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(B) The fundamental frequency of a closed organ pipe of length $L$ is given by $f = \frac{v}{4L}$.For the first pipe of length $l$,the frequency is $f

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$\frac{vx}{4l^2}$

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(B) The fundamental frequency of a closed organ pipe of length $L$ is given by $f = \frac{v}{4L}$.For the first pipe of length $l$,the frequency is $f_1 = \frac{v}{4l}$.For the second pipe of length $l + x$,the frequency is $f_2 = \frac{v}{4(l + x)}$.The beat frequency $f_b$ is the difference between these two frequencies: $f_b = f_1 - f_2 = \frac{v}{4l} - \frac{v}{4(l + x)}$.Factoring out $\frac{v}{4}$,we get $f_b = \frac{v}{4} \left( \frac{1}{l} - \frac{1}{l + x} \right) = \frac{v}{4} \left( \frac{l + x - l}{l(l + x)} \right) = \frac{v}{4} \left( \frac{x}{l(l + x)} \right)$.Since $x Therefore,the beat frequency is $f_b \approx \frac{vx}{4l^2}$.

$A$ closed organ pipe of length $l$ is sounded together with another closed organ pipe of length $l + x$ $(x << l)$,both in their fundamental mode. If $v$ is the speed of sound,the beat frequency heard is:

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