The second overtone of an open organ pipe is | vedclass

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(C) For an open organ pipe of length $L_o$,the frequency of the $n$-th overtone is given by $f_n = (n+1) \frac{v}{2L_o}$. The second overtone $(n=2)$

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$30$

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(C) For an open organ pipe of length $L_o$,the frequency of the $n$-th overtone is given by $f_n = (n+1) \frac{v}{2L_o}$. The second overtone $(n=2)$ is $f_2 = 3 \frac{v}{2L_o}$.For a closed organ pipe of length $L_c = 15 \ cm$,the frequency of the $m$-th overtone is given by $f'_m = (2m+1) \frac{v}{4L_c}$. The first overtone $(m=1)$ is $f'_1 = 3 \frac{v}{4L_c}$.Given that the frequencies are equal: $3 \frac{v}{2L_o} = 3 \frac{v}{4L_c}$.Simplifying the equation: $\frac{1}{2L_o} = \frac{1}{4L_c} \Rightarrow 2L_o = 4L_c \Rightarrow L_o = 2L_c$.Substituting $L_c = 15 \ cm$: $L_o = 2 	imes 15 \ cm = 30 \ cm$.

The second overtone of an open organ pipe is equal to the first overtone of a closed organ pipe. If the length of the closed organ pipe is $15 \ cm$,the length of the open organ pipe is: (in $cm$)

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