Some questions

[taiga]

How do induced magnetic poles work.

Say that there is a field going through a plane which is travelling at a certain velocity, which rule(s) do I apply to figure out where the induced pole is?

[taiga]

or if you could help me out with this.

answers are B and C, but i dont quite understand why, unless someone can prove them wrong :X

[3Xamz]

Q9) I'm saying D. As it hasn't passed the slit yet, there is no diffraction occuring. Hence the frequency doesn't matter because the speed of sound is constant through any one medium. However, it has travelled less distance to get to Y as opposed to X, hence it's intensity will be larger.

(If you take the loudspeaker to be the initial point, the radius to Y is shorter than the radius to X)
Since I = P/A, A would be larger for Point X. Hence lower intensity there.

[3Xamz]

Induced magnetic fields are created by induced currents.
And induced currents ALWAYS build up in a way so the induced magnetic field OPPOSES the magnetic field that initiated the whole thing.


Hence if you have a north pole moving towards a coil, it would try to push it away by creating a magnetic field that points a magnetic north at the on-coming magnet.

S[___]N ----->  ()

Imagine that, a north end of a magnet approaching a coil.
It will induce a current in the coil that creates a induced magnetic field as shown above.
The only way the induced magnetic field can be going left, is to have a current that is going clockwise. (Right hand grip rule).

Hence you would end up with;;

S[___]N ----->  N()S

I hope that made sense, if not I'll try and reword it.

[Chavi]

9)I think the answer is B, because of the wavelength and the spreading out of the sound waves.
The low frequency (I=100hz) has a much larger wavelength, and spreads out more, so it is louder at Y. The higher frequ (I=10000hz) is like a tiny beam that diffracts little, and travels mostly to X.
The larger wavelength has more spread, so I-100hz has a higher intensity at Y.

10) for the same reason outlined in 9) - higher frequencies are more effective at propagating long distances in straight lines (like a laser). Hence the higher frequency has a higher intensity at point Z.

[jackchan1993]

SOUND QUESTION:

There was a similar question on the CSE 2010 paper, regardless if there is a gap to diffract around, lower frequencies of sound (longer wavelengths) will SPREAD OUT more than higher frequencies of sound (shorter wavelengths), so evidently the 100Hz sound will spread out more than the 10kHz, and more of the 100Hz sound will reach point Y than the 10kHz sound, thus the intensity of 100Hz sound is more intense than the 10kHz sound at point Y, so B is the correct answer for 9.

Question 10 is a diffraction question, so we have to look back on our knowledge about diffraction, the 10kHz will have a shorter wavelength being the sound with the higher frequency, it will diffract less across the gap. While the 100Hz sound has a longer wavelength and will diffract more than the 10kHz. Thus essentially, the 100Hz sound will spread out more than the 10kHz sound due to diffraction, and in a sense the 10kHz will be more "concentrated" at point Z than the 100Hz sound since it spreads out less.

[3Xamz]

9)I think the answer is B, because of the wavelength and the spreading out of the sound waves.
The low frequency (I=100hz) has a much larger wavelength, and spreads out more, so it is louder at Y. The higher frequ (I=10000hz) is like a tiny beam that diffracts little, and travels mostly to X.
The larger wavelength has more spread, so I-100hz has a higher intensity at Y.

10) for the same reason outlined in 9) - higher frequencies are more effective at propagating long distances in straight lines (like a laser). Hence the higher frequency has a higher intensity at point Z.

Nope, it hasn't diffracted so they would both arrive at Y at the same time.
Or else listening to music or anything for that matter would be impossible.
We'd always hear different frequencies at different times, even though they were produced from the same point at the same time.

[jackchan1993]

9)I think the answer is B, because of the wavelength and the spreading out of the sound waves.
The low frequency (I=100hz) has a much larger wavelength, and spreads out more, so it is louder at Y. The higher frequ (I=10000hz) is like a tiny beam that diffracts little, and travels mostly to X.
The larger wavelength has more spread, so I-100hz has a higher intensity at Y.

10) for the same reason outlined in 9) - higher frequencies are more effective at propagating long distances in straight lines (like a laser). Hence the higher frequency has a higher intensity at point Z.

Nope, it hasn't diffracted so they would both arrive at Y at the same time.
Or else listening to music or anything for that matter would be impossible.
We'd always hear different frequencies at different times, even though they were produced from the same point at the same time.

That's what I thought at the start and I got that question wrong on the CSE 2010 paper. But yeah, sound with longer wavelengths spread out more in air.

[3Xamz]

what?! :|
they have the same speed though and it hasn't diffracted. It's not physically possible for one to be more intense than the other.

Can someone clarify this?!

[Chavi]

what?! :|
they have the same speed though and it hasn't diffracted. It's not physically possible for one to be more intense than the other.

Can someone clarify this?!

Because this has nothing to do with diffraction. Diffraction has not occurred at Y or X!!
It's just that as JAckie Chan pointed out above, sound with longer wavelengths spreads out more in air than sounds with shorter wavelengths). Also note that frequency is inversely proportional to wavelength.

[3Xamz]

Woah, really? I thought they all spread out equally until they reach a point of diffraction.

Then how can we apply;

I = P/(4pir^2)

It's the same intensity equation and its independent of frequency...

[Chavi]

Woah, really? I thought they all spread out equally until they reach a point of diffraction.

Then how can we apply;

I = P/(4pir^2)

It's the same intensity equation and its independent of frequency...

Right, so Intensity is inversely proportional to area. . .

[3Xamz]

Yeah so why would lower frequencies be more intense than high frequency at point Y?
It's the same area for both frequencies. Hence should be the same intensity?

[lachymm]

Q9 C

Q10 B

By the way i think that longer wavelength sounds dont spread out more in air but rather diffract more at the speaker, hence spread out more over all...

[lachymm]

Woah, really? I thought they all spread out equally until they reach a point of diffraction.

Then how can we apply;

I = P/(4pir^2)

It's the same intensity equation and its independent of frequency...

This formula assumes that it spreads out in a spherical shape but this is not the case for most speakers as there sound is stronger in front of the speaker especially for higher frequency's