on light and matter being "unified successfully" in quantum mechanics...leaving us in suspense (clearly a ploy to get more people to do physics at uni) :)
And exactly what I was thinking, how can a ball of energy diffract? Asked my teacher a few days ago, he said he doesn't even know.
Post by: NE2000 on June 26, 2009, 08:01:20 pm
Yeap, that's right.Yeah the Taylor's experiment with the diffracting photon, you will need to get into quantum mechanics to understand that; where wave/particle nature of light/matter is unified. For now I just think of photons as a discrete amount of energy, kind of like a ball of energy.
Yeah but how does a ball of energy diffract...
btw, do you use Jacaranda? It's just that from what I remember Jacaranda said something similar about the wave and particle theories on light and matter being "unified successfully" in quantum mechanics...leaving us in suspense (clearly a ploy to get more people to do physics at uni) :)
And exactly what I was thinking, how can a ball of energy diffract? Asked my teacher a few days ago, he said he doesn't even know.
Doesn't Scotch have this awesome physics teacher that inventing a diamond-making machine?
Post by: TrueTears on June 26, 2009, 08:03:36 pm
Post by: rhjc.1991 on June 26, 2009, 08:09:41 pm
Yeah [he was year 11 teacher, to be honest, he was pretty crap, he is definitely smart but has no idea how to teach.]Ah why does that line sound SO familar? lol
Post by: NE2000 on June 26, 2009, 08:11:48 pm
hmmm......why?? [sorry]Yeah [he was year 11 teacher, to be honest, he was pretty crap, he is definitely smart but has no idea how to teach.]Ah why does that line sound SO familar? lol
Post by: rhjc.1991 on June 26, 2009, 08:14:19 pm
I meant that I know/heard of so many teachers like that.Yeah [he was year 11 teacher, to be honest, he was pretty crap, he is definitely smart but has no idea how to teach.]Ah why does that line sound SO familar? lol
hmmm......why?? [sorry]
Post by: TrueTears on June 27, 2009, 01:45:22 am
Ahh so how would you answer the question before?
"what is the ratio of
-----------------------------------------------------------------------------------------------------------
A dark region in a 2-slit interference pattern is caused because
I choose "The photons or particles annihilate each other at that point"
but answer is "the photons or particles simply do not travel to that point"
and
A question says
light from the door to a very hot blast furnace
what describes the spectrum?
answer is "The spectrum is continuous: there is a spread of wavelength from the infra red right through to the visible range"
Why??
Thanks!
Post by: Mao on June 27, 2009, 01:51:22 am
Yeap, that's right.Yeah the Taylor's experiment with the diffracting photon, you will need to get into quantum mechanics to understand that; where wave/particle nature of light/matter is unified. For now I just think of photons as a discrete amount of energy, kind of like a ball of energy.
Yeah but how does a ball of energy diffract...
btw, do you use Jacaranda? It's just that from what I remember Jacaranda said something similar about the wave and particle theories on light and matter being "unified successfully" in quantum mechanics...leaving us in suspense (clearly a ploy to get more people to do physics at uni) :)
And exactly what I was thinking, how can a ball of energy diffract? Asked my teacher a few days ago, he said he doesn't even know.
A ball of energy does not diffract. But because the ball of energy also have a wave duality, due to some quantum stuff or other diffraction occurs no matter how you look at it.
A dark region in a 2-slit interference pattern is caused because
I choose "The photons or particles annihilate each other at that point"
but answer is "the photons or particles simply do not travel to that point"
A dark band exists because impact density there is zero. When you think about it in terms of waves, the superposition at that point gives you a flat line. A flat line (no displacement) is a wave with no energy, and since each photon has some energy, there must be no photons.
light from the door to a very hot blast furnace
what describes the spectrum?
answer is "The spectrum is continuous: there is a spread of wavelength from the infra red right through to the visible range"
In this case, we're talking about thermal emission. In heat, electrons accelerate/decelerate due to collisions with other particles, and in the process, release energy. Hence the light emitted is not a distinct set of frequencies, but rather a bunch of 'random' values that forms a continuous spectrum with peak related to temperature. Research 'black body radiation' for more. :)
Post by: TrueTears on June 27, 2009, 01:53:56 am
Post by: TrueTears on June 27, 2009, 02:22:34 am
But do we assume the graph we are given is the characteristics of ONE electron, ie the stopping voltage to stop ONE electron. Hence
Post by: mark_alec on June 27, 2009, 12:17:41 pm
Post by: kamil9876 on June 27, 2009, 12:30:22 pm
But do we assume the graph we are given is the characteristics of ONE electron, ie the stopping voltage to stop ONE electron. Henceright? [because if it's not for 1 electron then
would be unknown since the question doesn't state it]
THat is unnecesary. Since you are after Voltage, there is no talk of Energy here. Voltage is a rate, Energy/Charge so it's a measure of how much energy each charge carries. If you are given JUST VOLTAGE and no mention of energy, then you shouldn't worry about the number of electrons involved. E.g: Say the voltage is 2Joules/electron. If you have 4 electrons, you have 8Joules/4electrons. The rate doesn't change as long as it's the same material.
let n be the number of electrons. Let q_e be the charge of ONE electron, W be the work function of ONE electron and E be the energy of ONE electron:
but
voila, n cancels out and you get the same equation as if you had only one electron.
on a side note:
Post by: TrueTears on June 27, 2009, 04:38:41 pm
The light from a candle can be best described as :
A. coherent, arising from the vibrations of electrons.
B. incoherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
C. coherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
D. incoherent, arising from the vibrations of electrons.
Answer is D.
My books don't talk about coherent/incoherent or about how the vibrations of electrons are related with light. [I understand all the emission/absorption spectrum which is related with choices B and C, but don't understand anything about A or D -.-"]
Can someone please explain exactly what those 2 things are?
Thanks.
Post by: mark_alec on June 27, 2009, 05:09:42 pm
Post by: TrueTears on June 27, 2009, 05:24:49 pm
A coherent light source is one in which all the wavefronts are in sync. The typical example is a laser. An incoherent light source is one in which the wavefronts head in all direction in all phases. The typical example is a light bulb.Ah cool I see, so since a candle shoots light in all directions it would be a incoherent light source right?
And what about the business with vibrations with electrons?
Post by: Over9000 on June 27, 2009, 05:37:38 pm
Post by: mark_alec on June 27, 2009, 06:00:03 pm
Ah cool I see, so since a candle shoots light in all directions it would be a incoherent light source right?Yes, a candle is an incoherent light source. A coherent light source also produces light only one wavelength.
Quote
And what about the business with vibrations with electrons?All you need to know is that due to 'thermal vibrations', the energy is produced in a continuous band, not as discrete wavelengths.
Post by: mark_alec on June 27, 2009, 06:03:25 pm
Sorry to use ur thread privilages TT but I wish to ask, is it possible for photons to travel faster than the speed of light somehow, like if the photons got an extra amount of energy somehow coz that wud be interesting.It is not possible.
Firstly, you need to think of how you would give a photon extra energy. It doesn't have an electric charge, so you can't perform work on it in an electric field. But it does have a mass. In actual fact, gravity does effect light, affecting its energy in a similar way to a classical mass. Look up gravitational redshift if you are interested.
Post by: TrueTears on June 27, 2009, 06:07:14 pm
Also with thermal vibrations is that just adding heat to the electron causing it to vibrate and generate energy by shooting out photons?
But isn't this the same as emission spectrum, where you heat a metal and the electron absorb a discrete amount of energy and releases a photon of a discrete amount of energy.
So what's the difference between thermal vibration [how does it work?] and emission spectrum?
Thanks!
Post by: mark_alec on June 27, 2009, 06:59:17 pm
The mechanism for the continuous spectrum I suspect is due to a charged particle (the electron) being accelerated due to the thermal energy being given to the solid. The electrons in a metal inhabit almost continuous energy bands, which could also be part of it. To be honest, I don't know the exact mechanism by which 'thermal' radiation is produced, except that it can be modelled by oscillators moving in a quadratic potential that can only have an energy equal to
Post by: TrueTears on June 27, 2009, 07:07:01 pm
So is this interpretation correct?
The candle is an incoherent light source because the wavefronts head in all directions and they are in all kind of phases. A continuous spectrum is produced due to thermal vibrations. This means the electrons are accelerated due to the thermal energy of the flame of the candle and hence emit a continuous spectrum.
[And just as a side thing, thermal vibration is different from emission due to the fact that thermal vibration produces a continuous spectrum where as emission spectrum produces light of only certain wavelengths]
Thanks so much mark.
Post by: mark_alec on June 27, 2009, 08:41:34 pm
The candle is an incoherent light source because the wavefronts head in all directions and they are in all kind of phases. A continuous spectrum is produced due to thermal vibrations. This means the electrons are accelerated due to the thermal energy of the flame of the candle and hence emit a continuous spectrum.Scrap the bit about the light going in different directions, what is important is that it is out of phase. And I wouldn't bother writing the second sentence about the continuous spectrum, as I'm not sure of its validity.
Post by: TrueTears on June 27, 2009, 08:43:30 pm
Post by: TrueTears on June 27, 2009, 10:17:43 pm
Q1, These 2 formulas are from Nelson's textbook, there are just a few things I am unsure of. The diagram is attached
1st formula:
In a 2 slit Young's experiment:
The formula has
What I'm wondering is, is this formula is a general formula for all dark bands for a 2 slit experiment? Ie, if I wanted to find the angle the 3rd nodal line makes with the centre line then
But then, what if I wanted to find the angle an ANTI-nodal line makes with the centre line, would I change the formula to
Now, would this formula also work for single slit experiment? So instead of
2nd formula:
In a 2 slit experiment: Again, does this formula only work for nodal lines (hence dark bands)? Because the formula contains
In a single slit experiment: Could this formula work for a single slit experiment? So instead of
Post by: /0 on June 27, 2009, 10:38:48 pm
I don't think the formula works for single slits...
In the uni physics book there is no mention of a single-slit formula for antinodal points.
Post by: TrueTears on June 27, 2009, 10:40:19 pm
Post by: /0 on June 27, 2009, 10:48:14 pm
Yeah, I have asked kamil about it and he said his uni book does the same, but why does Nelson say it works for single slits? Further more, the proof comes from a single slit experiment in their book :|
soz lol do u know which page it's on?
Post by: TrueTears on June 27, 2009, 10:54:20 pm
A. 500 nm photons passing through a 0.05 mm slit
B.
I think the question is not correctly stated because I think both of the options would not show diffraction. My working is as follows:
For A: Using the ratio
For B before using the ratio we must first find out the wavelength of the electron. Using De Broglie's wavelength yields
Now using the ratio to find extent of diffraction:
Now clearly B has a much smaller ratio than A which means the extent of diffraction is insignificant, but A's ratio is way below 1. The question asks which is most likely to show diffraction effects but wouldn't niether of them show diffraction? Because A's ratio (0.01) is too much smaller than 1 diffraction is basically insignificant?
