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 $$ 484=n(330/330-15)$$ $$=n(330/315)$$ $$n=\frac{484*21}{22}$$

$$=462Hz$$