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(B) The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4\ell_c}$,where $\ell_c = 20\; cm$.The frequencies of an open organ p

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$120$

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(B) The fundamental frequency of a closed organ pipe is given by $f_c = \frac{v}{4\ell_c}$,where $\ell_c = 20\; cm$.The frequencies of an open organ pipe are given by $f_n = \frac{nv}{2\ell_o}$,where $n = 1, 2, 3, \dots$.The first overtone is $n=2$,and the second overtone is $n=3$.Thus,the second overtone frequency of an open organ pipe is $f_{o,2} = \frac{3v}{2\ell_o}$.According to the problem,the fundamental frequency of the closed pipe equals the second overtone of the open pipe:$\frac{v}{4\ell_c} = \frac{3v}{2\ell_o}$Canceling $v$ from both sides:$\frac{1}{4\ell_c} = \frac{3}{2\ell_o}$Rearranging to solve for $\ell_o$:$\ell_o = \frac{3 	imes 4\ell_c}{2} = 6\ell_c$Substituting $\ell_c = 20\; cm$:$\ell_o = 6 	imes 20\; cm = 120\; cm$.

The fundamental frequency of a closed organ pipe of length $20\; cm$ is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ...... $cm$.

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