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PhysicsSound WaveDopler's effectHard2 minPYQ_2020

PhysicsHardsingle choice

A whistle of frequency $500 Hz$ , tied to the end of a string of length $1.2 m$ , revolves at $400 rev / \min$ . A listener standing some distance away in the plane of rotation of the whistle, hears frequency in the range of (speed of sound = $340 {ms}^{- 1}$ )

Options:

Answer:

Solution:

$v_{s} = r ω$

$v_{s} = 1.2 \times 2 \times 3.14 \times (\frac{400}{60}) = 50 m s^{- 1}$

$v_{\min} = \frac{v}{v + v_{s}} ν$

$= \frac{340}{340 + 50} \times 500 = 436 Hz$

$v_{\max} = \frac{v}{v - v_{s}} v$

$= \frac{340}{340 - 50} \times 500$

$= 586 Hz$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Dopler's effect

⏱ 2mℹ️ Source: PYQ_2020

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