Approaching a platform at a speed of 54 km/h , a train sounds a...

Speed of the train, $v_s = 54 \text{ km/h} = 54 \times \dfrac{5}{18} \text{ m/s} = 15 \text{ m/s}$

Apparent frequency during approach, $f_1 = 1620 \text{ Hz}$

Speed of sound, $v = 332 \text{ m/s}$

By Doppler's effect, the apparent frequency when the train approaches the stationary observer is given by:

f_1 = f_0 \left(\dfrac{v}{v - v_s} \right)

$1620 = f_0 \left( \dfrac{332}{332 - 15} \right) = f_0 \left( \dfrac{332}{317} \right)$

$f_0 = 1620 \times \dfrac{317}{332}$

When the train crosses the platform and moves away, the apparent frequency heard by the observer is:

f_2 = f_0 \left(\dfrac{v}{v + v_s} \right)

Substituting the value of $f_0$:

$f_2 = 1620 \times \left( \dfrac{317}{332} \right) \times \left( \dfrac{332}{332 + 15} \right)$

$f_2 = 1620 \times \dfrac{317}{347}$

$f_2 \approx 1480 \text{ Hz}$