The third overtone of a closed pipe of length | vedclass

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

Vedclass · Accepted Answer

$7:8$

Vedclass · Answer

(B) For a closed pipe of length $L_c$, the frequency of the $n^{th}$ overtone is given by $f_c = (2n + 1) \frac{v}{4L_c}$, where $n$ is the overtone number. For the third overtone $(n = 3)$, $f_c = (2 	imes 3 + 1) \frac{v}{4L_c} = \frac{7v}{4L_c}$.For an open pipe of length $L_0$, the frequency of the $n^{th}$ overtone is given by $f_0 = (n + 1) \frac{v}{2L_0}$. For the third overtone $(n = 3)$, $f_0 = (3 + 1) \frac{v}{2L_0} = \frac{4v}{2L_0} = \frac{2v}{L_0}$.Given that the frequencies are equal, $\frac{7v}{4L_c} = \frac{2v}{L_0}$.Rearranging the terms to find the ratio $L_c : L_0$, we get $\frac{L_c}{L_0} = \frac{7}{4 	imes 2} = \frac{7}{8}$.Thus, the ratio $L_c : L_0$ is $7:8$.

The third overtone of a closed pipe of length '$L_c$' has the same frequency as the third overtone of the open pipe of length '$L_0$'. Both the pipes have same diameters. The ratio $L_c : L_0$ is equal to

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