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An open organ pipe has a fundamental frequency of $300 H z$ . The first overtone of a closed organ pipe has the same frequency as the first overtone of this open pipe. How long is each pipe?
(Speed of sound in air $= 330 m / s$ )

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$l_{o}$ $= 55 c m$

$l_{c} = 45.25 m$

$l_{o} = 42 . 5 c m$

$l_{c}$ $= 41.25 c m$

answer is A, D.

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Detailed Solution

Fundamental frequency of an open organ pipe,

$f_{1} = \frac{v}{2 L_{0}} \Rightarrow l_{o} = \frac{v}{2 f_{1}} = \frac{330}{2 \times 300}$

$= 0.55 m$

$= 55 c m$

Given, first overtone of closed organ pipe = first overtone of open organ pipe

Hence,

$3 (\frac{v}{4 l_{e}}) = 2 (\frac{v}{2 l_{o}})$

$l_{c} = \frac{3}{4} l_{e} = (\frac{3}{4}) (0.55)$

$= 0.4125 m$

$= 41.25 c m$

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