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?

 

[asteriatic]

I would approach this problem by first understanding the concept of phase difference and how it relates to the interference of sound waves. Phase difference is the difference in phase between two waves, and it can be calculated by finding the path difference divided by the wavelength.

In this scenario, we have two sound sources (S1 and S2) emitting waves with a frequency range of 1k Hz to 4k Hz. As the frequency increases, the wavelength decreases according to the equation v=fλ. This means that the path difference between the two waves will also decrease, resulting in a smaller phase difference.

To determine the number of minima at point M, we need to consider the number of wavelengths that fit into the path difference. Since the path difference is 28 cm, we can divide it by the wavelength to find the number of wavelengths. For example, at a frequency of 1k Hz, the wavelength is 33 cm, so there would be 28/33 = 0.848 wavelengths. This means that there would be no minima detected at point M.

However, as the frequency increases to 4k Hz, the wavelength decreases to 8.25 cm. Now, the number of wavelengths that fit into the path difference is 28/8.25 = 3.39. This means that there will be 3 minima detected at point M.

In general, as the frequency increases, the number of minima at point M will increase as well. This is because the wavelength decreases, resulting in a smaller phase difference and more wavelengths fitting into the path difference.

In conclusion, the number of minima detected at point M will depend on the frequency of the sound waves and the path difference between them. As a scientist, it is important to understand the relationship between these factors in order to accurately predict and explain phenomena related to sound interference.