The first overtone frequency of an open organ | vedclass

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(C) For an open organ pipe of length $L_1$,the fundamental frequency is $f_0 = \frac{v}{2L_1}$. The first overtone frequency is $f_1 = 2 	imes f_0 =

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$80$

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(C) For an open organ pipe of length $L_1$,the fundamental frequency is $f_0 = \frac{v}{2L_1}$. The first overtone frequency is $f_1 = 2 	imes f_0 = \frac{2v}{2L_1} = \frac{v}{L_1}$.For a closed organ pipe of length $L_2$,the fundamental frequency is $f_2 = \frac{v}{4L_2}$.Given that the first overtone frequency of the open pipe equals the fundamental frequency of the closed pipe,we have $f_1 = f_2$.Substituting the expressions: $\frac{v}{L_1} = \frac{v}{4L_2}$.This simplifies to $L_1 = 4L_2$.Given $L_2 = 20 \, cm$,we find $L_1 = 4 	imes 20 \, cm = 80 \, cm$.

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is $20 \, cm$,the length of the open organ pipe is ........ $cm$.

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