JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Waves & Sound

Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of

20 ms^{- 1}

with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source of car B? (speed of sound in air

= 340 ms^{- 1}

)

(A)2250 Hz✓ Correct
(B)2060 Hz
(C)2150 Hz
(D)2300 Hz

Answer: (A)

Solution

Both the source (car B) and the observer (car A) are receding from each other, so the pitch heard drops below the true pitch. With source and observer both moving away,

f = f_{0} \frac{v - v _{o}}{v + v _{s}}

, where

v = 340

v_{o} = v_{s} = 20

m/s. Then

2000 = f_{0} \frac{340 - 20}{340 + 20} = f_{0} \frac{320}{360}

, giving

f_{0} = 2000 \times \frac{360}{320} = 2250

Hz.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Waves & Sound

Solution

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