Question
The frequency changes by 10% as the source approaches a stationary observer with constant speed Vs. What should be the percentage change in frequency as the source recedes from the observer with the same speed? Given that Vs ≪ v(v is the speed of sound in air).
Answer: Option D
:
D
When the source approaches the observer,
f1=f(vv−vs)=f(1−vsv)−1≈f(1+vsv)or(f1−ff)×100=vsv×100=10.........(1)
In the second case, when the source recedes from the observer
f2=f(vv+vs)=f(1+vsv)−1=f(1−vsv)∴(f2−ff)×100=−vsv×100=−10
[from Eq.(i)]
In the first case, observed frequency increases by 10% while in the second case, observed frequency decreases by 10%
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:
D
When the source approaches the observer,
f1=f(vv−vs)=f(1−vsv)−1≈f(1+vsv)or(f1−ff)×100=vsv×100=10.........(1)
In the second case, when the source recedes from the observer
f2=f(vv+vs)=f(1+vsv)−1=f(1−vsv)∴(f2−ff)×100=−vsv×100=−10
[from Eq.(i)]
In the first case, observed frequency increases by 10% while in the second case, observed frequency decreases by 10%
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