Two sources of sound are moving in opposite directions with velocities V $_{1}$ and V $_{2}$ $(V_{1}$ > $V_{2}$ . Both are moving away from a stationary observer. The frequency of both the sources is 1700 Hz. If velocity of sound in air is 340 m/s and both V $_{1}$ and V $_{2}$ are much small compared to velocity of sound, then the value of V $_{1}$ -V $_{2}$ for beat frequency to be 10 Hz as heard by observer is

4 m / s

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3 m / s

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2 m / s

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1.5 m / s

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Solution

The correct option is D $2 m / s$
Let the frequencies heard by the observer be $f_{1}$ and $f_{2}$ from the sources moving with speeds $v_{1}$ and $v_{2}$ respectively.

$f_{1} = \frac{v}{v + v_{1}} f$

$f_{2} = \frac{v}{v + v_{2}} f$

where, $f = 1700 H z$ is the frequency of the sound emitted from the source and v is the velocity of sound.

$v_{1}$ and $v_{2}$ are the velocities of the sources respectively.

$f_{2} - f_{1} = \frac{v}{v + v_{2}} f - \frac{v}{v + v_{1}} f = f v (\frac{1}{v + v_{2}} - \frac{1}{v + v_{1}}) = f v \frac{(v_{1} - v_{2})}{(v + v_{1}) (v + v_{2})}$

Since $v_{1}$ << v and $v_{2}$ << v, we can ignore them in comparison to v.

$∴ f_{2} - f_{1} = f v \frac{v_{1} - v_{2})}{v^{2}} = f \frac{v_{1} - v_{2})}{v}$

Given: $f_{2} - f_{1} = 10$ and $f = 1700$ and $v = 340$

$∴ 10 = 1700 \times (v_{1} - v_{2}) \times \frac{1}{340}$

$\Rightarrow v_{1} - v_{2} = 2 m / s$

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