The Doppler Effect — Question 10

Question 10

An observed Doppler chirp includes the arrival-time compression

A source approaches a stationary observer collinearly in still air. It emits constant $f_s=400\,\mathrm{Hz}$, and at the relevant emission event has speed $v=34\,\mathrm{m/s}$ and acceleration $a=dv/dt_e=2.0\,\mathrm{m/s^2}$. Take $c=340\,\mathrm{m/s}$ and $dt_r/dt_e=1-v/c$.

1. Find the instantaneous received frequency.

2. Derive its rate of change with reception time and evaluate it. Explain why differentiating only with respect to emission time misses a factor.

Question 10 - Solution

An observed Doppler chirp includes the arrival-time compression

(a) Method and calculation

The local Doppler factor gives$$f_o=\frac{f_sc}{c-v}=\frac{400(340)}{306}=444.444\,\mathrm{Hz}.$$The speed is evaluated at emission, when the received cycle was generated.

(b) Interpretation and further calculation

Differentiate first with respect to $t_e$, then apply the chain rule:$$\frac{df_o}{dt_e}=\frac{f_sc a}{(c-v)^2},\qquad\frac{dt_e}{dt_r}=\frac c{c-v}.$$$$\frac{df_o}{dt_r}=\frac{f_sc^2 a}{(c-v)^3}=3.22763\,\mathrm{Hz/s}.$$Arrival intervals are compressed as well as the frequency being shifted. Ignoring the second factor underestimates the observed chirp rate.

Physical check

For $a=0$ the chirp vanishes. These expressions assume a slowly identifiable local frequency and subsonic approach before source passage.