JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Waves & Sound

A source of sound S is moving with the velocity of

50 m / s

towards a stationary observer. The observer measures the frequency of the sound as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him?
(Take velocity of sound in air is

350 m / s

)

(A)1143 Hz
(B)857 Hz
(C)750 Hz✓ Correct
(D)807 Hz

Answer: (C)

Solution

While approaching, the Doppler shift raises the pitch to

f_{a} = \frac{v}{v - v _{s}} f_{0} = 1000 Hz

. Once the source has passed and is receding, the same speed lowers it to

f_{a}^{'} = \frac{v}{v + v _{s}} f_{0}

. Taking the ratio cancels the true frequency,

\frac{f _{a}^{'}}{f _{a}} = \frac{v - v _{s}}{v + v _{s}} = \frac{350 - 50}{350 + 50} = \frac{3}{4}

, so the receding pitch is

\frac{3}{4} \times 1000 = 750 Hz

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Waves & Sound

Solution

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