In Doppler effect when the source is moving away from the observer at rest, the correct equation is:

f^{'} = f (\frac{v - v_{s}}{v + v_{s}})

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f^{'} = f (\frac{v_{s}}{v + v_{s}})

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f^{'} = f (\frac{v}{v + v_{s}})

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Solution

The correct option is C $f^{'} = f (\frac{v}{v + v_{s}})$
Source moving away from observer:
$f^{'} = f (\frac{v_{w a v e}}{v_{w a v e} + v_{s}})$

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Similar questions

When a source of sound of frequency f crosses a stationary observer with a speed $v_{s}$ (≪ speed of sound v), the apparent change in frequency $Δ$ f is given by

Q. When a source of sound of frequency

f

crosses a stationary observer with a speed

v_{s}

(

<<

speed of sound

v

), the apparent change in frequency

Δ f

is given by

Q. The Doppler shift is not the same for a moving observer as it is for a moving source. But show that, for

v_{S} << v

and

v_{O} << v

, the difference between the two effects becomes negligible.

Q. When a source of sound of frequency f crosses a stationary observer with a speed

v_{S}

(<< speed of sound v), the apparent change in frequency

△ f

is given by:

Q. 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).

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In Doppler effect when the source is moving away from the observer at rest, the correct equation is:

The correct option is C f′=f(vv+vs)Source moving away from observer:f′=f(vwavevwave+vs)

The correct option is C $f^{'} = f (\frac{v}{v + v_{s}})$
Source moving away from observer:
$f^{'} = f (\frac{v_{w a v e}}{v_{w a v e} + v_{s}})$