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(C) Let the length of the closed pipe be $l_c = 10 \, cm = 0.1 \, m$ and the length of the open pipe be $l_o$.The fundamental frequency of a closed pi

Vedclass · Accepted Answer

$30$

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(C) Let the length of the closed pipe be $l_c = 10 \, cm = 0.1 \, m$ and the length of the open pipe be $l_o$.The fundamental frequency of a closed pipe is given by $f_c = \frac{v}{4 l_c}$,where $v$ is the speed of sound.For an open pipe,the frequencies are given by $f_n = \frac{n v}{2 l_o}$,where $n = 1, 2, 3, \dots$. The first overtone is $n=2$ and the second overtone is $n=3$. Thus,the frequency of the second overtone of the open pipe is $f_{o,2} = \frac{3 v}{2 l_o}$.According to the problem,the fundamental frequency of the closed pipe is half that of the second overtone of the open pipe:$f_c = \frac{1}{2} f_{o,2}$Substituting the expressions:$\frac{v}{4 l_c} = \frac{1}{2} \left( \frac{3 v}{2 l_o} ight)$$\frac{v}{4 	imes 10} = \frac{3 v}{4 l_o}$$\frac{1}{40} = \frac{3}{4 l_o}$$4 l_o = 120$$l_o = 30 \, cm$.

$A$ closed pipe of length $10 \, cm$ has its fundamental frequency half that of the second overtone of an open pipe. The length of the open pipe is ......... $cm$.

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