A uniform narrow 1.95 , m long pipe is open a | vedclass

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(D) For an open pipe of length $L$,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $v$ is the speed of sound.Given $L

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$214.5$

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(D) For an open pipe of length $L$,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $v$ is the speed of sound.Given $L = 1.95 \, m$,and two successive harmonics are $f_n = 275 \, Hz$ and $f_{n+1} = 330 \, Hz$.The difference between two successive harmonics is the fundamental frequency $f_1 = \frac{v}{2L}$.$f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L} = 330 - 275 = 55 \, Hz$.Substituting the value of $L$:$\frac{v}{2 	imes 1.95} = 55$.$v = 55 	imes 3.9 = 214.5 \, m/s$.

$A$ uniform narrow $1.95 \, m$ long pipe is open at both ends. It resonates at two successive harmonics of frequency $275 \, Hz$ and $330 \, Hz$. The speed of sound in the tube is ...... $m/s$.

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