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(C) Let $l$ be the length of the open pipe and $L$ be the length of the closed pipe. Let $v$ be the speed of sound in air.For an open pipe,the frequen

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$2L$

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(C) Let $l$ be the length of the open pipe and $L$ be the length of the closed pipe. Let $v$ be the speed of sound in air.For an open pipe,the frequency of the $p$-th overtone is given by $f_p = (p+1) \frac{v}{2l}$. The second overtone corresponds to $p=2$,so $f_{o} = 3 \frac{v}{2l}$.For a closed pipe,the frequency of the $p$-th overtone is given by $f_p = (2p+1) \frac{v}{4L}$. The first overtone corresponds to $p=1$,so $f_{c} = 3 \frac{v}{4L}$.Given that the frequencies are equal: $f_{o} = f_{c}$.$\frac{3v}{2l} = \frac{3v}{4L}$.Canceling $3v$ from both sides,we get $\frac{1}{2l} = \frac{1}{4L}$.Therefore,$2l = 4L$,which simplifies to $l = 2L$.

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length $L$. The length of the open pipe will be

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