A whistle producing sound waves of frequencies 9500 Hz and 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 frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the whistle is:

Question

A whistle producing sound waves of frequencies 9500 Hz and 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 frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the whistle is: (A) 15√2 ms-1 (B) 15/√2 ms-1 (C) 15 ms-1 (D) 30 ms-1

Tardigrade Admin · Accepted Answer

Velocity of sound in air = 300 m/s If a source of sound is moving towards a stationaiy listener, the frequency heard by the listener would be different from the actual frequency of the source, this apparent frequency is given by fapp.=((vsound in air/vsound in air± vsource)), where symbols have their usual meanings. In the denominator +ve sign would be taken when source is receding away from the listener, while -ve sign would be taken when source is approaching the listener. Let v be the maximum value of source velocity for which the person is able to hear the sound, then 10000=fapp.=((300/300-v))×9500 ⇒ v=15 m/s

$152 m s^{- 1}$

$15/ 2 m s^{- 1}$

$15 m s^{- 1}$

$30 m s^{- 1}$

Q. A whistle producing sound waves of frequencies 9500Hz and above is approaching a stationary person with speed vms−1. The velocity of sound in air is 300ms−1. If the person can hear frequencies upto a maximum of 10,000Hz, the maximum value of v upto which he can hear the whistle is:

152​ms−1

15/2​ms−1

15ms−1

30ms−1

$152 m s^{- 1}$

$15/ 2 m s^{- 1}$

$15 m s^{- 1}$

$30 m s^{- 1}$