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(A) The fundamental frequency of a closed organ pipe is $f_c = \frac{v}{4l}$.The fundamental frequency of an open organ pipe is $f_o = \frac{v}{2l}$.G

Vedclass · Accepted Answer

$7$

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(A) The fundamental frequency of a closed organ pipe is $f_c = \frac{v}{4l}$.The fundamental frequency of an open organ pipe is $f_o = \frac{v}{2l}$.Given the initial beat frequency is $2 	ext{ Hz}$:$f_o - f_c = 2$$\frac{v}{2l} - \frac{v}{4l} = 2$$\frac{v}{4l} = 2 \implies \frac{v}{l} = 8$.After changing the lengths,the new length of the open pipe is $l' = \frac{l}{2}$ and the new length of the closed pipe is $l'' = 2l$.The new fundamental frequencies are:$f_o' = \frac{v}{2l'} = \frac{v}{2(l/2)} = \frac{v}{l} = 8 	ext{ Hz}$.$f_c' = \frac{v}{4l''} = \frac{v}{4(2l)} = \frac{v}{8l} = \frac{8}{8} = 1 	ext{ Hz}$.The new beat frequency is $|f_o' - f_c'| = |8 - 1| = 7 	ext{ beats } \sec^{-1}$.

$A$ closed organ pipe and an open organ pipe of the same length produce $2 \text{ beats } \sec^{-1}$ when they are set into vibrations together in their fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced will be

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