Post by: kamil9876 on June 27, 2009, 11:12:21 pm
For the first formula, yes I think it's the general formula, andis for anti-nodal lines.
is the path difference in each case, so making it equal
or
will give you nodal or antinodal lines respectively.
I don't think the formula works for single slits...gives the nodal lines in single slit.
In the uni physics book there is no mention of a single-slit formula for antinodal points.
Yes i agree with all that.
The book does explain why the formula for single slit antinodal lines is not a simple substitution of n for n+0.5 and I plaigarised that idea in this post:
http://vcenotes.com/forum/index.php/topic,9668.msg159100.html#msg159100
I also modified my proof in the post preceding that one just today(all this is found at the bottom of the post), for those interested. Apologies for the slight incompleteness in my first attempt at the proof.
Post by: dcc on June 28, 2009, 01:27:10 am
Post by: /0 on June 28, 2009, 05:24:36 pm
Thanks kamil ^^
Also, for the diffraction ratio
Or can't it ever equal to 1?
Thanks!
Hold on...
Does the formula break down immediately when
Post by: TrueTears on June 28, 2009, 05:33:19 pm
Well looking atThanks kamil ^^
Also, for the diffraction ratio
Or can't it ever equal to 1?
Thanks!
Hold on...but
Does the formula break down immediately when?
Post by: /0 on June 28, 2009, 05:37:45 pm
Post by: TrueTears on June 28, 2009, 05:39:39 pm
Not only that, but you knowTrue, haha, I have always ignored thefor constructive interference, eventually as you get far out and
becomes large,
for any values of
or
, so
will be undefined, even though it should still be
!
Post by: kamil9876 on June 28, 2009, 06:07:04 pm
Not only that, but you know
Im pretty sure it's because the path difference cannot be great enough.
Post by: TrueTears on June 28, 2009, 11:16:24 pm
Also what does it mean the direction that a wave travels is always perpendicular to the wavefront?
Post by: TrueTears on June 29, 2009, 12:18:40 am
However the sound intensity at the threshold of hearing is
Does this mean the
Thanks
Post by: Mao on June 29, 2009, 11:14:20 am
Post by: Mao on June 29, 2009, 11:19:17 am
When 2 rarefractions meet is this also an antinodal point?
yes
sound waves are longitudinal waves, hence the zero point is where pressure variation is nought. Rarefractions are below 'normal' pressure, hence two rarefractions will superposition to form a trough, antinode.
Post by: TrueTears on June 29, 2009, 05:03:54 pm
But when rarefaction occurs does this also produce the loudest sound?
Post by: NE2000 on June 29, 2009, 06:33:28 pm
When compression occurs this means the sound is very loud (ie it is produces loudest sound)
But when rarefaction occurs does this also produce the loudest sound?
That's the way I saw it. Because if you graph the pressure, those are both points of maximum amplitude.
Post by: kamil9876 on June 29, 2009, 06:37:43 pm
Post by: TrueTears on June 29, 2009, 06:41:48 pm
Post by: NE2000 on June 29, 2009, 06:43:20 pm
Post by: TrueTears on June 29, 2009, 08:18:32 pm
Post by: kamil9876 on June 29, 2009, 08:57:31 pm
However
The A and t cancel out and so:
and therefore:
Hence the difference in decibels is 10dB
Post by: NE2000 on June 29, 2009, 09:11:19 pm
Which now that I actually fully read kamil's post is basically just what kamil did. :laugh:
Post by: TrueTears on June 29, 2009, 09:36:34 pm
2 students pat and morgan are discussing an experiment testing the nature of sound. They imagine a speaker with a dust particle motionless in front of it and discuss what will happen to the particle when the speaker is turned on. pat says that since there is energy transferred by the wave the particle will gain energy and a succession of little impulses will push the particle continuously away from the speaker. Morgan agrees that energy is carried by the wave however the result of pressure variations will cause the dust particle to move back and forth about its original position.
Which one or more of the follow statements is true
A. Pat is correct
B. morgan is correct
C. A longitudinal pressure wave interacts with the dust particle
D. The particle will move from high pressure areas into low pressure areas
I know the answer is B C and D, but I'm wondering how exactly does compression and rarefaction in a sound wave cause the particle to oscillate?
Post by: TrueTears on June 30, 2009, 01:44:10 am
The threshold of hearing varies with frequency. To calculate intensity in decibels, I and Io must be at the same frequency. [the question will provide you with enough information to solve it]
Thanks Mao, but for sound intensity level, you know how if the frequency is lower then a higher sound intensity level is needed in order to have the same loudness with if the frequency was higher then a lower sound intensity is needed.
So
If the frequency decreases then the sound wave carries less energy so you will need a higher "I", but this I is at a different frequency to
Furthermore if the frequency increases then the sound wave carries more energy so a lower "I" is needed to produce same loudness, but again this "I" is at a different frequency to
Post by: Mao on June 30, 2009, 01:52:23 am
The threshold of hearing varies with frequency. To calculate intensity in decibels, I and Io must be at the same frequency. [the question will provide you with enough information to solve it]
Thanks Mao, but for sound intensity level, you know how if the frequency is lower then a higher sound intensity level is needed in order to have the same loudness with if the frequency was higher then a lower sound intensity is needed.
So
If the frequency decreases then the sound wave carries less energy so you will need a higher "I", but this I is at a different frequency toso how do you work out L?
Furthermore if the frequency increases then the sound wave carries more energy so a lower "I" is needed to produce same loudness, but again this "I" is at a different frequency to
Io changes with frequency. To calculate intensity at a frequency other than 1000hz, you will be given the respective threshold of hearing. (Or a graph to extrapolate the data from)
Another formula to remember for intensity is
Post by: TrueTears on June 30, 2009, 02:05:53 am
Also is the fundamental frequency always the one that resonates the strongest?
Post by: NE2000 on June 30, 2009, 09:30:32 am
Thanks.
2 students pat and morgan are discussing an experiment testing the nature of sound. They imagine a speaker with a dust particle motionless in front of it and discuss what will happen to the particle when the speaker is turned on. pat says that since there is energy transferred by the wave the particle will gain energy and a succession of little impulses will push the particle continuously away from the speaker. Morgan agrees that energy is carried by the wave however the result of pressure variations will cause the dust particle to move back and forth about its original position.
Which one or more of the follow statements is true
A. Pat is correct
B. morgan is correct
C. A longitudinal pressure wave interacts with the dust particle
D. The particle will move from high pressure areas into low pressure areas
I know the answer is B C and D, but I'm wondering how exactly does compression and rarefaction in a sound wave cause the particle to oscillate?
Just imagine those diagrams of compression and rarefaction. The whole idea of waves is a transfer of energy without 'net' movement of particles. But there is still particle movement. So if you have an animation or a diagram or something you can see all those small dots moving out (rarefaction) and in (compression). That's the oscillation. The dust particle just represents one of those small dots.
Post by: TrueTears on June 30, 2009, 06:09:19 pm
If you have a sound source of one frequency placed at the open end of a pipe that is closed the other end. Then only ONE standing wave will form right? [by ONE standing wave I mean a standing wave of only a certain frequency, not like heaps of frequencies mixed together]
Post by: NE2000 on June 30, 2009, 06:53:08 pm
Thanks.
If you have a sound source of one frequency placed at the open end of a pipe that is closed the other end. Then only ONE standing wave will form right? [by ONE standing wave I mean a standing wave of only a certain frequency, not like heaps of frequencies mixed together]
Based on what you said on the thread on multiple harmonics I would think yes. A sound wave with a node at the open end and an antinode at the closed end will form and the frequency will depend on the frequency of sound you put into the pipe.
Post by: TrueTears on July 01, 2009, 12:00:34 am
So only one standing wave will form if it is just one sound source with a certain frequency?Thanks.
If you have a sound source of one frequency placed at the open end of a pipe that is closed the other end. Then only ONE standing wave will form right? [by ONE standing wave I mean a standing wave of only a certain frequency, not like heaps of frequencies mixed together]
Based on what you said on the thread on multiple harmonics I would think yes. A sound wave with a node at the open end and an antinode at the closed end will form and the frequency will depend on the frequency of sound you put into the pipe.
Post by: TrueTears on July 01, 2009, 04:03:52 pm
Q2,
(http://img394.imageshack.us/img394/5307/candle.jpg)
When the loud speaker emits a sound with a frequency of 10Hz, the flame of the candle moves towards and away from the loudspeaker with a frequency of 10Hz. Which one or more of the following best explains the reason for the movement of the candle flame?
A. Sound is a longitudinal pressure wave.
B. The vibration of the air molecules is parallel to the table surface.
C. The candle is pushed by the transverse motion of the air pressure.
D. The candle always vibrates in phase with the loudspeaker.
I said A, B and D. But answer is A and B. Why is D not correct? Can't we just consider the flame as a particle, so when the loudspeaker emits sound at the flame there will be a period of compression followed by a period of rarefaction and this process keeps on going until the loudspeaker is turned off. So the flame (as a particle) is basically part of the process of compression/rarefaction particles, so when the air particles compress the candle should do the same, when the air particles undergoes rarefaction the candle should also do the same. Hence they are in phase?
Q3, A pedestrian tunnel near a busy road sometimes resonates at a low frequency when there is enough background noise. A particular tunnel is 3.1m long and has a resonant frequency close to 55 Hz. The traffic noise contains many different frequencies, but the resonating frequency is quite specific. Which of the following best explains this?
A. Waves from each speaker arrive at this point in phase
B. Waves from each speaker have a path difference of exactly 1 wavelength
C. Waves from each speaker have a path difference of exactly 0
D. The speakers are specially designed to keep in phase
I picked A. But answers says A and B.
I don’t see why B is right, the question stated “The traffic noise contains many different frequencies, but the resonating frequency is quite specific”. It never mentioned a specific resonant frequency such as first resonant frequency, 2nd/ 3rd and so on. So if you picked B, it does indeed describe why a resonant frequency is very loud (because of , constructive interference) but it is only the case for ONE resonant frequency, there is an infinite amount of resonant frequencies that could form in the tunnel. So shouldn’t the answer just be A? [Furthermore, if B was picked then should C also be picked since that also describes why a resonant frequency is loud. But again it only accounts for ONE certain resonant frequency.]
Q4, An empty circular cylinder (open at one end only) resonates weakly to frequencies of 1536 Hz and 2560 Hz. It does not resonate to frequencies in between these values. Which one of the following values is likely to correspond to a strong resonant frequency?
