A closed pipe is suddenly opened and changed | vedclass

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(C) Let the length of the pipe be $l$ and the speed of sound be $v$.The fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$.The $3^{rd}$

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$55$

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(C) Let the length of the pipe be $l$ and the speed of sound be $v$.The fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$.The $3^{rd}$ harmonic of the closed pipe is $f_{3c} = 3 	imes f_c = \frac{3v}{4l}$.The fundamental frequency of the open pipe is $f_o = \frac{v}{2l}$.According to the problem,the fundamental frequency of the open pipe is less than the $3^{rd}$ harmonic of the closed pipe by $55 ~Hz$:$f_{3c} - f_o = 55 ~Hz$$\frac{3v}{4l} - \frac{v}{2l} = 55$$\frac{3v - 2v}{4l} = 55$$\frac{v}{4l} = 55 ~Hz$.Since the fundamental frequency of the closed pipe is $f_c = \frac{v}{4l}$,the value is $55 ~Hz$.

$A$ closed pipe is suddenly opened and changed to an open pipe of the same length. The fundamental frequency of the resulting open pipe is less than that of the $3^{rd}$ harmonic of the earlier closed pipe by $55 ~Hz$. Then,the value of the fundamental frequency of the closed pipe is: (in $~Hz$)

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