A train A crosses a station with a speed of 40 m/s and whitles a short pulse of natural frequency `n_0=596Hz` another train B is approaching towards the same station with the same speed along a parallel track, Two track are `d=99m` apart. When train A whistles. train B is 152 m away from the station as shown in Fig. If velocity of sound in air is `v=300(m)/(s)`. calculate frequency of the pulse heard by driver of train B.

Explanation

When train A whistles sound pulse starts to travel in air from train A to train B. During this interval train B moves some distance towards the station. Let sound pulse take time t to travel from train A to train B. Distance moved by train B during this interval is 40 t. Therefore the distance of train B from station when its driver hears the pulse is `152-40t` Hence the distance travelled by the pulse is `sqrt((152-40t)^2+(99)^2)`. But it is equal to `vt=330t`.
`sqrt((152-40t)^2+(99)^2)=330t`
or `t=0.5s`
Therefore, driver of train B hears the pulse when his train is `152-40t=132` m from the station Hence path of pulse will be as shown in the figure Its inclination `theta` with track is given by `tantheta=(99)/(132)impliestheta=37^@`
velcocity component of train A along path of the pulse, `v_S=40cos37^@=32(m)/(s)`
Velocity component of train B along path of the pulse, `v_0=40cos37^@=32(m)/(s)` Hence, the frequency of pulse heard by driver of train B is
`n=n_0((v+v_0)/(v-v_S))=724Hz`