Q. A whistle producing sound waves of frequency 9500 Hz 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 10,000 Hz, the maximum value of v up to which he can hear the whistle is
A) $15\sqrt{2}ms^{- 1}$
B) $15/\sqrt{2}ms^{- 1}$
C) $15\ ms^{- 1}$
D) $30\ ms^{- 1}$
Explanation
(c) 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
$10000 = f_{app} = \left( \frac{300}{300 - v} \right) \times 9500$
$\Longrightarrow \ \ v = 15ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.