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(D) Let $L_1$ be the length of the closed organ pipe and $L_2$ be the length of the open organ pipe.The fundamental frequency of a closed organ pipe i

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$15$

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(D) Let $L_1$ be the length of the closed organ pipe and $L_2$ be the length of the open organ pipe.The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4L_1}$.The frequencies of an open organ pipe are given by $f_n = \frac{nv}{2L_2}$, where $n = 1, 2, 3, ...$.The first overtone of an open organ pipe corresponds to $n = 2$, so $f_{o1} = \frac{2v}{2L_2} = \frac{v}{L_2}$.According to the problem, the fundamental frequency of the closed pipe is equal to the first overtone frequency of the open pipe:$f_c = f_{o1}$$\frac{v}{4L_1} = \frac{v}{L_2}$$L_2 = 4L_1$Given $L_2 = 60 \,cm$, we have:$60 \,cm = 4L_1$$L_1 = \frac{60}{4} \,cm = 15 \,cm$.Therefore, the length of the closed pipe is $15 \,cm$.

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 will be: (in $\,cm$)

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