A train approching a hill at a speed of `40 km//hr` sounds a whitstle of frequency `600 Hz` when it is at a distance of `1 km` from a hill. A wind with a speed of `40 km//hr` is blowing in the direction of motion of the train. Find,
(a) The frequency of the whistle as heard by an observer on the hill.
(b) the distance from the hill at which the echo from the hill is heard by the driver and its frequency (Veocity of sound in air `= 1200 km//hr`)

Explanation

A train is moving towards a hill with speed `v_(s)` with respect to the ground. The speed to sound in air. i.e. the speed of the sound with respect to medium (air) is `c`. While air itself is blowing towards hill with velocity `v_(m)` (as observed from ground). for an observer standing on the ground. Which is the inertial frame. the speed of sound towards hill is given by
`v = c + v_(m)`
(a) The observer on the hill stationary while source is approching him. Hence, frequency of whistle. heared by him is
`f' = f(v)/(v - v_(s))`
for `f = 600 Hz, v_(s) = 40 km//hr`, and `v = (1200 + 40) km//hr`, we get
`f' = 600. (1240)/(1240 - 40) = 620 Hz`.
(b) the train sounds the whistie when it is at distance `x` from the hill. Sound moving with velocity `v` with respect to ground, takes time `t` to reach the hill, such that.
`t = (x)/(v) = (x)/(C + v_(m))`
After reflection from hill, sound waves move backwards, towards the train. the sound is now moving opposite to the wind direction. Hence, its velocity with respect to the ground is
`v' = c - v_(m)`
Suppose when this reflected sound (or echo) reaches the train, it is at distance `x'` from hill. The time taken by eacho to traval distance `'x`' is given by
`t' = (x')/(v) = (x')/(c - v_(m)) .....(ii)`
Thus, total time `(t + t')` elapes between sounding the whistle and echo reachaing back. In the same time, the train moves a distance `(x - x')` with constant speed `v_(s')` as observed from ground. That is, Substituting from `(i)` and `(ii)`, for `t'`, we find
`x - x' = (v_(s))/(c + v_(m)) x + (v_(s))/(c - v_(m)) x'` or, `(c + v_(m) - v_(s))/(c + v_(m)) x = (v_(s) + c- v_(m))/(c - v_(m)) x'`
For `x = 1 km, c = 1200 km//hr`, and `v_(m) = 40 km//hr`, we get
`(1200 + 40 - 40)/(1200 + 40)x'` or, `x' = (1160)/(1240) = 0.935 km`.
Thus, the echo is heard when train is `935 m` from the hill.
Now, for the observer moving along with train, echo is a sound produced by a stationary source, i.e. the all. Hence as observed from ground, source is stationary and observer is mofing towards source with speed `40 km//hr`. Hence `v_(O) = -40 km//hr`. On the other hand, reflected sound travels opposite to wind velocity. That is, velocity of echo with respect to ground is `v'`. Further, the source (hill) is emitting sound of frequncy of echo sa heard by observer on train, if given by
`f'' = f' (v' + v_(O))/(v') rArr f'' = ((1160 - (-40)))/(1160) xx 620 = 641 Hz`