A. 3072 Hz
B. 2048 Hz
C. 1024 Hz
D. 512 Hz
Clearly answer is D because since its only at 1 end only, then it follows that
So clearly
But what makes a resonant frequency resonate weakly and what makes it resonate strongly? So for example if the choices for the answer were like this:
A. 4608 Hz
B. 2048 Hz
C. 1024 Hz
D. 512 Hz
Clearly if we assume 512 Hz is the fundamental frequency then 4608 = 512 x 9, which is also another possible resonant frequency to occur in the cylinder. But then how do you know which of those 2 frequencies (the 4608 Hz or the 512 Hz) resonates louder? Is there a rule like the fundamental frequency always resonates the strongest etc? And how would you compare 2 resonant frequencies, excluding the fundamental frequency, as to which resonates stronger?
Q5,
(http://img34.imageshack.us/img34/5115/pipeg.jpg)
A very large pipe in a science museum is open at both ends, as shown. It is large enough for people to walk inside it. A large loudspeaker faces one end. Jin is walking along the pipe. During a demonstration one frequency resonates strongly in the pipe. Which of the following is the best explanation of the cause of this resonance?
A. A standing wave is generated in the loudspeaker
B. Waves moving to the left from the loudspeaker reflect from Jin
C. Waves moving to the left from the loudspeaker reflect from the far end
D. Odd harmonics are formed in the pipe
I know the answer is C, because since both ends are open, then a standing wave is formed due to reflections of compression/rarefactions at both open ends. But why is A not correct? Isn’t a standing wave formed? Or should A say “A standing wave is generated in the PIPE” ie, not in the loudspeaker, then it would be correct?
Q6,
(http://img526.imageshack.us/img526/9441/bugle.jpg)
Students are exploring why a bugle can be used to produce a range of notes, even if it is of fixed length. They model the bugle and player by using a length of pipe as shown with a sound source placed at S. They expect that this system will act as a pipe open at one end only. It is found that the sound emitted by a bugle normally consists of more than one frequency. Which of the following best explains how this is possible?
A. The resonant frequency can change from one of the harmonics to one of the others.
B. It is possible to excite several harmonics at the same time.
C. Reflections from the sides of the pipe are responsible for this phenomenon.
D. The player’s lips open and shut one end of the pipe as they vibrate.
The answer is A, but I don’t get exactly what it means. I know that, for example, if you have the first resonant frequency then it produces the 3rd harmonic (because one end is open another end is closed). But what does it mean when it says “can change from one of the harmonics to one of the others”. Does this mean the resonant frequency changes and hence the harmonics also change or does it mean the resonant frequency stays the same and the harmonics change [which is not possible because if the resonant frequency changes the harmonics must change as well]?
Q7,
(http://img43.imageshack.us/img43/1829/flutef.jpg)
Jack uses a tube closed at one end to model a wind instrument (a flute). By changing the frequency of a small loudspeaker very close to the open end, he creates resonances at several different frequencies. A flute soloist and a pianist travel to Nepal, where the speed of sound is 10% greater than at sea level, to perform. Which of the following statements are the best summary of the tuning implications for both musicians?
A. The tuning of both instruments will be unchanged
B. The flute will need to be lengthened to stay in pitch with the piano
C. The flute will need to be shortened to stay in pitch with the piano
D. The pitch of the flute will be unchanged but the piano will need to have its pitched reduced.
How do you go about this question?
Q8, (http://img5.imageshack.us/img5/7128/exam2003.jpg)
The ear can be modeled as a tube closed at one end. The length of the ear canal determines the length L of this model tube. The diagram below shows the modeled situation for a human.
(http://img193.imageshack.us/img193/8828/earr.jpg)
If the ear canal in a human is approximately 2 cm long, which of the following is closest to the approximate length of the ear canal of an elephant?
A. 10 cm
B. 11 cm
C. 12 cm
D. 13 cm
How would you go about this question?
Q9, Q10, In order to study resonance in air columns students use a narrow tube of length 0.5m that is closed at one end and open at the other. They use a signal generator and loudspeaker as shown in the diagram below.
(http://img26.imageshack.us/img26/1433/q10ofsound.jpg)
The students begin the experiment by using a sound of frequency 100 Hz. The students increase the frequency until the first resonance (first harmonic) is reached. Which of the following best describes what the students will hear that will enable them to identify this resonance frequency?
A. An increase in intensity to a maximum at the resonant frequency.
B. A decrease in intensity to a minimum at the resonant frequency.
C. An increase in sensitivity to a maximum at the resonant frequency.
D. A decrease in sensitivity to a minimum at the resonant frequency.
Because it is the first harmonic means the students increased the frequency of the loudspeaker until the fundamental frequency standing wave was created. I picked A and it is the correct answer. But I assumed that the fundamental frequency always resonates the strongest and that a stronger resonating frequency produces a larger intensity sound. So what I’m wondering is, does the fundamental frequency standing wave always resonate the strongest? If so, does that mean the stronger the resonating frequency resonates the louder the sound is (ie a higher intensity)? And how would you tell between 2 resonating frequencies which resonates stronger? So say I had the fundamental frequency (for a pipe closed at 1 end and open at another end) compared with its 4th resonant frequency, which one would resonate stronger? And which one would produce the louder sound?
Post by: mark_alec on July 01, 2009, 07:16:49 pm
What is the difference here between a standing wave and just a sound wave travelling "left". Do all standing waves (such as, open at both ends, open one end closed other end, a string tied to fix ends) move 'vertically'? ie, if we are required to draw the graph (pressure – distance) of a standing wave after (a quarter of a period) has passed we just change the amplitude. Where as for a sound wave that's actually MOVING we just have to shift it horizontally, ie left and right?A standing wave always has its nodes/antinodes in the same place. For a travelling wave, they travel.
Quote
I said A, B and D. But answer is A and B. Why is D not correct? Can't we just consider the flame as a particle, so when the loudspeaker emits sound at the flame there will be a period of compression followed by a period of rarefaction and this process keeps on going until the loudspeaker is turned off. So the flame (as a particle) is basically part of the process of compression/rarefaction particles, so when the air particles compress the candle should do the same, when the air particles undergoes rarefaction the candle should also do the same. Hence they are in phase?Can you conceive that there is a delay in the candle flame reacting to the pressure wave that passes over it? If there is any delay, then it is not in phase.
I'll get to the other questions later.
Post by: TrueTears on July 01, 2009, 09:55:09 pm
Ahh I get it now, thanks so much mark.What is the difference here between a standing wave and just a sound wave travelling "left". Do all standing waves (such as, open at both ends, open one end closed other end, a string tied to fix ends) move 'vertically'? ie, if we are required to draw the graph (pressure – distance) of a standing wave after (a quarter of a period) has passed we just change the amplitude. Where as for a sound wave that's actually MOVING we just have to shift it horizontally, ie left and right?A standing wave always has its nodes/antinodes in the same place. For a travelling wave, they travel.QuoteI said A, B and D. But answer is A and B. Why is D not correct? Can't we just consider the flame as a particle, so when the loudspeaker emits sound at the flame there will be a period of compression followed by a period of rarefaction and this process keeps on going until the loudspeaker is turned off. So the flame (as a particle) is basically part of the process of compression/rarefaction particles, so when the air particles compress the candle should do the same, when the air particles undergoes rarefaction the candle should also do the same. Hence they are in phase?Can you conceive that there is a delay in the candle flame reacting to the pressure wave that passes over it? If there is any delay, then it is not in phase.
I'll get to the other questions later.
Post by: mark_alec on July 01, 2009, 11:07:52 pm
The relative loudness of different harmonics is what gives different musical instruments their characteristic sound (a piano and a guitar playing middle C is the same fundamental frequency, but you don't need to be a musical genius to recognise that they don't sound identical.) You won't be required to know which harmonic resonates the most, so the question has no ambiguity with "strong" or "weak" resonance.
Q5
A isn't correct for the reasons you said. If it said "by the loudspeaker" or "in the pipe", then A would also be correct.
Q6
I would've thought B was the answer.
Q7
You can assume that the speed of sound doesn't change the frequency a piano produces, since the sound is generated in a wire. So given that the piano remains tuned, what can you find out about the length of tube that needs to make the same frequency, given that the velocity of sound is less (remember that frequency*wavelength = velocity.)
Q8
The length of the tube should be such that the fundamental frequency is the one the elephant is most sensitive to.
Q9
Your test is flawed is it is relying on what you hear, which as you should know, is not indicative of the actual sound energy (sensitivities change with frequency.) Given an ideal tube, with an energy loss not dependent on frequency, one would expect that all harmonics should resonate equally well.
Post by: TrueTears on July 02, 2009, 12:24:49 am
Post by: TrueTears on July 19, 2009, 10:42:00 pm
Post by: Mao on July 20, 2009, 03:01:06 pm
Ah a stupid q, what's the point of adding a number of coils around a wire that has a current in it for a DC motor? It increases the magnetic field force so the motor spins faster yeah? How though?
Try breaking a pencil with one hand
Now try breaking ten pencils with one hand at the same time.
Same principle, there are N coils, hence there are N streams of current running parallel, if the force on each current is F, the net force on the coil will hence be N*F.
Post by: TrueTears on July 20, 2009, 04:34:04 pm
Post by: TrueTears on August 06, 2009, 12:09:40 am
a) all in series.
b) all in parallel.
c) an 8 ohm and 12 ohm are in series, the rest are in parallel.
d) the 2 12 ohm resistors are in series and the 8 ohm and the 16 ohm are in series.
Thanks!
Post by: Mao on August 06, 2009, 02:17:33 pm
Post by: TrueTears on August 06, 2009, 03:28:27 pm
draw out each of the options as a circuit, calculate effective resistance, the one around 5.1 ohms will be the correct answer.Yeah that's what I did lol, none of the choices gives anywhere close to 5.1 ohms, I think they are wrong lols.
Post by: TonyHem on August 06, 2009, 03:44:39 pm
4 resistors are chosen from a box with values of 8, 12, 12 and 16 ohms. When they are connected together the total resistance is found to be close to 5.1 ohms. How are the resistors connected?
a) all in series.
b) all in parallel.
c) an 8 ohm and 12 ohm are in series, the rest are in parallel.
d) the 2 12 ohm resistors are in series and the 8 ohm and the 16 ohm are in series.
Thanks!
C
Add the 1st two together, 1/20 + 1/12 + 1/16 = 1/reff
Post by: TrueTears on August 30, 2009, 03:52:07 pm
For a pipe open at both ends then the longest wavelength you could have would be like 1
For a pipe closed at one end and open at another then the longest wavelength you could have would be like 2
But for a pipe closed at both ends then what would the longest wavelength look like? [I know this is not in the VCE course but a question poped up in a trial exam so just for the heck of it]
Also just to confirm the formulas for resonant frequencies etc for pipe closed at both ends are the same for pipe open at both ends right?
