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A closed organ pipe and an open organ pipe of same length produce 2 beats when they are set into vibration simultaneously in their fundamental mode. The length of the open organ pipe is now halved and of the closed organ pipe is doubled, the number of beats produced will be (When pipes are vibrating in their fundamental made)

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answer is 7.

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Detailed Solution

$\begin{array}{l} f_{0} - f_{c} = 2 \\ o r \frac{v}{2 l} - \frac{v}{4 l} = 2 o r \frac{v}{4 l} = 2 \\ o r \frac{v}{l} = 8 \end{array}$

When length of open organ pipe is halved and that of closed organ pipe is doubled, the beat frequency will be:
$f_{o}^{'} - f_{c}^{'} = \frac{v}{l} - \frac{v}{8 l} = \frac{7 v}{8 l} = \frac{7}{8} \times 8 = 7$

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