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Question: A source of sound is travelling with a velocity 40 \(km/hr\) towards an observer and emits sound of ...

A source of sound is travelling with a velocity 40 km/hrkm/hr towards an observer and emits sound of frequency 2000 Hz. If velocity of sound is 1220 km/hr, then what is the apparent frequency heard by an observer?
A. 2210 Hz\text{A}\text{. 2210 Hz}
B. 1920 Hz\text{B}\text{. 1920 Hz}
C. 2068 Hz\text{C}\text{. 2068 Hz}
D. 2086 Hz\text{D}\text{. 2086 Hz}

Explanation

Solution

When a source of wave is moving with respect to an observer, there is an apparent change in frequency of the wave depending on the direction of movement of the source. This effect is known as Doppler’s effect.
The shift in frequency can be calculated using Doppler’s shift formula.
Formula Used:
f=(v±vov±vs)f0f=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}

Complete answer:
Doppler effect, also known as Doppler shift, is the phenomenon that is observed when the source of a wave is moving with respect to an observer.
A train approaching you with blasting its horn is a common example of the Doppler Effect.
If f0{{f}_{0}} and v are the frequency and velocity of the wave produced by the source moving with velocity vs{{v}_{s}} and vo{{v}_{o}}is the velocity of the observer, then the apparent frequency is given by
f=(v±vov±vs)f0f=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}
vr{{v}_{r}} is taken positive if the observer moves towards the source and negative if it moves away from the source
vs{{v}_{s}} is positive if the source moves away from the observer and negative if it moves in the opposite direction.
In this question, the source of sound moves towards an observer with velocity, vs=40km/hr{{v}_{s}}=-40km/hr.
Observer is stationary, vo=0km/hr{{v}_{o}}=0km/hr
Velocity of the sound is v=1220km/hrv=1220km/hr
Frequency of the sound is f0=2000Hz{{f}_{0}}=2000Hz
We substitute these values in Doppler shift formula and get
f=(1220+0122040)(2000)=12201180×2000f=\left( \dfrac{1220+0}{1220-40} \right)(2000)=\dfrac{1220}{1180}\times 2000
f=2068Hz\Rightarrow f=2068Hz

Hence, option C is correct.

Note:
If the motion of the source of the wave relative to the observer is towards it then the Doppler shift is positive i.e. apparent frequency is higher than actual frequency. If the source moves in the opposite direction with respect to the observer, there is a downward shift in observed frequency.
It must be noted here that Doppler shift is not due to real shift in frequency of the source.