A closed organ pipe and an open pipe of same length produce $4$ beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths, the number of beats produced will be
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Let the lengths of the open pipe and closed pipe be $L$ and $L^{\prime}$ and their frequencies be $\nu$ and $\nu^{\prime}$

Speed of sound in air is $v m / s$

Frequency in the closedpipe $\quad \nu^{\prime}=\frac{(2 n-1) v}{4 L^{\prime}}$

Frequency in the open pipe $\quad \nu=\frac{m v}{2 L}$

Beat frequency $b=\nu-\nu^{\prime} \Longrightarrow \frac{m v}{2 L}-\frac{(2 n-1) v}{4 L^{\prime}}=4$

Now the lengths of both the pipes become double, beat frequency

$b^{\prime}=\frac{m v}{2(2 L)}-\frac{(2 n-1) v}{4\left(2 L^{\prime}\right)}$

$\Longrightarrow b=\frac{4}{2}=2$

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