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PhysicsSound WaveBeatsMedium2 minPYQ_2020

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Two identical piano strings of length $0.750$ $m$ are each, tuned exactly to $500$ $H z$ . The tension in one of the strings is then increased by $21 %$ . If they are now struck, what is the beat frequency between the fundamental mode of the two strings?

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Answer:

Solution:

$f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} = 500$
$f_{2} = \frac{1}{2 l} \sqrt{\frac{1.21 T}{μ}} = 1.1 (500)$
$f_{2} - f_{1} = 50$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Beats

⏱ 2mℹ️ Source: PYQ_2020

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