A train blowing the whistle moves with a constant velocity ' V ' away from an observer standing on the platform. The ratio of the natural frequency of the whistle "n' to the apparent frequency is 1 2: 1 . If the train is at rest and the observer moves away from it at the same velocity 'V', the ratio of 'n' to the apparent frequency is

Step-by-step solution

If the train is going away from the observer, the apparent frequency is v ₁= vu v + u = v 1+ u v It is observed that v =1.2 v ₁ (Given), In the second case the apparent frequency is v ₂= v ( v - u ) v = v (1- u v ) or v v ₂ = 1 1- M v Now, from equation (1) we have v v₁ =1+ u v or 1.2=1+ u v u =0.2 v That is, u v =0.2 Using this in equation (2), we get, v v₂ = 5 4 =1.25