Post by: mark_alec on August 30, 2009, 04:10:11 pm
But for a pipe closed at both ends then what would the longest wavelength look like? [I know this is not in the VCE course but a question poped up in a trial exam so just for the heck of itKeeping in line with what you have drawn (pressure anti-nodes at open ends), then in a pipe with two closed ends, you will have pressure nodes at the ends, and an anti-node in the middle (for the fundamental frequency).
Also just to confirm the formulas for resonant frequencies etc for pipe closed at both ends are the same for pipe open at both ends right?
The formula for resonant frequencies for a pipe closed at both ends in the same as one open at both ends.
Post by: TrueTears on August 30, 2009, 05:08:14 pm
Thanks I got it.But for a pipe closed at both ends then what would the longest wavelength look like? [I know this is not in the VCE course but a question poped up in a trial exam so just for the heck of itKeeping in line with what you have drawn (pressure anti-nodes at open ends), then in a pipe with two closed ends, you will have pressure nodes at the ends, and an anti-node in the middle (for the fundamental frequency).
Also just to confirm the formulas for resonant frequencies etc for pipe closed at both ends are the same for pipe open at both ends right?
The formula for resonant frequencies for a pipe closed at both ends in the same as one open at both ends.
Post by: TrueTears on August 30, 2009, 05:29:11 pm
1.
Let I be intensity and A be the amplitude of a sound wave
then
For this rule, is the amplitude referring to the amplitude of a
And does it hold for both standing waves and "moving" sound waves or just one of those?
[Reason I'm asking this question is cause I just saw this in the answer of a trial exam but I've never seen the rule anywhere in Jacaranda or in my notes so just want a bit of confirmation]
2.
http://img194.imageshack.us/img194/3058/soundsac2007.jpg
Any idea on question 8?
[Also the question says "...on another day James repeats the reading..." does this mean he is still 15 m away from the speaker when he takes the reading? =S]
Actually I had a go but I am not sure if this way is correct...
so at where James stands the intensity is
Which means the intensity at 15+r meters away from the speaker would be
so then
so wall is 30 m away from speaker.
But I'm wondering are you allowed to just minus the 2 intensities from each other? [Because they constructively interfere at where James is but it's the amplitude that undergoes addition of ordinates, does that mean the intensities do the same thing?]
3.
Just something trivial but for a pipe open at both ends, the 2nd resonant frequency is called the 1st overtone right?
Then for a pipe closed one end and open the other, you can only have odd resonant frequencies, so the 1st resonant freq(the fundamental freq) is called the 1st harmonic, then is the next resonant freq called the 3rd resonant freq or the 3rd harmonic? Or is it the 2nd resonant frequency = 3rd harmonic?
And would this 3rd harmonic be called the 1st overtone?
Post by: TrueTears on August 30, 2009, 06:12:39 pm
Is it defined as "The point where 2 sound waves constructively interfere" or is it "The point where 2 sound waves constructively interfere to give a MAXIMUM variation in air pressure"
Eg, consider the following graph of a fundamental frequency in a pipe open at both ends.
(http://img291.imageshack.us/img291/6239/physicspipe2.jpg)
At A in the first diagram, that is obviously a pressure antinode.
However in the 2nd diagram, at B, would that also be a pressure antinode?
Because if pressure antinode is defined as "The point where 2 sound waves constructively interfere to give a MAXIMUM variation in air pressure" then B wouldn't be a pressure antinode would it? Since only A produces the MAXIMUM variation in air pressure, B still constructively interferes but it doesn't produce a MAXIMUM variation in pressure.
But if pressure antinode was defined as "The point where 2 sound waves constructively interfere" then that would mean EVERYWHERE on the standing wave (except at the 2 ends) can be called a pressure antinode, because they are all formed from constructive interference.
Or is an antinode "the maximum variation of a sound wave at that instant in time". So say you had that standing wave in the first diagram, then at A is the antinode at THAT moment in time, but after 1/8 of a period, the antinode is now at B, which is the antinode again at THAT specific moment in time.
So what is actually the definition of a pressure antinode?
Thanks!
Post by: TrueTears on August 30, 2009, 07:19:47 pm
True or False
"The fundamental transverse wave on a guitar string travels at approx 340
And...
For Tweeter, woofer loudspeakers what are the approx freq ranges they work from?
Like I know tweeter is used for high freq and woofer loudspeakers are for low freq, but what is the approx ranges for "low" and "high" freq ranges?
Also what is the approx. freq range for a midrange loudspeaker??
Post by: NE2000 on August 30, 2009, 08:10:01 pm
Post by: TrueTears on August 30, 2009, 08:25:24 pm
How about, the pressure antinode is a point on the standing wave where the two waves constructively interfere the most regardless of what the time is, this antinode remains the point where the two waves constructively interfere the most (creating a maximal variation in air pressure at that particular point). That is, the antinode is where there is a maximum air pressure variation in relation to the rest of the standing wave at any particular time. The only problem with this definition is the zero variation instant of a standing wave...Thanks man!
Post by: TrueTears on August 30, 2009, 09:17:38 pm
2. A student plays a high frequency note on the trumpet. A second student stands 5 m directly in front of the trumpet and measures the sound intensity level. Explain why the measured sound level is lower when the student plays a low frequency note of the same power.
Post by: mark_alec on August 30, 2009, 11:13:25 pm
Post by: TrueTears on August 30, 2009, 11:25:19 pm
lower freq = bigger wavelength = more diffraction. The lower freq wave thus records a lower sound intensity since it spreads out to a much better area.
Post by: Mao on August 30, 2009, 11:56:09 pm
But for a pipe closed at both ends then what would the longest wavelength look like? [I know this is not in the VCE course but a question poped up in a trial exam so just for the heck of itKeeping in line with what you have drawn (pressure anti-nodes at open ends), then in a pipe with two closed ends, you will have pressure nodes at the ends, and an anti-node in the middle (for the fundamental frequency).
Also just to confirm the formulas for resonant frequencies etc for pipe closed at both ends are the same for pipe open at both ends right?
The formula for resonant frequencies for a pipe closed at both ends in the same as one open at both ends.
Isn't it pressure nodes at open ends and pressure anti-nodes at closed ends?
Post by: TrueTears on August 31, 2009, 07:12:55 pm
I did
So when
Hence intensity should be 4 times greater than before.
However the answers were the intensity was 2 times greater.
But isn't that
Post by: NE2000 on August 31, 2009, 07:45:33 pm
Um again regarding the question of. I checked with my teacher and he said that that law can be used for ANY waves, not just sound, but he also said for VCE they won't ask you anything related to that formula. So what I'm wondering is this: I met a question that said what would happen to the intensity of a sound wave if the amplitude is doubled.
I did
So when
Hence intensity should be 4 times greater than before.
However the answers were the intensity was 2 times greater.
But isn't that, not
Just my opinion:
If it appears it may be safer to forget the squared. The reason being that we are given in our books this idea that in sound waves energy is proportional to the amplitude of the wave. It might be one of those instances where there is a difference between what is right and what is right.
Post by: /0 on August 31, 2009, 07:48:25 pm
The energy of an oscillating particle is
Since intensity is proportional to the rate at which energy is transferred through a medium, we have
Post by: TrueTears on August 31, 2009, 07:49:39 pm
It isYeap, that's what kamil said as well...
The energy of an oscillating particle is, where
is the 'spring' constant and
is the amplitude.
Since intensity is proportional to the rate at which energy is transferred through a medium, we havewhere
depends on the type of wave.
So in an exam they shouldn't take a mark off right...? Because it is technically correct...
Post by: /0 on August 31, 2009, 07:58:06 pm
It isYeap, that's what kamil said as well...
The energy of an oscillating particle is
Since intensity is proportional to the rate at which energy is transferred through a medium, we have
So in an exam they shouldn't take a mark off right...? Because it is technically correct...
I think it's always safer to use the 'more' correct answer. Dr. B said the folks over at the examining panel are quite capable physicists in their own right, so they should recognise a correct answer when they see one (though they suck at writing exams).
Post by: TrueTears on August 31, 2009, 07:58:42 pm
LOL what if he is drunk man, I wouldn't wanna lose some marks coz of his drunknessIt isYeap, that's what kamil said as well...
The energy of an oscillating particle is
Since intensity is proportional to the rate at which energy is transferred through a medium, we have
So in an exam they shouldn't take a mark off right...? Because it is technically correct...
I think it's always safer to use the 'more' correct answer. Dr. B said the folks over at the examining panel are quite capable physicists in their own right, so they should recognise a correct answer when they see one (though they suck at writing exams).
Post by: Mao on August 31, 2009, 09:47:28 pm
Post by: TrueTears on August 31, 2009, 09:47:51 pm
The examination panel have a meeting prior to the release to the marking scheme, so all examiners may have their input on what should be accepted as a correct answer. Amongst these teachers are very capable academics, so any answers that are technically correct with reasoning beyond the VCE level will be accepted as correct (that is, if they differ from the VCE answers).I see, thanks!
Post by: TrueTears on September 01, 2009, 12:33:29 am
How about, the pressure antinode is a point on the standing wave where the two waves constructively interfere the most regardless of what the time is, this antinode remains the point where the two waves constructively interfere the most (creating a maximal variation in air pressure at that particular point). That is, the antinode is where there is a maximum air pressure variation in relation to the rest of the standing wave at any particular time. The only problem with this definition is the zero variation instant of a standing wave...Ahh I just cleared it up with teacher day, the antinode is actually the PLACE where the maximum variation of air pressure occurs, so every POINT on that "PLACE" is an antinode, so when the standing wave is at its quarter of a period (straight line on a pipe open at both ends) then it still has an antinode in the middle.
Post by: TrueTears on October 11, 2009, 10:45:31 pm
For double slit experiments regarding the diffraction of light:
1.
Nodal points:
Where:
2.
Where:
Antinodal points:
Simply change the above formulas, instead of
For single slit experiments regarding the diffraction of light:
Nodal points:
1.
Where:
2.
Where:
Antinodal points:
Does not exist for single slit.
Post by: naved_s9994 on October 12, 2009, 05:47:27 pm
Thanks :)
Post by: TrueTears on October 12, 2009, 05:56:26 pm
Thus
If one variable increases, the other must decrease in order to compensate.
Is that what you're looking for?
Post by: naved_s9994 on October 12, 2009, 06:02:01 pm
One more thing. Describe why the wave model of light can not account for
the experimental results produced by the photoelectric effect.
Thanks!
Post by: TrueTears on October 12, 2009, 06:08:06 pm
Yea, preety much. Thanks! :)Several reasons.
One more thing. Describe why the wave model of light can not account for
the experimental results produced by the photoelectric effect.
Thanks!
1. Higher intensity light would produce electrons with higher kinetic energies which would mean the stopping voltage depends on light intensity, which it doesn't.
