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(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$

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halved

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(A) The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4(L + 0.6r)}$,where $v$ is the speed of sound,$L$ is the length,and $r$ is the radius.Since $L \gg r$,the end correction $0.6r$ is negligible compared to $L$,so the frequency is approximately $n \approx \frac{v}{4L}$.If the length $L$ is doubled $(L' = 2L)$ and the radius $r$ is halved $(r' = r/2)$,the new frequency $n'$ is:$n' = \frac{v}{4(L' + 0.6r')} = \frac{v}{4(2L + 0.6(r/2))} = \frac{v}{4(2L + 0.3r)}$.Since $L \gg r$,$0.3r$ is still negligible,so $n' \approx \frac{v}{4(2L)} = \frac{1}{2} \left( \frac{v}{4L} \right) = \frac{n}{2}$.Thus,the frequency is halved.

$A$ closed organ pipe has a fundamental frequency $n$. If its length is doubled and its radius is halved,its fundamental frequency nearly becomes-

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