A closed organ pipe and an open organ pipe have their first overtone identical in frequency. Their lengths are in the ratio

1 : 2

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2 : 3

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3 : 4

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4 : 5

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Solution

The correct option is B $3 : 4$
Frequency of first overtone of closed pipe
$= \frac{3 v}{4 l_{1}} = n_{1}$
Frequency of first overtone of open pipe
$= \frac{2 v}{2 l_{2}} = n_{2}$

As $n_{1} = n_{2}$

$∴ \frac{3 v}{4 l_{1}} = \frac{2 v}{2 l_{2}}$

$∴ \frac{l_{1}}{l_{2}} = \frac{6}{8} = \frac{3}{4}$

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