An organ pipe P_1 closed at one end is vibrat | vedclass

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(B) For a closed organ pipe $P_1$ of length $L_1$,the frequency of the $n$-th overtone is given by $f_n = (2n+1) \frac{V}{4L_1}$.The first overtone co

Vedclass · Accepted Answer

$0.37$

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(B) For a closed organ pipe $P_1$ of length $L_1$,the frequency of the $n$-th overtone is given by $f_n = (2n+1) \frac{V}{4L_1}$.The first overtone corresponds to $n=1$,so $f_1 = (2(1)+1) \frac{V}{4L_1} = \frac{3V}{4L_1}$.For an open organ pipe $P_2$ of length $L_2$,the frequency of the $n$-th overtone is given by $f_n = (n+1) \frac{V}{2L_2}$.The third overtone corresponds to $n=3$,so $f_3 = (3+1) \frac{V}{2L_2} = \frac{4V}{2L_2} = \frac{2V}{L_2}$.Since both pipes are in resonance with the same tuning fork,their frequencies are equal:$\frac{3V}{4L_1} = \frac{2V}{L_2}$.Rearranging for the ratio $\frac{L_1}{L_2}$:$\frac{L_1}{L_2} = \frac{3}{4 	imes 2} = \frac{3}{8} = 0.375$.

An organ pipe $P_1$ closed at one end is vibrating in its first overtone. Another pipe $P_2$ open at both ends is vibrating in its third overtone. They are in resonance with a given tuning fork. The ratio of the length of $P_1$ to that of $P_2$ is

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