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Question

Two sources of sound are moving in opposite directions with velocities V1 and V2 (V1 > V2. Both are moving away from a stationary observer. The frequency of both the sources is 1700 Hz. If velocity of sound in air is 340 m/s and both V1 and V2 are much small compared to velocity of sound, then the value of V1-V2 for beat frequency to be 10 Hz as heard by observer is

A
4m/s
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B
3m/s
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C
2m/s
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D
1.5m/s
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Solution

The correct option is D 2m/s
Let the frequencies heard by the observer be f1 and f2 from the sources moving with speeds v1 and v2 respectively.

f1=vv+v1f

f2=vv+v2f

where, f=1700Hz is the frequency of the sound emitted from the source and v is the velocity of sound.

v1 and v2 are the velocities of the sources respectively.

f2f1=vv+v2fvv+v1f=fv(1v+v21v+v1)=fv(v1v2)(v+v1)(v+v2)

Since v1 << v and v2 << v, we can ignore them in comparison to v.

f2f1=fvv1v2)v2=fv1v2)v

Given: f2f1=10 and f=1700 and v=340

10=1700×(v1v2)×1340

v1v2=2m/s

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