Question

A train moving with speed 72 km h-1 emits a whistle of frequency 600 Hz. A person is standing stationary on the platform. Calculate the frequency heard by the person if the train (i) approaches and (ii) recedes away from the listener.

Answer 1

This is the Doppler's shift.
 
Apparent frequency= f.
v = velocity of sound = 330 m/s
Frequency of the sound source = f₀ = 600 Hz
Velocity of the source (train) = Vs = + 72 kmph = +20 m/s
velocity of the observer = Vo = 0 

$f=f_0\ \frac{v\underline{+}v_o}{v\bar{+}v_s}\\\\(1)\ f=600*\frac{330+0}{330-20}=638.71\ Hz\\\\(2)\ f=600*\frac{330+0}{330+20}=565.71\ Hz$

When the train approaches, the frequency will be more and it will be less as the train goes away.