The frequency of the fourth overtone of a clo | vedclass

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(B) For a closed pipe of length $L_c$,the frequency of the $n^{th}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number

Vedclass · Accepted Answer

$3: 4$

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(B) For a closed pipe of length $L_c$,the frequency of the $n^{th}$ overtone is given by $f_c = \frac{(2n+1)v}{4L_c}$,where $n$ is the overtone number. For the fourth overtone $(n=4)$,$f_c = \frac{(2(4)+1)v}{4L_c} = \frac{9v}{4L_c}$.For an open pipe of length $L_o$,the frequency of the $m^{th}$ overtone is given by $f_o = \frac{(m+1)v}{2L_o}$,where $m$ is the overtone number. For the fifth overtone $(m=5)$,$f_o = \frac{(5+1)v}{2L_o} = \frac{6v}{2L_o} = \frac{3v}{L_o}$.Given that the frequencies are in unison,$f_c = f_o$,so $\frac{9v}{4L_c} = \frac{3v}{L_o}$.Simplifying the equation: $\frac{9}{4L_c} = \frac{3}{L_o} \implies \frac{L_c}{L_o} = \frac{9}{4 	imes 3} = \frac{9}{12} = \frac{3}{4}$.Thus,the ratio of the length of the closed pipe to the open pipe is $3: 4$.

The frequency of the fourth overtone of a closed pipe is in unison with the fifth overtone of an open pipe. The ratio of the length of the closed pipe to that of the open pipe is

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