A stationary source is emitting sound at a fixed frequency , which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of . What is the difference in the speeds of the cars (in k

Question

A stationary source is emitting sound at a fixed frequency $f_{0}$ , which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of $f_{0}$ . What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ${ms}^{- 1}$ . [2010]

Smart Achievers · Accepted Answer

(7)

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.

Frequency of sound reflected by the car $f_{0}$

$f_{1} = f_{0} (\frac{v + v_{1}}{v - v_{1}})$

And frequency of sound reflected by the car $C_{2}$

$f_{2} = f_{0} (\frac{v + v_{2}}{v - v_{2}})$

$∴$ $f_{1} - f_{2} = (\frac{1.2}{100}) f_{0} = f_{0} [\frac{v + v_{1}}{v - v_{1}} - \frac{v + v_{2}}{v - v_{2}}]$

$or$ $(\frac{1.2}{100}) f_{0} = \frac{2 v (v_{1} - v_{2})}{(v - v_{1}) (v - v_{2})} f_{0}$

$(v - v_{1}) = (v - v_{2}) \approx v as v_{1} and v_{2} are very very less than v .$

$∴$ $(\frac{1.2}{100}) f_{0} = \frac{2 (v_{1} - v_{2})}{v} f_{0}$

$or$ $(v_{1} - v_{2}) = \frac{v - 1.2}{200} = \frac{330 \times 1.2}{200} = 1.98 {m s}^{- 1} \approx 7 {km h}^{- 1}$

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