A sound source S is moving along a straight track with speed $v$, and is emitting sound of frequency $f_0$. An observer is standing at a finite distance, at the point O, from the track. The time variation of frequency heard by the observer is best represented by: ($t_0$ represents the instant when the distance between the source and observer is minimum)
Answer: (B) see figure
Only the component of the source velocity along the line joining source and observer causes a Doppler shift: $f = f_0\dfrac{v_s}{v_s - v\cos\theta}$, where $\theta$ is the angle between the source velocity and the line from source to observer.
Far away on the approach side, $\theta \approx 0$ and $f$ is highest. As the source comes closer, $\theta$ grows, so $f$ falls smoothly. At the closest approach $\theta = 90^\circ$ and $f = f_0$. After that $\cos\theta < 0$ and $f$ keeps falling below $f_0$, approaching its lowest value far away.
So the frequency decreases continuously and smoothly, passing through $f_0$ at about $t_0$.
Solution by Sreeraj P, M.Sc Physics