Solution:
Given,
Speed of car (vs) = 20 m/s
Frequency of sound (f) = 600 Hz
Velocity of sound (v) = 330 m/s
Change in the frequency (Δf) = ?
When the car approaches towards stationary observer,
\(f'=\frac v{v-v_s}\cdot f=\frac{330}{330-20}\times600=638.7\) Hz
When the car passes the stationary observer,
\(f''=\frac v{v+v_s}\cdot f=\frac{330}{330+20}\times600=565.7\) Hz
Change in frequency heard by the observer as the car approaches and passes him is,
Δf = f' - f'' = 638.7 - 565.7 = 73 Hz
