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)

14.3 %

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20 %

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10.0 %

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8.5 %

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Solution

The correct option is D 8.5 %
$n_{1} = n [\frac{V}{V - V_{s}}]$
$\Rightarrow \frac{11}{10} = \frac{V}{V - V_{s}} \Rightarrow V_{s} = \frac{V}{11}$
$n_{2} = n [\frac{V}{V + V_{s}}]$
which gives the percentage as 8.5%.

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Similar questions

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. 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).

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:

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

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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 vs<v. (v = speed of sound in air)

The correct option is D 8.5 %n1=n[VV−Vs] ⇒1110=VV−Vs⇒Vs=V11 n2=n[VV+Vs] which gives the percentage as 8.5%.

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)

The correct option is D 8.5 %
$n_{1} = n [\frac{V}{V - V_{s}}]$
$\Rightarrow \frac{11}{10} = \frac{V}{V - V_{s}} \Rightarrow V_{s} = \frac{V}{11}$
$n_{2} = n [\frac{V}{V + V_{s}}]$
which gives the percentage as 8.5%.