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

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(B) For an organ pipe closed at one end,the frequency of the $n$-th overtone is given by $f = \frac{(2n+1)v}{4L_1}$,where $n$ is the overtone number.

Vedclass · Accepted Answer

$3/8$

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(B) For an organ pipe closed at one end,the frequency of the $n$-th overtone is given by $f = \frac{(2n+1)v}{4L_1}$,where $n$ is the overtone number. For the first overtone $(n=1)$,$f = \frac{3v}{4L_1}$.For an organ pipe open at both ends,the frequency of the $n$-th overtone is given by $f = \frac{(n+1)v}{2L_2}$. For the third overtone $(n=3)$,$f = \frac{4v}{2L_2} = \frac{2v}{L_2}$.Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $\frac{3v}{4L_1} = \frac{2v}{L_2}$.Rearranging the terms to find the ratio $L_1/L_2$: $\frac{L_1}{L_2} = \frac{3v}{4} 	imes \frac{1}{2v} = \frac{3}{8}$.Thus,the ratio of the length of $P_1$ to that of $P_2$ is $3/8$.

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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