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PhysicsSound WaveStanding waveMedium2 minPYQ_2024

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The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 cm$ , the length of the closed pipe will be :

Options:

Answer:

Solution:

$\frac{λ}{4} = L_{1} 2 (\frac{λ}{2}) = L_{2}$

$As, v = f λ \Rightarrow f_{2} = \frac{2 v}{2 L_{2}}$

$\Rightarrow v = f_{1} (4 L_{1}) \Rightarrow f_{2} = \frac{v}{L_{2}}$

$\Rightarrow f_{1} = \frac{v}{4 L_{1}}$

As per question, $f_{1} = f_{2}$

$\frac{v}{4 L_{1}} = \frac{v}{L_{2}}$

$\Rightarrow L_{2} = 4 L_{1}$

$\Rightarrow 60 = 4 \times L_{1}$

$\Rightarrow L_{1} = 15 cm$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Standing wave

⏱ 2mℹ️ Source: PYQ_2024

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