Two trains A and B simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration `2(m)/(s^2)` from rest, while B with the same acceleration but with initial velocity of 40 m/s. Twenty seconds after the start, passenger of A hears whistle of B. If frequency of whistle is 1194 Hz and velocity of sound in air is 322 m/s, calculate frequency observed by the passenger.

Two trains A and B simultaneously start moving along parallel tracks f

1.	Two trains A and B simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration `2(m)/(s^2)` from rest, while B with the same acceleration but with initial velocity of 40 m/s. Twenty seconds after the start, passenger of A hears whistle of B. If frequency of whistle is 1194 Hz and velocity of sound in air is 322 m/s, calculate frequency observed by the passenger.
Answer» Since both the trains move in same direction with same acceleration, therefore, their relative acceleration is zero. Initial velocity of train A is zero while that of train B is 40 m/s. Therefore, train A of observer is behind the train B or source when whistle is heard by passenger of A. Let this sound be produced at time t. Considering motion of train B (sound source) up to this instant, `u=40(m)/(s)`,`a=2(m)/(s^2)`,`time=t` `v=v_b=?` `s=s_B=?` ..(i) We have `v=u+at`, `v_B=40+2t` `s=ut+(1)/(2)at^2` `s_B=40t+t^2` But this sound is heard by the observer at `t=20`s. Considering his motion (motion of train A) up to this instant, `u=0`,`a=2(m)/(s^2)`,`t=20s` `v=v_B=?`,`s=s_B=?` We have, `v=u+at` `v_A=40(m)/(s)` `s=ut+(1)/(2)at^2` `s_A=400m` When sound was produced, train B (source) was at a distance `s_B` from initial point while observer receives this sound at `t=20s` and at a distance `s_A` from initial point. It means that sound waves travel a distance `(s_B-s_A)` in air and take time `(20-t)` to travel it. `s_B-s_A=(20-t)v` ..(iii) Where `v=322(m)/(s)` (speed of sound). Substituting values of `s_A` and `s_B` in Eq.(iii), `t=18s` When sound was produced source was moving with velocity `v_B=(40+2t)=76(m)/(s)` (away from observer) while observer receive these sound waves at `t=20s`, when he was moving with velocity `v_A=40(m)/(s)` (towards the source). Hence, the observed frequency is `n=n_0(v+v_A)/(v+v_B)` Where `n_0=1194Hz` (natural frequency of source) and so `n=1086Hz`.

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