2. There would be a time delay while enough shared energy is accumulated for electrons to escape the late. Thus this delay would be shorter/longer for higher intensity light. Obviously this is not observed in the photoelectric effect.
3. The waiting time for electrons to emerge would be longer using lower freq light (since they have lower energy) which means all there would be a photocurrent no matter what happens as long as you keep the light shining on the plate for a long time. Obviously this does not happen.
Post by: dcc on October 12, 2009, 06:40:56 pm
1. Higher intensity light would produce electrons with higher kinetic energies which would mean the stopping voltage depends on light intensity, which it doesn't.
2. There would be a time delay while enough shared energy is accumulated for electrons to escape the late. Thus this delay would be shorter/longer for higher intensity light. Obviously this is not observed in the photoelectric effect.
3. The waiting time for electrons to emerge would be longer using lower freq light (since they have lower energy) which means all there would be a photocurrent no matter what happens as long as you keep the light shining on the plate for a long time. Obviously this does not happen.
Your explanation of #1 misses the point (that is, explaining the inadequacies of the wave model of light). Perhaps it would be best if some explanation was provided as to why the wave model of light predicts that more intense light frees up electrons with higher kinetic energies?
Post by: NE2000 on October 12, 2009, 06:44:46 pm
Yea, preety much. Thanks! :)Several reasons.
One more thing. Describe why the wave model of light can not account for
the experimental results produced by the photoelectric effect.
Thanks!
1. Higher intensity light would produce electrons with higher kinetic energies which would mean the stopping voltage depends on light intensity, which it doesn't.
2. There would be a time delay while enough shared energy is accumulated for electrons to escape the late. Thus this delay would be shorter/longer for higher intensity light. Obviously this is not observed in the photoelectric effect.
3. The waiting time for electrons to emerge would be longer using lower freq light (since they have lower energy) which means all there would be a photocurrent no matter what happens as long as you keep the light shining on the plate for a long time. Obviously this does not happen.
I think you may have a slight misconception here TT (based on what you wrote on point 3). Under the wave model of light (based on modelling water waves for example) it is the amplitude of the wave that determines the energy it carries while the frequency would determine simply how much energy is transferred to the electrons per second. So the reason low frequency would mean greater waiting time is it takes longer for adequate energy to accumulate.
Post by: TrueTears on October 12, 2009, 06:45:34 pm
Sorry I was referring to the photoelectric experiment/effect for #1.1. Higher intensity light would produce electrons with higher kinetic energies which would mean the stopping voltage depends on light intensity, which it doesn't.
2. There would be a time delay while enough shared energy is accumulated for electrons to escape the late. Thus this delay would be shorter/longer for higher intensity light. Obviously this is not observed in the photoelectric effect.
3. The waiting time for electrons to emerge would be longer using lower freq light (since they have lower energy) which means all there would be a photocurrent no matter what happens as long as you keep the light shining on the plate for a long time. Obviously this does not happen.
Your explanation of #1 misses the point (that is, explaining the inadequacies of the wave model of light). Perhaps it would be best if some explanation was provided as to why the wave model of light predicts that more intense light frees up electrons with higher kinetic energies?
To elaborate more, since the energy in a wave is shared, that means a higher intensity light would deliver energy at a greater rate.
Post by: TrueTears on October 12, 2009, 06:48:01 pm
Post by: naved_s9994 on October 12, 2009, 06:51:01 pm
Obviously, when ans required, in eVs, 4.14, but otherwise..?
Post by: Flaming_Arrow on October 12, 2009, 06:55:16 pm
When to use 6.63, when to use 4.14.
Obviously, when ans required, in eVs, 4.14, but otherwise..?
if its kinetic energy(J) then u use 6.63
Post by: NE2000 on October 12, 2009, 07:01:47 pm
Post by: naved_s9994 on October 12, 2009, 07:34:10 pm
It's always safer to use 6.63 in everything other than where you know eV or V are explicitly involved
Thanks :)
Post by: TrueTears on October 14, 2009, 04:49:12 pm
Post by: /0 on October 14, 2009, 08:03:53 pm
(i.e. they weren't giving the units, they were giving the 'symbol')
Post by: TrueTears on October 14, 2009, 08:06:51 pm
I think they probably meantlol I thought they meant Voltage Second.with 's' as a subscript
(i.e. they weren't giving the units, they were giving the 'symbol')
That was stupid of them...
Post by: naved_s9994 on October 15, 2009, 07:40:12 pm
Just a few clarifications of some formulas which are not on the course but still pretty fun/useful. Can someone check if they're right or not? I derived them long time ago but can't really remember it now lol.
For double slit experiments regarding the diffraction of light:
1.
Nodal points:
Where:is the horizontal distance of the
nodal point to the centre maximum point.
is the width of the the 2 slits (ie the distance from the middle of 1 slit to the other one)
is the distance from the centre of the 2 slits to the
2.
Where:
is the angle between the
Antinodal points:
Simply change the above formulas, instead of, it will just be
For single slit experiments regarding the diffraction of light:
Nodal points:
1.
Where:
2.
Where:
Antinodal points:
Does not exist for single slit.
So do these work? Anyone gone ahead and tested them?
Post by: TrueTears on October 15, 2009, 07:42:37 pm
Post by: kamil9876 on October 15, 2009, 08:02:55 pm
Quote
So do these work? Anyone gone ahead and tested them?http://en.wikipedia.org/wiki/Thomas_Young_(scientist)
Post by: naved_s9994 on October 15, 2009, 08:07:46 pm
How do you calculate that?
EDIT: 500 Posts :D ;) ;D :coolsmiley:
Post by: TrueTears on October 17, 2009, 01:53:42 am
(http://img39.imageshack.us/img39/7522/tssm2008physics.jpg)
First two are: NO CHANGE, DECREASE respectively.
What's the third one? Rather how do you work it out?
Post by: Mao on October 17, 2009, 02:02:51 am
Post by: TrueTears on October 17, 2009, 02:09:22 am
DC motor: constant voltage. no changeThat's what I thought too, but the answer says increase.
Apparently their reasoning is: "Increased (zero at horizontal)"
huh?
Post by: kamil9876 on October 17, 2009, 02:16:11 am
Post by: kurrymuncher on October 17, 2009, 02:19:05 am
DC motor: constant voltage. no changeThat's what I thought too, but the answer says increase.
Apparently their reasoning is: "Increased (zero at horizontal)"
huh?
What! how could the current be zero when it is a motor lol?
what brand is this?
Post by: TrueTears on October 17, 2009, 02:21:35 am
TSSM 2008, they have a habit of making some weirdo questions...DC motor: constant voltage. no changeThat's what I thought too, but the answer says increase.
Apparently their reasoning is: "Increased (zero at horizontal)"
huh?
What! how could the current be zero when it is a motor lol?
what brand is this?
Post by: TrueTears on October 17, 2009, 02:22:31 am
A coherent light source is said to be:
"When interfering, two waves can add together to create a larger wave (constructive interference) or subtract from each other to create a smaller wave (destructive interference), depending on their relative phase. Two waves are said to be coherent if they have a constant relative phase."
But what if a question just asked is Laser a coherent light source?
I thought for coherency you need two light sources so they can undergo superimposition, how can one light source be coherent?
Post by: kamil9876 on October 17, 2009, 02:23:19 am
Post by: kurrymuncher on October 17, 2009, 02:32:58 am
Anyway another question:
A coherent light source is said to be:
"When interfering, two waves can add together to create a larger wave (constructive interference) or subtract from each other to create a smaller wave (destructive interference), depending on their relative phase. Two waves are said to be coherent if they have a constant relative phase."
But what if a question just asked is Laser a coherent light source?
I thought for coherency you need two light sources so they can undergo superimposition, how can one light source be coherent?
Coherent just means that light is produced with a constant frequency and phase. A laser source is coherent as it produces light that has a constant frequency, wavelength
Coherent= a light source that produces light with a constant frequency, wavelength
Post by: TrueTears on October 17, 2009, 02:36:02 am
Awesome bro, thanks!Anyway another question:
A coherent light source is said to be:
"When interfering, two waves can add together to create a larger wave (constructive interference) or subtract from each other to create a smaller wave (destructive interference), depending on their relative phase. Two waves are said to be coherent if they have a constant relative phase."
But what if a question just asked is Laser a coherent light source?
I thought for coherency you need two light sources so they can undergo superimposition, how can one light source be coherent?
Coherent just means that light is produced with a constant frequency and phase. A laser source is coherent as it produces light that has a constant frequency, wavelength an phase. A light bulb in our room would not be coherent as it would produce light with all sorts of frequencies and phases.
Coherent= a light source that produces light with a constant frequency, wavelength and phase
Also, what do you mean by constant phase? How do you have a single light source to be in phase with itself? I thought it only applies two 2 more or more sources.
Post by: kurrymuncher on October 17, 2009, 02:42:07 am
Awesome bro, thanks!Anyway another question:
A coherent light source is said to be:
"When interfering, two waves can add together to create a larger wave (constructive interference) or subtract from each other to create a smaller wave (destructive interference), depending on their relative phase. Two waves are said to be coherent if they have a constant relative phase."
But what if a question just asked is Laser a coherent light source?
I thought for coherency you need two light sources so they can undergo superimposition, how can one light source be coherent?
Coherent just means that light is produced with a constant frequency and phase. A laser source is coherent as it produces light that has a constant frequency, wavelength an phase. A light bulb in our room would not be coherent as it would produce light with all sorts of frequencies and phases.
Coherent= a light source that produces light with a constant frequency, wavelength and phase
Also, what do you mean by constant phase? How do you have a single light source to be in phase with itself? I thought it only applies two 2 more or more sources.
Oh yeah, my bad dude, I wasn't meant to say that lol
Post by: Mao on October 17, 2009, 09:18:51 am
Hence, a single wave always in phase with itself, and is always coherent (same freq, wavelength, and the phase always defer by exact multiples of a period, and wave formula evaluated at any point will coincide with the wave formula evaluated at any other point).
It isn't very hard to have multiple sources generating the same frequency and wavelength (metal vapour lamps, or any type of atomic emission). However, to get the same phase, it is very very difficult.
How laser work is you have multiple atoms emitting light, and you have multiple waves with the same frequency and wavelength but different phase. The ends of a laser are 100% and 99.9% reflective mirrors, and waves travel a huge distance before escaping the cavity. Over a huge distance, the waves can be thought to superimpose and form a single wave, and hence is coherent.
Post by: TrueTears on October 17, 2009, 03:52:56 pm
(http://img382.imageshack.us/img382/9854/debroglien3.jpg)
Thank you!
Post by: TrueTears on October 17, 2009, 03:54:43 pm
Thanks
Post by: bem9 on October 17, 2009, 05:29:44 pm
For n = 3, the "standing wave" formed by the electron in the energy level, does it look like this?
