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The first overtone of an open pipe has frequency n. The first overtone of a closed pipe of the same length will have frequency

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$\frac{n}{2}$

$2 n$

$\frac{3 n}{4}$

$\frac{4 n}{3}$

answer is C.

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Detailed Solution

First over tone of open pipe $n_{0} = \frac{2 V}{2 l_{0}}$
First over tone of closed pipe $n_{c} = \frac{3 V}{2 l_{c}}$
$l_{0} = l_{c} ∴ n_{c} = \frac{3}{4} n_{o}$

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