A source of sound of frequency 360 Hz is moving towards a stationary listener with speed v -10 m - s- where speed of sound in still air is v- Then- the apparent frequency received by listener will beA- 450 HzB- 430 HzC- 400 HzD- 390 Hz

The correct option is C 400 Hzf_app=f_0 ( v v_2/v v_s ) =360 ( v 0/v v/10 ) nbsp; =400 Hz ...

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Standard XII

Physics

Apparent Frequency:Source Moving

A source of s...

Question

A source of sound of frequency 360 Hz is moving towards a stationary listener with speed $\frac{v}{10} m / s$ , where speed of sound in still air is $v$ . Then, the apparent frequency received by listener will be

450 H z

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430 H z

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400 H z

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390 H z

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Solution

The correct option is C $400 H z$
$f_{a p p} = f_{0} (\frac{v - v_{2}}{v - v_{s}})$

$= 360 (\frac{v - 0}{v - \frac{v}{10}})$

$= 400 H z$

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A source of sound of frequency 360 Hz is moving towards a stationary listener with speed v10 m/s, where speed of sound in still air is v. Then, the apparent frequency received by listener will be

The correct option is C 400 Hzfapp=f0(v−v2v−vs) =360(v−0v−v10) =400 Hz

A source of sound of frequency 360 Hz is moving towards a stationary listener with speed $\frac{v}{10} m / s$ , where speed of sound in still air is $v$ . Then, the apparent frequency received by listener will be

The correct option is C $400 H z$
$f_{a p p} = f_{0} (\frac{v - v_{2}}{v - v_{s}})$

$= 360 (\frac{v - 0}{v - \frac{v}{10}})$

$= 400 H z$