A stationary source is emitted sound at a fixed frequency f 0 , which is reflected by two cars approaching the source. The difference between the is frequencies of sound reflected from the car is 1 .

Question

A stationary source is emitted sound at a fixed frequency f 0 , which is reflected by two cars approaching the source. The difference between the is frequencies of sound reflected from the car is 1 . 2 % of f 0 . What is the difference in the speed of the cars (in m s - 1 )? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m s - 1

TestHub · Accepted Answer

Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.

Hence,

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

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

$∴ f_{1} - f_{2} = (\frac{1.2}{100}) f_{o}$

$= f_{o} [\frac{v + v}{v + v_{1}} - \frac{v + v_{2}}{v - v_{2}}]$

Or

$(\frac{1.2}{100}) f_{o} = \frac{2 v (v_{1} - v_{2})}{(v - v_{1}) (v - v_{2})} f_{o}$

$a s v_{1} a n d v_{2}$ are very very less than v.

We can write, $(v - v_{1}) o r (v - v_{2}) \approx v .$

$∴ (\frac{1.2}{100}) f_{o} = \frac{2 (v_{1} - v_{2})}{v} f_{o}$

$O r (v_{1} - v_{2}) = \frac{v \times 1.2}{200}$

$= \frac{300 \times 1 .2}{200} = 1 .98 m s^{- 1}$

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