The Doppler Effect — Question 5

Question 5

An echo from a moving target has two Doppler transformations

A stationary source and receiver transmit $f_s=40000\,\mathrm{Hz}$ toward a target receding at $u=10.0\,\mathrm{m/s}$ in still air with $c=340\,\mathrm{m/s}$. The target reflects backward without changing frequency in its own moving frame.

1. Derive the returned echo frequency.

2. Find the beat rate between the echo and the transmitted reference. Compare with the low-speed approximation $2f_su/c$.

Question 5 - Solution

An echo from a moving target has two Doppler transformations

(a) Method and calculation

The target first acts as a receding observer:$$f_T=f_s\frac{c-u}{c}.$$It then acts as a source moving away from the original receiver, emitting backward at $f_T$. Hence$$f_e=f_T\frac c{c+u}=f_s\frac{c-u}{c+u}=37714.286\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The echo-reference difference is$$f_b=f_s-f_e=\frac{2f_su}{c+u}=2285.714\,\mathrm{Hz}.$$The approximation gives $2(40000)(10)/340=2352.941\,\mathrm{Hz}$, about $2.94\%$ above the exact value. It drops the $u$ in the denominator and is appropriate only when $u/c\ll1$.

Physical check

A receding target lowers the returned frequency twice. The echo tends to the transmitted tone as $u\to0$.