In a closed organ pipe of length 105 ,cm,stan | vedclass

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(B) For a closed organ pipe of length $L$,the frequency of the $n$-th overtone is given by $f_n = \frac{(2n+1)v}{4L}$,where $n$ is the overtone number

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$15$

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(B) For a closed organ pipe of length $L$,the frequency of the $n$-th overtone is given by $f_n = \frac{(2n+1)v}{4L}$,where $n$ is the overtone number.For the third overtone,$n=3$,so the frequency is $f_3 = \frac{(2 	imes 3 + 1)v}{4L} = \frac{7v}{4L}$.The wavelength is $\lambda = \frac{v}{f_3} = \frac{4L}{7}$.Given $L = 105 \,cm$,we have $\lambda = \frac{4 	imes 105}{7} = 60 \,cm$.In a standing wave,pressure nodes are points where the pressure variation is zero. In a closed pipe,a displacement node (which is a pressure antinode) is at the closed end,and a displacement antinode (which is a pressure node) is at the open end.The positions of pressure nodes (displacement antinodes) from the closed end are given by $x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \dots$The first pressure node from the closed end is at $x = \frac{\lambda}{4} = \frac{60}{4} = 15 \,cm$.

In a closed organ pipe of length $105 \,cm$,standing waves are set up corresponding to the third overtone. What is the distance from the closed end where a pressure node is formed? (in $cm$)

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