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Physics - Sound Wave Question with Solution | TestHub

PhysicsSound WaveBeatsEasy2 minPYQ_2020

PhysicsEasysingle choice

Each of the two strings of length $51.6 cm$ and $49.1 cm$ are tensioned separately by $20 N$ force. Mass per unit length of both the strings is same and equal to $1 g m^{- 1}$ . When both the strings vibrate simultaneously in fundamental mode, the number of beats is

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Answer:

Solution:

$l_{1} = 0.516 m, l_{2} = 0. 491 m, T = 20 N$

Mass per unit length, $μ = {0.001 kg m}^{- 1}$

$Frequency, υ = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$υ_{1} = \frac{1}{2 \times 0 \cdot 516} \sqrt{\frac{20}{0 \cdot 001}}$

$υ_{2} = \frac{1}{2 \times 0 \cdot 491} \sqrt{\frac{20}{0 \cdot 001}}$

∴ Number of beats $= υ_{1} - υ_{2} = 7$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Sound WaveSubtopic:Beats

⏱ 2mℹ️ Source: PYQ_2020

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