A sound source is falling under gravity. At some time t=0 the detector lies vertically below source at a height H as shown . If v is velocity of sound and f0 is frequency of the source then the apparent frequency recorded after t=2 second is:
A
f0
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B
f0v(v+2g)
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C
f0(v+2g)v
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D
f0(vv−2g)
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Solution
The correct option is Cf0(vv−2g) vs=0+g(2)=2g and fapp=f0vv−vs=f0(vv−2g)