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.

Options

A. 620Hz
B. 724Hz
C. 830Hz
D. 532Hz

Answer

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 takes time t to travel from train A to train B. Distance moved by train B during this interval is 40t. 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))`. so it will be 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=132m from the station. hence, path of pulse will be as shown in the figure. its inclination `theta` with tack is given by
`tantheta=(99)/(132) implies theta=37^(@)`
Veloity component of train A along path of the pulse,
`V_(S)=40cos37^(@)=32m//s`
Velocity component of train B along path of the pulse,
`V_(0)=40cos37^(@)=32m//s`
Hence, the frequency of pulse heard by driver by driver of train B is
`n=n_(0)((V+V_(0))/(V-V_(S)))=724Hz`.