(http://img382.imageshack.us/img382/9854/debroglien3.jpg)
Thank you!
yep im pretty sure it does,
i drew this for n=2, 3, 4, 5, 6 except my dotted lines and solid lines are the wrong way around in some, hope it helps :)
Post by: dejan91 on October 17, 2009, 05:50:10 pm
Wavelength of electron = 6.2x10-12
Wavelength of X-Ray = 110x10-12
For 'Moderate' diffraction to occur, ratio of
For 'Significant' diffraction to occur, ratio of
The only option where this is possible is 2.0x10-12 = C.
i.e 6.2x10-12 / 2.0x10-12 = 3.1
and 110x10-12 / 2.0x10-12 = 55
Post by: TrueTears on October 17, 2009, 06:21:01 pm
Here's my reasoning (not sure if I'm right though...):Awesome thanks dejan! It's what I planned to do until I thought of something, maybe I'm just confusing myself but anyways:
Wavelength of electron = 6.2x10-12
Wavelength of X-Ray = 110x10-12
For 'Moderate' diffraction to occur, ratio ofapproxitately (don't know how to do the approximately sign in latex :P)
For 'Significant' diffraction to occur, ratio ofmust be much greater than one.
The only option where this is possible is 2.0x10-12 = C.
i.e 6.2x10-12 / 2.0x10-12 = 3.1
and 110x10-12 / 2.0x10-12 = 55
Say you got the double slit experiment and you got the first antinodal point, ie n = 1.
Now
Let's say it's moderate diffraction,
Thus
which would mean the first antinodal point is way out there at 'infinity' (Sketch a diagram to be sure)
So how can
[ \approx for approximately equal to in LaTeX :P]
Post by: TrueTears on October 17, 2009, 06:25:17 pm
Thanks bem, those graphs helps :PFor n = 3, the "standing wave" formed by the electron in the energy level, does it look like this?
(http://img382.imageshack.us/img382/9854/debroglien3.jpg)
Thank you!
yep im pretty sure it does,
i drew this for n=2, 3, 4, 5, 6 except my dotted lines and solid lines are the wrong way around in some, hope it helps :)
Post by: dejan91 on October 17, 2009, 06:41:56 pm
Maybe the angle of diffraction isn't being measured parallel to the slits (like the left side of pic). Maybe it actually diffracts along a perpendicular line through the middle of the slits (right side of pic)?
I haven't seen this anywhere just a thought that came up now.
Not sure of the mathematical reasoning behind this because according to that formula, it doesn't work. As you said,
Post by: TrueTears on October 17, 2009, 06:45:02 pm
So when
Which means the antinodal point will be way out there in infinity lol
Hmm...
Post by: dejan91 on October 17, 2009, 06:49:40 pm
Yeah that's confusing lol I hate not knowing why things are the way they are, but I guess that's just part of the limitations of VCE physics.
Post by: TrueTears on October 17, 2009, 06:51:23 pm
Well I'm an idiot for not knowing that was the actual angle...you learn something new everyday!Yeah haha, I guess I'll just ignore it lol.
Yeah that's confusing lol I hate not knowing why things are the way they are, but I guess that's just part of the limitations of VCE physics.
Thanks for your help again!
Post by: dejan91 on October 17, 2009, 07:03:26 pm
Post by: TrueTears on October 17, 2009, 10:17:51 pm
Is that right?
Post by: Mao on October 18, 2009, 12:28:07 pm
Just a quick question, the brushes in a DC motor basically just allows charge to flow from the power supply to the coil via the commutator and it also allows the commutator to turn more smoothly.
Is that right?
The brushes ensure the commutator stays electrically connected to the power supply.
Post by: TrueTears on October 18, 2009, 04:33:29 pm
Thanks again Mao XDJust a quick question, the brushes in a DC motor basically just allows charge to flow from the power supply to the coil via the commutator and it also allows the commutator to turn more smoothly.
Is that right?
The brushes ensure the commutator stays electrically connected to the power supply.
Post by: homghomg1 on October 18, 2009, 08:21:57 pm
Another question:
(http://img39.imageshack.us/img39/7522/tssm2008physics.jpg)
First two are: NO CHANGE, DECREASE respectively.
What's the third one? Rather how do you work it out?
I was quite confused when i first did that question and thought they were wrong, but the reason is that for a split second, when the commutator is reversing the direction of current in the coil, and the ends of the coil are not touching the metal part, there is zero current.
It's a pretty good question, they should've given a better explanation though, pretty useless if they just state the answer. Specially since these exams don't hesitate to put all sorts of wrong answers which contradict with answers from other papers.
Post by: TrueTears on October 18, 2009, 09:02:08 pm
lol that is so sly. But still it's quite flawed... how do you know what the cummutator is touching when it's at it's horizontal position, it could be touching the brushes and then when it reaches the vertical position, it'd be off the brushes for a split second which means it should be decreased. Anyhow, weird question.Another question:
(http://img39.imageshack.us/img39/7522/tssm2008physics.jpg)
First two are: NO CHANGE, DECREASE respectively.
What's the third one? Rather how do you work it out?
I was quite confused when i first did that question and thought they were wrong, but the reason is that for a split second, when the commutator is reversing the direction of current in the coil, and the ends of the coil are not touching the metal part, there is zero current.
It's a pretty good question, they should've given a better explanation though, pretty useless if they just state the answer. Specially since these exams don't hesitate to put all sorts of wrong answers which contradict with answers from other papers.
Thanks for your help!
Post by: ngRISING on October 18, 2009, 11:54:09 pm
Post by: TrueTears on October 18, 2009, 11:57:23 pm
Post by: kurrymuncher on October 18, 2009, 11:59:04 pm
Also another quick question, standing wave formulas are the same for strings tied at both ends and a pipe open at both ends right?
Yes
Post by: TrueTears on October 19, 2009, 12:01:42 am
Thanks :)Also another quick question, standing wave formulas are the same for strings tied at both ends and a pipe open at both ends right?
Yes
Post by: appianway on October 19, 2009, 03:31:23 pm
Post by: TrueTears on October 19, 2009, 05:21:05 pm
Just remember that the wavelength in air for a wave produced in a string will be different to the wavelength produced in the string, unless the velocity of the wave in the string is the same as the speed of sound. :)Yeap, thanks for that :P
Post by: TrueTears on October 19, 2009, 05:34:15 pm
(http://img26.imageshack.us/img26/3411/neap05questionphysics.jpg)
Answer is D.
My reasoning: As the magnet falls through the loop, the magnetic is pointing upwards (north to south). So the flux is increasing upwards as it falls, thus to oppose this the loop sets up a magnetic field downwards, using RHG rule, the current is clockwise.
As the magnet falls through, the magnetic field is going upwards (out from north and into south). So the flux is decreasing upwards as it falls out, so the loop sets up a magnetic field upwards to compensate for this loss. Thus the current goes anticlockwise.
Is my reasoning right?
The reason I'm just confirming is because I do it the vector way but I thought it'd be good if I can understand this method as well.
Thanks!
Post by: appianway on October 19, 2009, 05:50:28 pm
But you never know. I could be wrong... I'm doing other homework at the moment, so maybe I misread the diagram or did something careless :)
Post by: TrueTears on October 19, 2009, 05:55:10 pm
Did you get the correct answer? I would've said that the initially the magnetic field was increasing DOWNWARDS, as the magnetic field is drawn for N into S. The initial induced current must thus produce an upwards flux, and is therefore anticlockwise.Yeah, it is the right answer, I tried vector method as well and got the same thing.
But you never know. I could be wrong... I'm doing other homework at the moment, so maybe I misread the diagram or did something careless :)
I think it's a bit tricky because it has the south pole on the bottom which means the magnetic field is coming IN to the south and going through the magnetic to north, ie, the magnetic field is going up rather than down. If you had north pole on the bottom,then it'd be going downwards as its coming OUT of the north and into the south.
Post by: appianway on October 19, 2009, 06:06:55 pm
Post by: TrueTears on October 19, 2009, 07:36:07 pm
Ahh that would explain it. Maybe I should have spent more time looking at it... >.< In that case, your explanation's perfect :)Awesome, thanks appianway!
Post by: TrueTears on October 19, 2009, 09:24:33 pm
Like I know why a sodium lamp emits light of discrete wavelength because each electron jumps a shell and emits a photon which corresponds to the change in energy between the 2 energy levels, what happens in a LED?
Thanks
Post by: appianway on October 19, 2009, 09:29:35 pm
Post by: TrueTears on October 19, 2009, 09:32:33 pm
Well, according to the reliable source of Wikipedia, the photons emitted correspond to the band gap energy between the P and N junctions. As I'm sure you already know from unit 3, the electrons in LEDs "fill" holes and hence drop energy levels. I'm not sure precisely why the LED doesn't emit light with a discrete wavelength though (it sounds like it should be!), but presumably the wavelength's within a certain range.Hmm yeah I thought the same, I thought they'd emit a range of wavelength but each photon emitted should have a "discrete" wavelength? Not sure heh.
Post by: TrueTears on October 19, 2009, 10:51:49 pm
ty
Post by: /0 on October 19, 2009, 11:02:21 pm
If we assume we're generating a fundamental standing wave,
Using a line of best fit on the data (i cheat with regression):
Ok, nevermind... maybe this was meant to be a second harmonic........ is this from insight :/
Post by: TrueTears on October 19, 2009, 11:04:14 pm
I did exactly what you did /0... funny thing is it says "loudest" which meant I thought it was fundamental, but 671.2 lol too fast for speed of sound =.=
If we assume we're generating a fundamental standing wave,, with gradient
Using a line of best fit on the data (i cheat with regression):
Ok, nevermind... maybe this was meant to be a second harmonic........ is this from insight :/
Post by: TrueTears on October 20, 2009, 12:47:26 am
(http://img80.imageshack.us/img80/1894/lisachem2008physics.jpg)
I thought answer was C but according to the solutions it's B, are they wrong?
2.
(http://img23.imageshack.us/img23/320/lisachem2008physics2.jpg)
I said A was blue and B was red, but answer is vice versa, are they wrong?
3. I have also noticed that when a question asks "find the minimum energy of a photon that will cause an electron to escape" and a graph of E_k - freq is provided. There are 2 ways of doing it, one is to read off the y intercept, the Work function, but the other way is to do f_0h and find f_0 from the x axis. However in 2 exams, they've all done the x f_0h method and the Work function method tends to give slightly different answers ~0.4-0.5 off (in eV). So what method do you do? Would both be considered right?
Many thanks guys!
Post by: /0 on October 20, 2009, 12:53:46 am
Post by: kurrymuncher on October 20, 2009, 12:57:04 am
2. Red light has a larger wavelength than blue, so it should have larger fringe spacing, So I would also say B is red and A is blue. hmmm. which exam is this from?
