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  4. A whistle producing sound waves of frequency 9500Hz above is approaching a stationary person with speed v ms- 1 . The velocity of sound in air is 300 ms- 1 . If the person can hear frequency up to a maximum of 10,000Hz , the maximum value of v up to which he can hear the whistle is

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Q. A whistle producing sound waves of frequency $9500Hz$ above is approaching a stationary person with speed $v \, ms^{- 1}$ . The velocity of sound in air is $300 \, ms^{- 1}$ . If the person can hear frequency up to a maximum of $10,000Hz$ , the maximum value of $v$ up to which he can hear the whistle is

NTA AbhyasNTA Abhyas 2022

Solution:

The velocity of sound in air $=300 \, ms^{- 1}$
Let v be the maximum value of source velocity for which the person is able to hear the sound, then
$10000=f_{a p p}=\left(\frac{300}{300 - v}\right)\times 9500$
$\Longrightarrow \, \, v=15ms^{- 1}$