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Standard X

Physics

Doppler Effect

A sound sourc...

Question

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 $f_{0}$ is frequency of the source then the apparent frequency recorded after $t = 2$ second is:

f_{0}

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f_{0} \frac{v}{(v + 2 g)}

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f_{0} \frac{(v + 2 g)}{v}

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f_{0} (\frac{v}{v - 2 g})

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Solution

The correct option is C $f_{0} (\frac{v}{v - 2 g})$
$v_{s} = 0 + g (2) = 2 g$
and $f_{a p p} = f_{0} \frac{v}{v - v_{s}} = f_{0} (\frac{v}{v - 2 g})$

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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:

The correct option is C f0(vv−2g)vs=0+g(2)=2gand fapp=f0vv−vs=f0(vv−2g)

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 $f_{0}$ is frequency of the source then the apparent frequency recorded after $t = 2$ second is:

The correct option is C $f_{0} (\frac{v}{v - 2 g})$
$v_{s} = 0 + g (2) = 2 g$
and $f_{a p p} = f_{0} \frac{v}{v - v_{s}} = f_{0} (\frac{v}{v - 2 g})$