A train approaching a hill at a speed of 40 km/hr sounds a whistle of frequency 580 Hz when it is at a distance of 1 km from the hill. Wind with a speed of 40 km/hr is blowing in the direction of motion of the train. Find
(i) The frequency of the whistle as heard by an observer on the hill. (ii) The distance from the hill at which echo from the hill is heard by the driver and its frequency. (Velocity of sound in air = 1200 km/hr)

Answer

(i) `599.33 Hz`
(ii) `0.935 km,621.43 Hz`

Explanation

(i) `f_("Hill")=f_(0)((v+v_(w))/(v+v_(w)-v_(s)))`
`=580((1200+40)/(1200+40-40))=599Hz`

`11240 t_(1) = 1rArr t_(1) = (1)/(1240)hr`
where `t_(1)` = time the sound to reach the hill
Let `t_(2)` = time for the echo to reach the train
`v_("echo")`=speed of echo
`=1200-40=1160km//hr`
`:.V_("train")+V_("echo")=d-V_("train")t_(1)`
`rArr (40-1160)t_(2)=1-(40)/(1240)=(1200)/(1240)`
`rArr t_(2)=(1)/(1240)hr`
`:.`Distance from the hill where echo reaches the train
`=d-v(t_(1)+t_(2))=1(40xx2)/(1240)=0.935km`
Frequency reaching the hill
`f_(1)=f_(0)((v+v_(w))/(v+v_(w)-v_(f)))`
Frequency of echo
`f_(1)=f_(1)=((v-v_(w)+v_(1))/(v-v_(w)))`
`=f_(0)((v+v_(w))/(v+v_(w)-v_(t)))((v-v_(w)+vt)/(v-v_(w)))`
`=580xx(1240)/(1200)xx(1200)/(1160)=620Hz`