A train passes through a station with constant speed. A stationary observer at the station platform measures the tone of the train whistle as $$484Hz$$ when it approaches the station and $$442Hz$$ when it leaves the station. If the sound velocity is $$330m/s$$, then the tone of the whistle and the speed of the train are
(a) $$462hz, 54km/h$$
(b) $$463Hz, 52Km/h$$
(c) $$463Hz, 56Km/h$$
(d) $$464Hz, 52Knm/h$$

Solution:

When train approaches the station, the frequency heard by the observer
$$n_1=n\frac{v}{v-v_s}=n(\frac{330}{330-v_s})$$
Here, $$v=330m/s$$
n is the actual frequency of the whistle
$$484 =n(330/330-v_s)$$….. (i)
When the train leaves the station $$n_2=n\frac{v}{v+v_s}=n(\frac{330}{330+v_s})$$
$$442=n(\frac{330}{330+v_s})$$…. (ii)
Divide Eqs (i) by (ii), we get
$$\frac{484}{442}=330+v_s/330-v_s$$
$$1.09=(330+v_s)/(330-v_s)$$
$$330+v_s=1.09(330-v_s)$$
$$v_s=\frac{31.35}{2.09}$$$$=15m/s$$
Substituting $$v_s$$ in Eqn (i) gives $$484=n(330/330-15)$$ $$=n(330/315)$$ $$n=\frac{484*21}{22}$$
$$=462Hz$$