Post by: TrueTears on October 20, 2009, 12:57:50 am
Anyways, thanks heaps for your help guys!
Post by: TrueTears on October 20, 2009, 01:17:21 am
What's the question even asking? I have no idea what it even means. The answer is C btw.
Post by: kurrymuncher on October 20, 2009, 01:25:09 am
Post by: TrueTears on October 20, 2009, 01:26:31 am
LOL that is a shit question. Its almost like they are asking you to state the range and domain of that graph. lemme guess....Lisachem??it's a fucking joke... god... why is lisaCHEM doing trial exams for physics?
Post by: TrueTears on October 20, 2009, 04:09:29 pm
The wire is coming out of the page.
Now since AC is used the wire will move up the page and then move down the page and oscillate like this. But the problem is determining the period. Is the period of the wire where it moves up to its furthest from the original position and then come back to original? Or is the period when the wire moves up to the furthest position, return to original and then move downwards to its furthest position and then return to normal?
Post by: NE2000 on October 20, 2009, 04:19:03 pm
Hence the period of the wire's motion is the same as the period of the current.
Post by: TrueTears on October 20, 2009, 04:24:50 pm
As the current is increasing in positive direction (out of the page), the wire is being forced up. The magnitude of this force increases with the magnitude of the current. Then the current will be decreasing but will still be positive. So the force is still up. So the wire reaches its maximum displacement when the current is at T/2 in its AC sinusoidal waveform. Then the current will be negative and the force will reverse, the magnitude of the force will increase and decrease but the wire will continue travelling down. Hence after one period of the AC current, the wire has returned to its original position. It will not go below this position, but will quickly start going up again.Yeap thanks for that, but the answer says 20Hz which gives a period of 0.05 s but since it's meant to be the same with the current, shouldn't it be 0.1s?
Hence the period of the wire's motion is the same as the period of the current.
Post by: /0 on October 20, 2009, 04:31:04 pm
Post by: TrueTears on October 20, 2009, 04:31:44 pm
I think I can predict what will happen next... the answers are wrong again:knuppel2: grrrrrr
Post by: kamil9876 on October 20, 2009, 06:28:34 pm
assume
The big coefficient at the front is a constant, while the term in parenthesis is ALWAYS negative. Therefore the sign of the velocity should never change and so it shouldn't oscillate me thinks. I had my doubts about this question from just a simple intuitive argument but I guess I included this since the three ppl invovled all do spec. The intuitive argument is that even though force reverses, velocity does not neccesarily, in fact the force(when reversed) does most of it's work slowing the object down to 0 and then once at zero the force reverses back to positive, hence speeding the object back up.
This all gives me doubts about this question, only possible resolution is that B isn't constant (weaker as the thing moves up) or the current is NOT sinusoidal. :-\
Post by: TrueTears on October 20, 2009, 06:32:48 pm
Wow thanks there, so what's the period?(Bieng a bit naughty with the upper terminal but in my defence this is only physics)
assume
The big coefficient at the front is a constant, while the term in parenthesis is ALWAYS negative. Therefore the sign of the velocity should never change and so it shouldn't oscillate me thinks. I had my doubts about this question from just a simple intuitive argument but I guess I included this since the three ppl invovled all do spec. The intuitive argument is that even though force reverses, velocity does not neccesarily, in fact the force(when reversed) does most of it's work slowing the object down to 0 and then once at zero the force reverses back to positive, hence speeding the object back up.
This all gives me doubts about this question, only possible resolution is that B isn't constant (weaker as the thing moves up) or the current is NOT sinusoidal. :-\
Post by: kamil9876 on October 20, 2009, 07:02:13 pm
Should be the same as the current frequency, since if you antidiffirentiate the speed once again you get a term of the form
Actually, I think the problem I adressed wouldn't be a problem unless the initial speed WAS NOT zero, but the maximum speed such that the -1 term cancels.
Post by: /0 on October 20, 2009, 07:09:57 pm
Post by: kamil9876 on October 20, 2009, 07:14:10 pm
Quote
Actually, I think the problem I adressed wouldn't be a problem unless the initial speed WAS NOT zero, but the maximum speed such that the -1 term cancels.
otherwise it wouldn't osscilate. Just realised that. So there is a chance it is osccilating and in the case it does, the
Post by: NE2000 on October 20, 2009, 08:12:36 pm
assume
The big coefficient at the front is a constant, while the term in parenthesis is ALWAYS negative. Therefore the sign of the velocity should never change and so it shouldn't oscillate me thinks. I had my doubts about this question from just a simple intuitive argument but I guess I included this since the three ppl invovled all do spec. The intuitive argument is that even though force reverses, velocity does not neccesarily, in fact the force(when reversed) does most of it's work slowing the object down to 0 and then once at zero the force reverses back to positive, hence speeding the object back up.
This all gives me doubts about this question, only possible resolution is that B isn't constant (weaker as the thing moves up) or the current is NOT sinusoidal. :-\
Aah, didn't think of that...forgetting Unit 3 motion (concept of force, acceleration, velocity etc.) amidst all this magnetism lol. Thanks.
Practically, if the current is initially running out of the page then the wire will move up and continue moving up until it is out of the significant field of the magnet :), but if we assume that the magnet is infinite width then the period ought to be the time for the wire to come to rest again after moving upwards right? And that should be the period of the current in the wire.
But that's not really oscillation is it, what am I missing here?
Post by: kamil9876 on October 20, 2009, 08:25:18 pm
and notice that iff
Post by: TrueTears on October 20, 2009, 08:28:57 pm
I did my integral under the assumption that the initial velocity was 0. If instead it wasRight, and your point is? It doesn't oscillate or does it?then it would be:
and notice that iffthen the -1 term cancels out and so displacement in this case would just be some sinusoidal function. Wheras if the -1 term didn't cancel then the function would be of the form
and the linear part shows that as
gets big the thing just flies away from the vicinity :P
Post by: TrueTears on October 20, 2009, 08:56:31 pm
I said that incandescent light source emits light due to thermal vibration and thus has a continuous spectrum, so it emits light of different wavelengths. However a sodium lamp emits light of discrete wavelengths since electrons can only jump certain shells and thus emitting light with the difference in energy between 2 shells.
However what I don't understand is why doesn't a globe emit discrete wavelength of light? I know it emits a spectrum of wavelengths but doesn't each light that make up the spectrum have a discrete wavelength?
Thanks!
Post by: /0 on October 20, 2009, 09:14:27 pm
I did my integral under the assumption that the initial velocity was 0. If instead it wasRight, and your point is? It doesn't oscillate or does it?
and notice that iff
Whether or not it oscillates depends on the initial velocity, as kamil explained.
So what's the difference between a incandescent light source (such as a globe) and a sodium lamp.
I said that incandescent light source emits light due to thermal vibration and thus has a continuous spectrum, so it emits light of different wavelengths. However a sodium lamp emits light of discrete wavelengths since electrons can only jump certain shells and thus emitting light with the difference in energy between 2 shells.
However what I don't understand is why doesn't a globe emit discrete wavelength of light? I know it emits a spectrum of wavelengths but doesn't each light that make up the spectrum have a discrete wavelength?
Thanks!
Yeah, and when you sum it all up you get a continuous spectrum. Like adding infinitesimally thin strips in calculus.
Post by: kamil9876 on October 20, 2009, 09:19:08 pm
I did my integral under the assumption that the initial velocity was 0. If instead it wasRight, and your point is? It doesn't oscillate or does it?
and notice that iff
Post by: TrueTears on October 20, 2009, 09:23:00 pm
Post by: Mao on October 20, 2009, 11:59:12 pm
So, then shouldn't a globe emit discrete wavelengths?
No. In a wire, light is emitted through thermal excitation via a current. The current directly pass through the tungsten, in the form of delocalized electrons, which lose energy through collisions, and random amounts of energy are released as photons.
In a metal-vapour lamp, however, the electrons are not delocalized, and hence the only energy transitions allowed are between distinct energy levels between shells.
Post by: mark_alec on October 24, 2009, 12:47:55 am
No. In a wire, light is emitted through thermal excitation via a current.One can be even more general. When things are 'hot', they emit a continuous spectrum - the current in the wire just serves to make the filament very hot.
Post by: TrueTears on November 08, 2009, 01:28:10 am
Post by: /0 on November 08, 2009, 01:36:35 am
It's the same thing in terms of conservation of energy. The rod is generating electrical energy, so it's kinetic energy must be reduced.
If you want to find the terminal velocity of the rod, let it be
The emf generated by the falling rod is
If the rod has a resistance
Since the magnetic force on the rod is
So the terminal velocity occurs when
Post by: TrueTears on November 08, 2009, 01:40:15 am
but I'm wondering why the force INCREASES as the speed INCREASES?
What's the formula?
Only one that I know that involves speed is F = bqv but it's not that... since that is the force no each electron (getting pushed to the sides) doesn't explain why the magnetic force opposing it increases.
Post by: /0 on November 08, 2009, 01:45:40 am
Post by: TrueTears on November 08, 2009, 01:48:15 am
Sorry I just finished editing my last postI love you.
Post by: TrueTears on November 08, 2009, 04:10:44 pm
It diffracts through the slit but it also interferes with itself... so what is the cause of it?
This is a MC question btw it's got both diffraction and interference, I picked diffraction but the answer is interference... Shouldn't it both?
2. Also in the double slit diffraction experiment, if you decrease the distance of the 2 slits then the spacing is much wider, so on an intensity vs distance (from the central maximum) the peaks would be much further apart.
However, does the intensity of the respective antinodes/nodes before changing the slit decrease after the change? Or does it stay the same?
This is just a general question, not from an exam or anything. I think it would decrease since it's further from the source and since intensity is inversely proportional to the distance, it decreases... can anyone confirm?
Thus not only does the intensity vs distance graph have a wider spacing b/w the peaks, the peaks are also now lower?
Post by: krzysiek on November 09, 2009, 07:15:27 pm
Sorry, I have not opened a physics book in a while and plan on doing so shortly - but I thought the relationship of intensity decreasing with the inverse of the distance only applied to the intensity of sound, and not light? If this were true, then the peaks would remain the same - but if it does decrease with the distance, then if distance increases, it only makes sense for the peaks to decrease.
Post by: appianway on November 09, 2009, 07:32:03 pm
Post by: TrueTears on November 10, 2009, 12:29:54 am
What a funny question, isn't A, C, D all correct?
Post by: /0 on November 10, 2009, 12:39:25 am
If you have one wave with period 1 and one going in the opposite direction with period 2 then they will form antinodes... i think
ok that's just a guess I can't do superposition in my head
Post by: TrueTears on November 10, 2009, 02:25:23 am
Isn't it A and D? You don't need two waves travelling in the same direction for a standing wave.Exactly you DON'T need them to travel in same direction so it's NOT necessary. So A, C and D =SActually I think B is also correct.
