Sound, phase difference and number of minima

[thoradicus]

Homework Statement

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20%289702%29/9702_s09_qp_2.pdf

number 5b

The speed of sound is 330ms-1

The frequency of sound from S1 and S2 is increased. Determine the number of minima that will be detected at M as the frequency is increased from 1k to 4k Hz.

Homework Equations

v=fλ
(n+1)λ

The Attempt at a Solution

Okay, first have to find the path difference, which first we have to find the hypotenuse, 128cm.
and then, 128-100=28 cm, that is the path difference
330=(1000)λ
330=(4000)λ
λ is found to be 33 cm and 8.25 cm respectively.

After that, I am not too sure what to do. ..

 

[Doc Al]

How many wavelengths does that path difference correspond to? As the wavelength is slowly decreased, at what points will there be destructive interference?

 

[thoradicus]

How many wavelengths does that path difference correspond to? As the wavelength is slowly decreased, at what points will there be destructive interference?

28=1/2λ, so λ is 56cm.


So its like 5/2λ=28, 7/2λ=28 and so on?
So path difference is constant? for all frequencies?
minima will be detected at 18.7 and 11.2 cm. So 2 minima are detected

 

[Doc Al]

28=1/2λ, so λ is 56cm.

You already found the range of wavelengths. Now make use of that by figuring out how many wavelengths fit into that path difference.

If the path difference ends up being a whole number of wavelengths, what does that mean? Under what conditions will there be destructive interference?