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 `1km` from a hill. A wind with a speed of `40km//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 the echo from the hill is heard by the driver and its frequency.
(Velocity of sound in air `= 1, 200 km//hr`)

Answer

(i) 599Hz, (ii) 621 Hz

Explanation

(485.7 Hz,257.3 Hz)
(i) 599 Hz (ii) 621 Hz
(i) The frequency of the whistle as heard by observer on the hill
`f'=f_(0). [((v+v_w)-v_0)/((v+v_w)-v_s)]`
`=580[(1200+40)/(1200+40+40)]=599 Hz`
(ii) Let echo from the hill is heard by the driver at B which is at x from the hill.The sound produced whent the source was at a distance 1 km from hill
The time taken by the driver to reach from A to B
`t_1=(1-x)/40` ....(i)
The time taken by the echo to reach hill
`t_2=t_(AH)+t_(HB)`
`t_2=(1)/((1200+40))+(x)/((1200-40))`....(ii)
where `t_(AH)=` time taken by sound from A to H with velocity (1200+40)
`t_(HB)=` time taken by sound from H to B with velocity (1200-40)
From (i) and(ii) `t_1=t_2`
`implies (1-x)/40=(1)/((1200+40))+(x)/((1200-40))`
Which gives x=0.902m
The frequency of echo as heard by the driver can be calculated by considering that the source is the acoustic image
`f''=f_0[((v-v_w)(-v_0))/((v-v_w)-v_s)]=f_0[((v-v_w)+v_0)/((v-v_w)-v_s)]`
`=580[((1200-40)+40)/((1200-40)-40)]=621 Hz`