Doppler effect of an ambulance

Problem

An ambulance sounds its siren at $f_0 = 700\;\mathrm{Hz}$ and approaches you at $v_s = 30\;\mathrm{m/s}$. Calculate the perceived frequency while approaching and after passing. Speed of sound $343\;\mathrm{m/s}$.

Solution

Setup. Observer at rest, source in motion. Doppler effect:

$f' = f_0\,\frac{c - v_R}{c - v_E}$

with $v_R = 0$ (observer at rest) and $v_E$ the component of the source’s velocity towards the observer.

Approaching (the source moves towards you, $v_E = +30$): the wavefronts compress, the frequency rises.

$f'_{\text{av}} = 700\cdot\frac{343}{343 - 30} = 700\cdot\frac{343}{313} = 700\cdot 1{,}096$

$\ev{f'_{\text{av}} \approx 767\;\mathrm{Hz}}$

Receding (after passing, $v_E = -30$): the wavefronts stretch out, the frequency drops.

$f'_{\text{al}} = 700\cdot\frac{343}{343 + 30} = 700\cdot\frac{343}{373} = 700\cdot 0{,}920$

$\ev{f'_{\text{al}} \approx 643\;\mathrm{Hz}}$

The audible pitch jump at the moment of passing ($767 \to 643\;\mathrm{Hz}$, roughly a minor third) is the Doppler effect everyone hears as an ambulance speeds past.

Linked atoms

Topics: Onde Concepts: Effetto Doppler Skills: Analisi dimensionale