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 fr

Question

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

TestHub · Accepted Answer

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

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