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(C) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_o = \frac{v}{2L}$ and the fundamental frequency of a cl

Vedclass · Accepted Answer

$7$

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(C) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_o = \frac{v}{2L}$ and the fundamental frequency of a closed organ pipe is $f_c = \frac{v}{4L}$.Given that the beat frequency is $2$,so $|f_o - f_c| = 2$.$|\frac{v}{2L} - \frac{v}{4L}| = 2 \implies \frac{v}{4L} = 2 \implies \frac{v}{L} = 8$.Now,the length of the open pipe becomes $L' = \frac{L}{2}$ and the length of the closed pipe becomes $L'' = 2L$.The new fundamental frequency of the open pipe is $f_o' = \frac{v}{2L'} = \frac{v}{2(L/2)} = \frac{v}{L} = 8 	ext{ Hz}$.The new fundamental frequency of the closed pipe is $f_c' = \frac{v}{4L''} = \frac{v}{4(2L)} = \frac{v}{8L} = \frac{1}{2} 	imes (\frac{v}{4L}) = \frac{1}{2} 	imes 2 = 1 	ext{ Hz}$.The number of beats produced per second is $|f_o' - f_c'| = |8 - 1| = 7 	ext{ Hz}$.

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