Worked exercise — The cyclist and the acoustic mirror

Problem

A cyclist moves towards a wall at speed $v = 30\;\text{m/s}$ emitting a sound of frequency $f_S = 440\;\text{Hz}$. The wall acts as an “acoustic mirror”, reflecting the wave at the same frequency at which it received it. Find (a) the frequency the wall “hears”; (b) the frequency the cyclist hears returning after the reflection. Speed of sound: $v_0 = 340\;\text{m/s}$.

Solution

Setting up. The problem is a double Doppler effect: first the cyclist (moving source) towards the wall (stationary receiver), then the wall (stationary source) towards the cyclist (moving receiver).

(a) First Doppler — the wall receives. Source approaching, stationary receiver:

$f_\text{muro} = f_S\,\frac{v_0}{v_0 - v} = 440\cdot\frac{340}{340-30} = 440\cdot\frac{340}{310}$

$\ev{f_\text{muro} \approx 482{,}6\;\mathrm{Hz}}$

(b) Second Doppler — the cyclist receives. The wall re-emits at $f_\text{muro}$ as a stationary source; the cyclist is now an approaching receiver:

$f_\text{cic} = f_\text{muro}\,\frac{v_0 + v}{v_0} = 482{,}6\cdot\frac{340+30}{340} = 482{,}6\cdot\frac{370}{340}$

$\ev{f_\text{cic} \approx 525\;\mathrm{Hz}}$

The cyclist hears his own sound return to him almost a tone and a half higher (from $440$ to $525$ Hz). This is the principle behind sonar and Doppler ultrasound: the target’s speed is worked out from the double shift.

Inverse exercise. If the cyclist heard the reflection at $f_\text{rif} = 700\;\text{Hz}$, combining the two consecutive Doppler shifts $f_\text{rif}/f_S = (v_0+v)/(v_0-v)$, from which:

$v = v_0\,\frac{f_\text{rif}-f_S}{f_\text{rif}+f_S} = 340\cdot\frac{700-440}{700+440} = 340\cdot\frac{260}{1140} \approx 77{,}5\;\mathrm{m/s}$

About $280$ km/h: a record-breaking cyclist!

Collegamenti

Argomenti: Waves Concetti: Doppler effect Competenze: Analysis of limiting cases and thought-experiment physics