The fundamental frequency of a closed pipe of | vedclass

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(D) For a closed pipe of length $L$,the fundamental frequency is given by $f_c = \frac{v}{4L}$.For an open pipe of length $L' = XL$,the frequencies ar

Vedclass · Accepted Answer

$6$

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(D) For a closed pipe of length $L$,the fundamental frequency is given by $f_c = \frac{v}{4L}$.For an open pipe of length $L' = XL$,the frequencies are given by $f_n = \frac{n v}{2L'}$,where $n = 1, 2, 3, \dots$.The first overtone is $n=2$ and the second overtone is $n=3$.Therefore,the second overtone frequency of the open pipe is $f_o = \frac{3v}{2(XL)}$.According to the problem,$f_c = f_o$,so $\frac{v}{4L} = \frac{3v}{2XL}$.Canceling $v$ and $L$ from both sides,we get $\frac{1}{4} = \frac{3}{2X}$.Solving for $X$,we get $2X = 12$,which implies $X = 6$.

The fundamental frequency of a closed pipe of length $L$ is equal to the second overtone of a pipe open at both the ends of length $(XL)$. The value of $X$ is (Neglect end correction).

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