Physics - Sound Wave Question with Solution | TestHub

PhysicsSound WaveStanding waveEasy2 minPYQ_2020

PhysicsEasysingle choice

The fundamental frequency of a closed organ pipe of length $20 cm$ is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is

Options:

Answer:

Solution:

For closed organ pipe,

$n_{c} = \frac{V}{4 l}$

For open organ pipe,

$n_{0} = \frac{V}{2 l^{'}}$

For the second overtone of an open organ pipe,

$n^{'} = 3 n_{0} = \frac{3 V}{2 l^{'}}$

$n_{c} = n^{'}$

$\frac{V}{4 l} = \frac{3 V}{2 l^{'}}$

$l^{'} = 6 l = 6 \times 20$

$l^{'} = 120 c m$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Standing wave

⏱ 2mℹ️ Source: PYQ_2020

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