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TrueTears

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Re: TrueTears question thread
« Reply #300 on: July 01, 2009, 12:00:34 am »
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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.
So only one standing wave will form if it is just one sound source with a certain frequency?
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Re: TrueTears question thread
« Reply #301 on: July 01, 2009, 04:03:52 pm »
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Q1,

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?

Q2, A burning candle is placed on a table in front of a loudspeaker as shown in the diagram.

 
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 where is the resonant frequency and is the fundamental frequency.

So clearly  and  

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,


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,


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,


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,

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.



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.



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?
« Last Edit: July 01, 2009, 07:31:56 pm by TrueTears »
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Re: TrueTears question thread
« Reply #302 on: July 01, 2009, 07:16:49 pm »
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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.

TrueTears

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Re: TrueTears question thread
« Reply #303 on: July 01, 2009, 09:55:09 pm »
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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.
Ahh I get it now, thanks so much mark.
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Re: TrueTears question thread
« Reply #304 on: July 01, 2009, 11:07:52 pm »
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Q4

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.

TrueTears

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Re: TrueTears question thread
« Reply #305 on: July 02, 2009, 12:24:49 am »
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Ahh that makes it a lot more clear, thanks mark!
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TrueTears

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Re: TrueTears question thread
« Reply #306 on: July 19, 2009, 10:42:00 pm »
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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?
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Re: TrueTears question thread
« Reply #307 on: July 20, 2009, 03:01:06 pm »
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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.
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Re: TrueTears question thread
« Reply #308 on: July 20, 2009, 04:34:04 pm »
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Thanks.
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Re: TrueTears question thread
« Reply #309 on: August 06, 2009, 12:09:40 am »
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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!
« Last Edit: August 06, 2009, 12:16:00 am by TrueTears »
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Re: TrueTears question thread
« Reply #310 on: August 06, 2009, 02:17:33 pm »
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draw out each of the options as a circuit, calculate effective resistance, the one around 5.1 ohms will be the correct answer.
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Re: TrueTears question thread
« Reply #311 on: August 06, 2009, 03:28:27 pm »
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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.
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Re: TrueTears question thread
« Reply #312 on: August 06, 2009, 03:44:39 pm »
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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
« Last Edit: August 06, 2009, 03:46:21 pm by TonyHem »

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Re: TrueTears question thread
« Reply #313 on: August 30, 2009, 03:52:07 pm »
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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?


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Re: TrueTears question thread
« Reply #314 on: August 30, 2009, 04:10:11 pm »
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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?
Keeping 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).

The formula for resonant frequencies for a pipe closed at both ends in the same as one open at both ends.