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

Question

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 (A) 15√2ms- 1 (B) 15 m s- 1 (C) 30 m s- 1 (D) none of these

Tardigrade Admin · Accepted Answer

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=fa p p=((300/300 - v))× 9500 Longrightarrow v=15ms- 1

$152 m s^{- 1}$

$15 m s^{- 1}$

$30 m s^{- 1}$

none of these

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

Q. A whistle producing sound waves of frequency 9500Hz above is approaching a stationary person with speed vms−1 . The velocity of sound in air is 300ms−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

152​ms−1

15ms−1

30ms−1

none of these

The velocity of sound in air =300ms−1 Let v be the maximum value of source velocity for which the person is able to hear the sound, then 10000=fapp​=(300−v300​)×9500 ⟹v=15ms−1

$152 m s^{- 1}$

$15 m s^{- 1}$

$30 m s^{- 1}$

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