JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Waves & Sound

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A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is

345 m / s

, then the speed of the car is:

(A) $24 km / hr$
(B) $36 km / hr$
(C) $18 km / hr$
(D) $54 km / hr$ ✓ Correct

Answer: (D)

Solution

The horn's sound undergoes two Doppler shifts. The wall first receives it from the approaching car, a moving source with a stationary receiver, so

f_{1} = f_{0} \frac{v _{s}}{v _{s} - v _{c}}

. The wall then re-radiates this, and the car, now a moving observer, closes on it, so the driver hears

f_{2} = f_{1} \frac{v _{s} + v _{c}}{v _{s}} = f_{0} \frac{v _{s} + v _{c}}{v _{s} - v _{c}}

. With

f_{0} = 440

Hz and

f_{2} = 480

Hz,

\frac{480}{440} = \frac{345 + v _{c}}{345 - v _{c}}

, i.e.

\frac{12}{11} = \frac{345 + v _{c}}{345 - v _{c}}

. Cross-multiplying,

12 (345 - v_{c}) = 11 (345 + v_{c})

, so

23 v_{c} = 345

and

v_{c} = 15

m/s

= 15 \times \frac{18}{5} = 54

km/h.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Waves & Sound

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