A stationary source is emitting sound at a fixed frequency `f_(o)`, 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_(o)`. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? Thee 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`.