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Standard XIII

Physics

Source and Listener Moving in Different Lines

A source of s...

Question

A source of sound S is travelling at $\frac{100}{3} m / s$ along a road, towards a point A. When the source is 3 m away from A, a person standing at a point O on a road perpendicular to AS hears a sound of frequency v’. The distance of O from A at that time is 4m. If the original frequency is 640 Hz, then the value of v’ is (velocity of sound is 340 m/s)

620 Hz

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680 Hz

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720 Hz

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840 Hz

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Solution

The correct option is B 680 Hz
Effective value of velocity of source is

$v_{s} = \frac{100}{3} c o s θ$
$= \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$v^{'} = \frac{v}{v - v_{s}} v$
$v^{'} = \frac{340}{340 - 20} \times 640 H z = 680 H z$

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The correct option is B 680 HzEffective value of velocity of source is vs=1003 cosθ =1003×35=20 m/s v′=vv−vsv v′=340340−20×640 Hz=680 Hz

The correct option is B 680 Hz
Effective value of velocity of source is

$v_{s} = \frac{100}{3} c o s θ$
$= \frac{100}{3} \times \frac{3}{5} = 20 m / s$
$v^{'} = \frac{v}{v - v_{s}} v$
$v^{'} = \frac{340}{340 - 20} \times 640 H z = 680 H z$