A closed organ pipe of radius r_1 and an open | vedclass

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Question

(C) For a closed organ pipe with end correction,the frequency of the $n^{th}$ mode is given by $
u_{n}^{\prime} = \frac{(2n-1)v}{4(L+0.6r_1)}$,where

Vedclass · Accepted Answer

$r_2 - 2r_1 = 2.5L$

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(C) For a closed organ pipe with end correction,the frequency of the $n^{th}$ mode is given by $
u_{n}^{\prime} = \frac{(2n-1)v}{4(L+0.6r_1)}$,where $0.6r_1$ is the end correction.For the fundamental mode $(n=1)$,$
u_{1}^{\prime} = \frac{v}{4(L+0.6r_1)}$.For an open organ pipe with end correction,the frequency of the $m^{th}$ mode is given by $
u_{m} = \frac{mv}{2(L+2 	imes 0.6r_2)}$,where $0.6r_2$ is the end correction at each end.For the first overtone $(m=2)$,$
u_{2} = \frac{2v}{2(L+1.2r_2)} = \frac{v}{L+1.2r_2}$.Since both pipes resonate with the same tuning fork,$
u_{1}^{\prime} = 
u_{2}$.Therefore,$\frac{v}{4(L+0.6r_1)} = \frac{v}{L+1.2r_2}$.$L + 1.2r_2 = 4L + 2.4r_1$.$1.2r_2 - 2.4r_1 = 3L$.Dividing by $1.2$,we get $r_2 - 2r_1 = 2.5L$.

$A$ closed organ pipe of radius $r_1$ and an open organ pipe of radius $r_2$ having the same length $L$ resonate when excited with a given tuning fork. The closed organ pipe resonates in its fundamental mode,whereas the open organ pipe resonates in its first overtone. Then:

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