The Doppler Effect — Question 8

Question 8

Zero shift from a passing source arrives after closest approach

A source moves at $v=34\,\mathrm{m/s}$ along $(x,y)=(vt,h)$ with $h=340\,\mathrm m$. A stationary observer is at the origin, and $c=340\,\mathrm{m/s}$. The source is closest at emission time $t_e=0$.

1. Find when the sound emitted at closest approach reaches the observer and the source position at that reception time.

2. Use $t_r=t_e+\sqrt{v^2t_e^2+h^2}/c$ to show that the instantaneous frequency received at that time equals the emitted frequency.

Question 8 - Solution

Zero shift from a passing source arrives after closest approach

(a) Method and calculation

The closest-approach sound travels distance $h$, so$$t_r=h/c=1.00\,\mathrm s.$$During its travel the source advances $vt_r=34\,\mathrm m$, and is at $(34,340)\,\mathrm m$ when this sound is received.

(b) Interpretation and further calculation

Differentiate the retarded-time relation:$$\frac{dt_r}{dt_e}=1+\frac{v^2t_e}{c\sqrt{v^2t_e^2+h^2}}.$$At $t_e=0$ this derivative is one, so $f_o=f_s/(dt_r/dt_e)=f_s$. Thus the zero-shift instant occurs one second after geometric closest approach in reception time. A transverse current line of sight is not the required condition.

Physical check

The source remains subsonic, so a unique emitted time corresponds to each received event. The delay is a propagation effect.