An open organ pipe of length ' l ' is sounded together with another open organ pipe of length ( l + l 1 ) in their fundamental modes. Speed of sound in air is ' V '. The beat fre

Question

An open organ pipe of length ' l ' is sounded together with another open organ pipe of length ( l + l 1 ​ ) in their fundamental modes. Speed of sound in air is ' V '. The beat frequency heard will be ( l 1 ​ ≪ l )

TestHub · Accepted Answer

Fundamental frequency of an open organ pipe of length $L$ is $f = \frac{V}{2 L}$ .

For the first pipe of length $l$ , $f_{1} = \frac{V}{2 l}$ .

For the second pipe of length $l + l_{1}$ , $f_{2} = \frac{V}{2 ( l + l _{1} )}$ .

The beat frequency is $f_{B} = f_{1} - f_{2}$ .

$f_{B} = \frac{V}{2 l} - \frac{V}{2 ( l + l _{1} )}$

$f_{B} = \frac{V}{2} (\frac{l + l _{1} - l}{l ( l + l _{1} )})$

$f_{B} = \frac{V l _{1}}{2 l ( l + l _{1} )}$

Since $l_{1} ≪ l$ , we can approximate $l + l_{1} \approx l$ .

$f_{B} \approx \frac{V l _{1}}{2 l ^{2}}$

Answer: $\frac{V l _{1}}{2 l ^{2}}$

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