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

$\frac{2 f v_{s}}{v}$

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$2 f v v_{s}$

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$\frac{2 f v}{v_{s}}$

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$\frac{f v_{s}}{v}$

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Solution

The correct option is A
$\frac{2 f v_{s}}{v}$

$Δ f = f_{1} - f_{2}$

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

$= f [{(1 - \frac{v_{s}}{v})}^{- 1} - {(1 + \frac{v_{s}}{v})}^{- 1}]$

$\approx f [(1 + \frac{v_{s}}{v}) - (1 - \frac{v_{s}}{v})] = \frac{2 f v_{s}}{v}$

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Similar questions

When a source of sound of frequency f crosses a stationery 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

△ f

is given by:

Q. The frequency changes by

10 %

as a sound source approaches a stationary observer with constant speed

V_{S} .

What would be the percentage change in frequency as the source recedes the observer with the same speed. Given that

V_{S}

< V

Q. When a source of sound approaches a stationary observer with velocity

v_{s}

, then the apparent frequency observed by the observer will be

(v

= velocity of sound

)

Q. The frequency changes by 10% as a sound source approaches a stationary observer with constant speed vs. what would be the percentage change in frequency as the source recedes the observer with the same speed. Given that

v_{s} < v

. (v = speed of sound in air)

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

The correct option is A 2fvsv Δf=f1−f2 =f(vv−vs)−f(vv+vs) =f[(1−vsv)−1−(1+vsv)−1] ≈f[(1+vsv)−(1−vsv)]=2fvsv

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

The correct option is A
$\frac{2 f v_{s}}{v}$

$Δ f = f_{1} - f_{2}$

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

$= f [{(1 - \frac{v_{s}}{v})}^{- 1} - {(1 + \frac{v_{s}}{v})}^{- 1}]$

$\approx f [(1 + \frac{v_{s}}{v}) - (1 - \frac{v_{s}}{v})] = \frac{2 f v_{s}}{v}$