Consider a source of sound S, and an observer/detector D. The source emits a sound wave of frequency f0. The frequency observed by D is found to be
A
f1≠f2≠f3
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B
f1<f2
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C
f3<f0
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D
f1<f3<f2
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Solution
The correct option is Cf3<f0 Let relative velocity be v and the speed sound be v0. Then, f1=v0−(−v)v0×f0=v0+vv0f0f2=v0v0−v×f0f3=v0+v2v0−v2×f0
It is clear from above that f1≠f2≠f3,f3>f0 and we can prove that f2>f3>f1.