Physics - Sound Wave Question with Solution | TestHub

PhysicsSound WaveDopler's effectEasy2 minPYQ_2020

PhysicsEasysingle choice

A whistle producing sound waves of frequencies $9500 Hz$ and above is approaching a stationary person with speed of $v m s^{- 1}$ . The velocity of sound in air is $300 m s^{- 1}$ . If the person can hear frequencies up to a maximum of $10000 Hz$ , the maximum value of $v$ up to which he can hear the whistle is

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Answer:

Solution:

$\frac{ν^{'}}{ν} = \frac{v_{s}}{v_{s} - v}$

where $v_{s}$ is the velocity of sound in air.

$\frac{10000}{9500} = \frac{300}{300 - v}$

$\Rightarrow (300 - v) = 285 \Rightarrow v = 15 m s^{- 1}$

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

⏱ 2mℹ️ Source: PYQ_2020

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