Two closed organ pipes of length 100,cm and 1 | vedclass

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(C) The fundamental frequency of a closed organ pipe is given by $f = \frac{v}{4L}$,where $v$ is the velocity of sound and $L$ is the length of the pi

Vedclass · Accepted Answer

$323.2$

Vedclass · Answer

(C) The fundamental frequency of a closed organ pipe is given by $f = \frac{v}{4L}$,where $v$ is the velocity of sound and $L$ is the length of the pipe.The beat frequency is the difference between the fundamental frequencies of the two pipes: $f_{beat} = |f_1 - f_2| = \frac{16}{20} = 0.8\,Hz$.Given $L_1 = 100\,cm = 1\,m$ and $L_2 = 101\,cm = 1.01\,m$,we have:$f_1 = \frac{v}{4 	imes 1}$ and $f_2 = \frac{v}{4 	imes 1.01}$.$f_1 - f_2 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.01} ight) = 0.8$.$\frac{v}{4} \left( \frac{1.01 - 1}{1.01} ight) = 0.8$.$\frac{v}{4} \left( \frac{0.01}{1.01} ight) = 0.8$.$v = \frac{0.8 	imes 4 	imes 1.01}{0.01} = 80 	imes 4.04 = 323.2\,m/s$.

Two closed organ pipes of length $100\,cm$ and $101\,cm$ long give $16$ beats in $20\,sec$ when each pipe is sounded in fundamental mode. Calculate the velocity of sound in $m/s$.

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