The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, the length of the closed pipe is 3(x) cm. Then x is

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answer is 5.

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Detailed Solution

We know the fundamental frequency of a closed organ pipe is $\frac{v}{4 l_{1}}$ and first overtone frequency of open pipe is $\frac{v}{l_{2}}$ . Thus we have
$\begin{array}{l} \frac{v}{4 l_{1}} = \frac{v}{60} \\ l_{1} = \frac{1}{4} \times 60 c m = 15 c m \end{array}$

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