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)`.

Answer

7

Explanation

Let speed of cars are `V_(1)` and `V_(2)`
frequency recivied by car `f_(1) = ((v)/(v - v_(1)))f_(0)`
frequency reflected by car `f_(2) = ((v + v_(1))/(v))((v)/(v - v_(1)))f_(0)`
`Deltaf = f_(2^(')) - f_(2) = ((v + v_(2))/(v - v_(2))(v + v_(1))/(v - v_(1)))f_(0)`
`Deltaf = Deltaf = (((v+v_(2))(v-v_(1))-(v+v_(1))(v-v_(2)))/((v-v_(1))(v-v_(2))))f_(0) : Deltaf = (2v(v_(2)-v_(1)))/((v-v_(1))(v-v_(2)))approx(2(v_(2)-v_(1)f_(0)))/(v))`
Given `(2(v_(2)-v_(1)f_(0)))/(v) = (1.2)/(100)f_(0) rArr v_(2) - v_(1) = 7.126`
Answer in nearest integer is `7`.