In a closed-end pipe of length 105 , cm,stand | vedclass

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(D) For a closed pipe of length $L = 105 \, cm$,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.The third overtone corr

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$45$

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(D) For a closed pipe of length $L = 105 \, cm$,the frequency of the $n^{th}$ overtone is given by $f_n = (2n+1) \frac{v}{4L}$.The third overtone corresponds to $n = 3$,so the length $L$ is related to the wavelength $\lambda$ by $L = \frac{(2n+1)\lambda}{4} = \frac{7\lambda}{4}$.Given $L = 105 \, cm$,we have $105 = \frac{7\lambda}{4}$,which gives $\lambda = \frac{105 	imes 4}{7} = 60 \, cm$.In a closed pipe,the closed end is a displacement node (pressure antinode) and the open end is a displacement antinode (pressure node).Pressure nodes occur at distances $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots$ from the closed end.For $\lambda = 60 \, cm$,the positions of pressure nodes are $x = 15 \, cm, 45 \, cm, 75 \, cm, \dots$Comparing with the given options,$45 \, cm$ is a pressure node.

In a closed-end pipe of length $105 \, cm$,standing waves are set up corresponding to the third overtone. What distance from the closed end,amongst the following,is a pressure node in $cm$?

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