If you have one wave with period 1 and one going in the opposite direction with period 2 then they will form antinodes... i think
ok that's just a guess I can't do superposition in my head
Post by: naved_s9994 on November 10, 2009, 10:22:04 am
2. Also in the double slit diffraction experiment, if you decrease the distance of the 2 slits then the spacing is much wider, so on an intensity vs distance (from the central maximum) the peaks would be much further apart.
However, does the intensity of the respective antinodes/nodes before changing the slit decrease after the change? Or does it stay the same?
This is just a general question, not from an exam or anything. I think it would decrease since it's further from the source and since intensity is inversely proportional to the distance, it decreases... can anyone confirm?
Thus not only does the intensity vs distance graph have a wider spacing b/w the peaks, the peaks are also now lower?
You're right. Think about energy conservation as well. If the light is concentrated in a small area, it will have higher intensity than if it is spread out over a large interference/diffraction pattern. So if you decrease d, the intensity of all the maxima will decrease. Intensity v distance does play a part as well - the maxima out to the side will be dimmer than those in the centre because they are more distant from the slits. Lastly, and this is beyond the course probably, for the two slit experiment the intensity of the maxima will also depend on the width of the slit. If the light does not diffract much from each slit then there will not be much light out to the side to interfere.
Post by: moekamo on November 10, 2009, 12:24:02 pm
(http://img29.imageshack.us/img29/5419/asdfeh.jpg)
What a funny question, isn't A, C, D all correct?
Remember this is for standing waves on a string
Heres the definition of standing wave in my text book:
"Standing waves are the result of superposition of two waves of equal amplitude and frequency, travelling in opposite directions in the same medium."
So,
You do NEED equal amplitude
You NEED equal frequency
You NEED the waves to travel in opposite directions
Hence the only unnecesary one is that they travel in the same direction
So B is the answer
Post by: IntoTheNewWorld on November 10, 2009, 12:30:06 pm
(http://img29.imageshack.us/img29/5419/asdfeh.jpg)
What a funny question, isn't A, C, D all correct?
Remember this is for standing waves on a string
Heres the definition of standing wave in my text book:
"Standing waves are the result of superposition of two waves of equal amplitude and frequency, travelling in opposite directions in the same medium."
So,
You do NEED equal amplitude
You NEED equal frequency
You NEED the waves to travel in opposite directions
Hence the only unnecesary one is that they travel in the same direction
So B is the answer
You mean C right?
Post by: moekamo on November 10, 2009, 12:38:23 pm
Post by: ngRISING on November 10, 2009, 01:27:04 pm
Post by: Politik23 on November 10, 2009, 02:28:14 pm
Post by: moekamo on November 10, 2009, 02:57:26 pm
Post by: TrueTears on November 10, 2009, 03:36:16 pm
Post by: TrueTears on November 10, 2009, 05:03:29 pm
(Question has a mistake it should say 3rd harmonic not 2nd!)
Answer is C, but I always thought the fundamental frequency sounded the loudest, so how can there be an increase in volume for the 3rd harmonic if the fundamental the is loudest?
Post by: /0 on November 10, 2009, 05:07:43 pm
Post by: NE2000 on November 10, 2009, 05:08:07 pm
EDIT: /0 beats me to it
Post by: NE2000 on November 10, 2009, 05:09:00 pm
That's weird... my book says you don't need equal amplitude.
I think you do. Try drawing it with one having double the amplitude. Remember the key is that the nodes and antinodes have to be fixed. I don't think that's the case when the amplitude is not equal.
Post by: TrueTears on November 10, 2009, 05:11:17 pm
Post by: /0 on November 10, 2009, 05:14:06 pm
Post by: TrueTears on November 10, 2009, 09:50:54 pm
ii) and iii) what do you guys get?
Post by: Fireworks on November 10, 2009, 09:52:38 pm
Post by: TrueTears on November 10, 2009, 09:57:47 pm
Me and /0 both got 5.5
The current flows through only 1 wire goes to the transformer then returns to the power supply through the other wire.
Thus the power loss
R should only be 2 and not 4.
Post by: /0 on November 10, 2009, 10:00:55 pm
Post by: TrueTears on November 10, 2009, 10:02:01 pm
Yeah if it was asking "what is the voltage delivered at the input to the primary coil of the transformer" then that would just beExactly that's what I thought as well...right? not
Let's ask BW tomorrow?
Post by: Fireworks on November 10, 2009, 10:03:25 pm
when I do these calculations, you must include resistance of
both of the wires, therefore a total of 4 ohms.
Voltage drop across the line = I x R = 0.5 x 4 = 2vdrop.
Therefore at V1, P=VI = 10*0.5 = 5w
Sorry, I only got a B+ on mid year so I'm not all that
great at physics but that's my understanding..
Post by: TrueTears on November 10, 2009, 10:04:03 pm
Post by: dejan91 on November 10, 2009, 10:06:21 pm
EDIT: yeah haha after looking into it deeper, I get what you mean, because voltage is being measured from the one wire, so resistance is only 2.
Post by: TrueTears on November 10, 2009, 10:08:00 pm
You can't just neglect a wire like that, can you?I know but it doesn't make sense to add the resistance of both. The current can travel whereever from the left terminal to the right or right to the left, but it only passes through ONE wire then the next step in its journey is the transformer then it returns to the other terminal through the OTHER wire. Thus at the transformer it should only have dropped some power over the resistance of ONE wire and not both.
Post by: /0 on November 10, 2009, 10:10:17 pm
________after wire resist______stepped down________after globe_______step up_____after wire resistance
12V---------11V------------------2.2V-----------------0.2V---------------1V----------------0V
If you're looking for the input to the primary coil of the transformer I think it would be 11V
Post by: TrueTears on November 10, 2009, 10:11:26 pm
You can think of it like this:Yeah I had 11 V as well, but the exam says 10 V rofl, let's check tomoz.
________after wire resist______stepped down________after globe_______step up_____after wire resistance
12V---------11V------------------2.2V-----------------0.2V---------------1V----------------0V
If you're looking for the input to the primary coil of the transformer I think it would be 11V
Post by: dejan91 on November 10, 2009, 10:11:40 pm
You can't just neglect a wire like that, can you?I know but it doesn't make sense to add the resistance of both. The current can travel whereever from the left terminal to the right or right to the left, but it only passes through ONE wire then the next step in its journey is the transformer then it returns to the other terminal through the OTHER wire. Thus at the transformer it should only have dropped some power over the resistance of ONE wire and not both.
Exactly what I just thought. I think what VCAA are trying to say is that because current is traveling in both directions, it experiences resistance and thus power loss in both wires, and then the total voltage loss is the sum of these. It doesn't really make sense now when you think about it. It's like a parallel circuit - voltage is distributed evenly...
Post by: almostatrap on November 10, 2009, 10:16:00 pm
Post by: TrueTears on November 10, 2009, 10:17:56 pm
the circuit has to 'know' not to use all the power/voltage so the current can make it home. ahah electricity is crazyYeah ok that works well for the voltage, what about power then? It didn't use up all the power after passing through 1 wire heh
Post by: littlecherry25 on November 10, 2009, 10:23:47 pm
can someone explain to me Question 12 of the VCAA 2008 paper
the question about the flood light?
the assesor's report sayd "the potential difference across the floodlight was in the same ratio as the resistance = (3/4) x 12 = 9"
Huh? where did the 3/4 come from ?
man i've forgotten basic circuit theory stuff, didn't know it was relevant to unit 4 :S
Post by: TrueTears on November 10, 2009, 10:25:04 pm
Logic disappears when you are dealing with electricity.
Post by: naved_s9994 on November 10, 2009, 10:34:27 pm
Why not 5.5 for ii and iii?
Me and /0 both got 5.5
The current flows through only 1 wire goes to the transformer then returns to the power supply through the other wire.
Thus the power loss
R should only be 2 and not 4.
DC
Post by: TrueTears on November 10, 2009, 10:38:17 pm
????????????????????????????????????????????????????????????????????????Why not 5.5 for ii and iii?
Me and /0 both got 5.5
The current flows through only 1 wire goes to the transformer then returns to the power supply through the other wire.
Thus the power loss
R should only be 2 and not 4.
DC
...
I'm sure transformers can function with DC.
Post by: naved_s9994 on November 10, 2009, 10:38:49 pm
Why not 5.5 for ii and iii?
Me and /0 both got 5.5
The current flows through only 1 wire goes to the transformer then returns to the power supply through the other wire.
Thus the power loss
R should only be 2 and not 4.
SOrry cant fix this thing -- but my answer is below
i) The power supplied by the 12V power supply = Voltage of the supply x current through the supply
I.e 12 x 0.5 = 6W
ii) The power consumed by the primary coil of the transformer = voltage across the coil x current through the coil
I.e 10 x 0.5 = 5W
iii) The transformer is ideal. Hence same as ii) 5W
Post by: arthurk on November 10, 2009, 10:46:01 pm
i was like rage when i listened to him and got it wrong in prac exam
Post by: jules on November 10, 2009, 11:27:07 pm
if a south end of a magnet enters a coil first, do we say the flux is negative?
(this question may be linked with draw a flux time graph)
Post by: Mao on November 10, 2009, 11:39:03 pm
I'm sure transformers can function with DC.
Transformers do NOT function with DC as the primary.
Transformers work of the principle of electromagnetic induction. The change in flux generated by the primary coil (due to alternating current) induces a current in the secondary coil.
Hence when the primary is DC, change in flux is zero, and secondary coil has no current induced.
Post by: TrueTears on November 10, 2009, 11:39:42 pm
lol I was being sarcastic... =.=I'm sure transformers can function with DC.
Transformers do NOT function with DC as the primary.
Transformers work of the principle of electromagnetic induction. The change in flux generated by the primary coil (due to alternating current) induces a current in the secondary coil.
Hence when the primary is DC, change in flux is zero, and secondary coil has no current induced.
Post by: Mao on November 10, 2009, 11:41:44 pm
lol I was being sarcastic... =.=I'm sure transformers can function with DC.
Transformers do NOT function with DC as the primary.
Transformers work of the principle of electromagnetic induction. The change in flux generated by the primary coil (due to alternating current) induces a current in the secondary coil.
Hence when the primary is DC, change in flux is zero, and secondary coil has no current induced.
oh oops, didn't follow the thread very closely.
GL for tomorrow :)
Post by: kakar0t on December 16, 2009, 11:09:16 pm
Question 7. please helpppp
Post by: TrueTears on December 16, 2009, 11:35:59 pm
http://i50.tinypic.com/zc1a9.jpgYou have to find out the vertical component of the velocity and then use trig to work out the angle. You're given the hypotenuse and you can work out the vertical. So it'd be involving
Question 7. please helpppp