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Question: A police car with a siren of frequency 8kHz is moving with a uniform velocity 36\(\dfrac{{km}}{{hr}}...

A police car with a siren of frequency 8kHz is moving with a uniform velocity 36kmhr\dfrac{{km}}{{hr}} towards a tall building which reflects the sound waves. The speed of sound in air is 320ms\dfrac{m}{s}. The frequency of the siren heard by the car driver is:
A.A. 8.50kHz
B.B. 8.25kHz
C.C. 7.75kHz
D.D. 7.50kHz

Explanation

Solution

HINT- This question is based on the concept of Doppler’s effect which is the change in frequency of a wave in relation to an observer who is moving relative to the wave source. We have to keep in mind about the pair of source and observer than using the given formula- f=f[ccv]f' = f[\dfrac{c}{{c - v}}] and f=f[c+vc]f' = f[\dfrac{{c + v}}{c}].

Step-By-Step answer:
Now from the question, we have
Frequency of car siren = 8kHz
Velocity of car (v) = 36kmhr=36×518=10ms36\dfrac{{km}}{{hr}} = 36 \times \dfrac{5}{{18}} = 10\dfrac{m}{s}
Speed of sound in air (c) = 320ms\dfrac{m}{s}
Now, frequency received by the building = f=f[ccv]f' = f[\dfrac{c}{{c - v}}]
f=8[32032010]×8×103×[320+10320]=8.5kHz\Rightarrow f' = 8[\dfrac{{320}}{{320 - 10}}] \times 8 \times {10^3} \times [\dfrac{{320 + 10}}{{320}}] = 8.5kHz
f=8.5kHz\Rightarrow f' = 8.5kHz, therefore, option A.A. is the correct option.

NOTE- DOPPLER’S effect
It is named after the Austrian physicist Christian Doppler, who described the phenomenon in 1842.
The reason for the Doppler effect is that when the source of the waves is moving towards the observer, each successive wave crest is emitted from a position closer to the observer than the crest of the previous wave. Therefore, each wave takes slightly less time to reach the observer than the previous wave. Hence, the time between the arrivals of successive wave crests at the observer is reduced, causing an increase in the frequency. While they are travelling, the distance between successive wave fronts is reduced, so the waves “bunch together”. Conversely, if the source of waves is moving away from the observer, each wave is emitted from a position farther from the observer than the previous wave, so the arrival time between successive waves is increased, reducing the frequency. The distance between successive waves fronts is then increased, so the waves “spread out”.