question_answer

A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed\[v\text{ }m{{s}^{-1}}\]. The velocity of sound in air is\[300\tilde{\ }m{{s}^{-1}}\]. If the person can hear frequencies up to a  maximum of 10000 Hz, the maximum value of v upto which he can hear the whistle is           [AIEEE 2006]

A)  \[15\sqrt{2}m{{s}^{-1}}\]       

B)       \[15/\sqrt{2}m{{s}^{-1}}\]

C)  \[15m{{s}^{-1}}\]                              

D)  \[30\,m{{s}^{-1}}\]

Correct Answer: C

Solution :

[c] Velocity of sound in air = 300 m/s.

If a source of sound is moving towards a stationary listener, the frequency heard by the listener would be different from the actual frequency of the source, this   apparent   frequency is given by

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