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Two sound waves emitted bt one sound source is approaching at a point through two path. when path diff is 36cm or 60cm, then points are silence point if speed of sound in air is 330m/sec. find the frequency of sound source
Most Upvoted Answer
Two sound waves emitted bt one sound source is approaching at a point ...
Concept :
Path difference between consecutive:        Max-Max or Min-Min = wavelength(a) 
And between Max-Min or Min-Max = a/2

Silence point/Destructive/Min
It's a case of consecutive silent point ie Min-Min 

So, Wavelength(a) = 60-36= 24cm =0.24 m
V = na 
330 =n*0.24
1375 Hz =n


#I have changed notation of wavelength as can't type lamda here.. So, I'm using 'a' instead of lamda to denote wavelength. 
Community Answer
Two sound waves emitted bt one sound source is approaching at a point ...
Explanation:

Introduction:

When two sound waves are emitted by one sound source and they approach a point through two different paths, they interfere with each other. Depending on the path difference between the waves, the interference can be constructive or destructive. When the path difference is such that the waves interfere destructively, the point is known as a silence point or a node.

Given:

- Path difference between the two waves: 36 cm and 60 cm

- Speed of sound in air: 330 m/s

Solution:

To find the frequency of the sound source, we need to use the formula for the path difference between two waves:

path difference = (m + ½)λ

Where λ is the wavelength of the sound wave and m is an integer representing the number of half wavelengths between the two waves.

We can rearrange this formula to solve for the wavelength:

wavelength = 2 (path difference - mλ)

We know that at the silence point, the path difference between the two waves is either 36 cm or 60 cm. Let's assume that the wavelength of the sound wave is the same in both cases (since we don't know the frequency yet). We can then solve for the wavelength using the above formula:

For path difference of 36 cm:
wavelength = 2 (36/100 - mλ)
For path difference of 60 cm:
wavelength = 2 (60/100 - mλ)

Since the wavelength must be the same in both cases, we can equate the two expressions and solve for λ:

2 (36/100 - mλ) = 2 (60/100 - mλ)

36/100 - mλ = 60/100 - mλ

36/100 = 60/100

This is clearly not true, so we made an incorrect assumption. The wavelength of the sound wave must be different in the two cases. Let's call the wavelength for the 36 cm path difference λ1 and the wavelength for the 60 cm path difference λ2. We can then solve for the frequencies of the two waves using the formula:

frequency = speed of sound / wavelength

For the 36 cm path difference:
λ1 = 2 (36/100 - mλ1)
λ1 = 72/100 - 2mλ1
2mλ1 + λ1 = 72/100
λ1 = 72/300m
frequency1 = 330 / 72/300 = 137.5 Hz

For the 60 cm path difference:
λ2 = 2 (60/100 - mλ2)
λ2 = 120/100 - 2mλ2
2mλ2 + λ2 = 120/100
λ2 = 120/300m
frequency2 = 330 / 120/300 = 137.5 Hz

Therefore, the frequency of the sound source is 137.5 Hz.
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Two sound waves emitted bt one sound source is approaching at a point through two path. when path diff is 36cm or 60cm, then points are silence point if speed of sound in air is 330m/sec. find the frequency of sound source
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Two sound waves emitted bt one sound source is approaching at a point through two path. when path diff is 36cm or 60cm, then points are silence point if speed of sound in air is 330m/sec. find the frequency of sound source for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two sound waves emitted bt one sound source is approaching at a point through two path. when path diff is 36cm or 60cm, then points are silence point if speed of sound in air is 330m/sec. find the frequency of sound source covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two sound waves emitted bt one sound source is approaching at a point through two path. when path diff is 36cm or 60cm, then points are silence point if speed of sound in air is 330m/sec. find the frequency of sound source.
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