A level Physics Help needed with a question

[shabihsherjeel]

I am really stumped by this question. Waves just doesn't make place in my head. I don't know why.


The question is from May/June 2009 Physics 9702 A levels.
Question No. 5.b.

Here is the direct link to the past paper on XtremePaper's server.
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

The frequency of vibration can be varied. The two sources always vibrate in phase but have
different amplitudes of vibration.
A microphone M is situated a distance 100 cm from S1
along a line that is normal to S1
S2
(b) The speed of sound in air is 330 m s–1
The frequency of the sound from S1
and S2
is increased. Determine the number of
minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz.

I have tried following the marking scheme but it just doesn't make complete sense to me. Please help me with this question. A simple pointer in the right direction is good too.

I have found the path difference to be 28 cm. I even tried drawing the waves at different frequencies. But I just don't get it.

 

[shabihsherjeel]

I am amazed. Just figured out the question when I read other material online. That was simple.

Minima means destructive interference.
So for destructive interference you have this equation:
Path Difference = 0.5 * λ or P.D = 1.5 * λ or P.D. = 2.5 * λ

Using this you can find the wavelengths that will produce destructive interference with this phase difference.
For the phase difference you use geometry to find the distance from S2 to M. Using the equation H^2 = B^2 + P^2
Then subtract the distance of S2 to M from the distance of S1 to M. That gives you the difference in length of when the wave from S1 reaches M before the one from S2. Phase Difference = 28 cm.
Once you have the phase difference you find the change in wavelength when the frequency changes. So you have v=frequency * λ. Find the two λ's for the two given frequencies. The two wavelengths you'll get will be the limits. You will get 33 cm to 8.25 cm.

Now use the equation of phase difference to find wavelengths at which destructive interference occurs. First with P.D. = 28 = 0.5 * λ. Magic of Algebra and you get λ = 56 cm. But that is way above your limit. Do the same with 28 = 1.5 * λ. You get 18.7 cm. Bingo. That's in your limits. Keep doing this until you cross your limit. This results in two wavelengths at which destructive interference occurs. Thus two minimas is the answer.

 

[shabihsherjeel]

Requesting L&A on this thread.

 

[Ziki629]

Thanks alot!

 

[TheJDOG]

That extreme necro though, this is sooo old xD

 

[Ziki629]

Ha xD well mate my as is in 2 weeks so old or not it helped me :3