**Solution:**

Given,

Speed of car (v_{s}) = 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