The fundamental frequency in an open organ pipe is two times the third overtone of a closed organ pipe. Find ratio of $l_{1}$ and $l_{2}$ where $l_{1}$ and $l_{2}$ are respectively the lengths of open and closed organ pipe.

\frac{2}{3}

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\frac{3}{5}

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\frac{1}{7}

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\frac{4}{7}

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Solution

The correct option is C $\frac{1}{7}$
Fundamental frequency of open organ pipe, $n_{o_{(o p e n p i p e)}} = \frac{v}{2 l_{1}}$

The frequency of $p^{t h}$ overtone is $(2 p + 1) n_{0}^{'}$ , where $n_{0}^{'}$ is the fundamental frequency of closed pipe.
In a closed pipe, $n_{0}^{'} = \frac{v}{4 l_{2}}$
$∴$ 3rd overtone, $n_{7}^{'}$ = $\frac{7 v}{4 l_{2}}$

$∵$ $n_{o_{(o p e n p i p e)}} = 2 \times n_{7}^{'}$
On solving we get, $l_{1} : l_{2} = 1 : 7$

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The fundamental frequency in an open organ pipe is two times the third overtone of a closed organ pipe. Find ratio of l1 and l2 where l1 and l2 are respectively the lengths of open and closed organ pipe.

The fundamental frequency in an open organ pipe is two times the third overtone of a closed organ pipe. Find ratio of $l_{1}$ and $l_{2}$ where $l_{1}$ and $l_{2}$ are respectively the lengths of open and closed organ pipe.