Question

A train approaching a hill at a speed of 40 km/hr sounds a whistle 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. The distance from the hill at which the echo from the hill is heard by the driver and its frequency (Velocity of sound in air = 1200 km/hr.)

Answer 1

speed of sound in air is c and as the air is also moving the velocity of sound is $c+v_a$ where $v_a$ is the speed of air

The train sounds the whistle 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

$t=\frac{x}{v}=\frac{x}{c+v_a}$

after the reflection from the sound waves travels towards the train hence now the velocity of sound is $c-v_a$

let the distance travel by reflected wave be $x^{{}^{\prime }}$ therefore time taken by it $t^{{}^{\prime }}=\frac{x^{{}^{\prime }}}{c-v_a}$

the total time between echoing the whistle and its reflection $t+t^{{}^{\prime }}$ meanwhile the train travels the distance $x-x^{{}^{\prime }}$ with constant speed $v_s$

$x-x^{{}^{\prime }}=(t+t^{{}^{\prime }})v_s$

$x-x^{{}^{\prime }}=\frac{v_s}{c+v_a}x+\frac{v_s}{c-v_a}{x}^{{}^{\prime }}$

$x\frac{c+v_a-v_s}{c+v_A}=x^{{}^{\prime }}\frac{c-v_a+v_s}{c-v_a}$

$x^{{}^{\prime }}=x\frac{(c-v_a)\times (c+v_a-v_s)}{(c+v_a)\times (c-v_a+v_s)}$

$x^{{}^{\prime }}=\frac{(1200-40)\times (1200+40-40)}{(1200+40)\times (1200+40-40)}=.935Km$

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 hill. Hence as observed from ground, source is stationary and observer is moving towards source with speed 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) emitting sound of frequency f which is the frequency observed by the hill.

Thus, frequency of echo as heard by observer on train, is given by

$f^{{}^{\prime \prime }}=f^{{}^{\prime }}\frac{v^{{}^{\prime }}+v_o}{v^{{}^{\prime }}}=\frac{1160-(-40)}{1160}\times 620=641Hz$