Classical Doppler effect for sound ($f_o=f_s(c-u_o)/(c-u_s)$)

Work in the material's rest reference frame, with constant speed of sound $c>0$. For one-dimensional propagation, choose the positive $x$ direction along the travelling wave from its emission point toward the receiver. Define $u_s=d{x}_s/dt$ and $u_o=d{x}_o/dt$ as the signed source and receiver velocities along that same axis. Both use the same sign convention: positive $u_s$ moves the source toward the receiver ahead, whereas positive $u_o$ moves that receiver away from the incoming wave. Assume $|u_s|<c$ and $|u_o|<c$, constant velocities, and no source-receiver crossing during the interval considered. Then

To derive this, let $T_s=1/f_s$ be the emission wave period. During $T_s$, one wave crest travels $c{T}_s$ while the source advances $u_s{T}_s$. Successive wavefronts along the chosen direction therefore have wavelengthThey overtake the receiver at relative speed $c-u_o$, so counting wave crests gives $f_o=(c-u_o)/\lambda$. This separates the source's effect on wavelength from the receiver's effect on crossing rate. Even with a moving source, a stationary receiver obeys $f_o\lambda =c$ because the wave speed through the material remains $c$.

For a stationary receiver and a source approaching at speed $v$, set $u_o=0$ and $u_s=v$: $f_o=f_{s}c/(c-v)>f_s$. For recession choose the positive axis along the departing wave toward the receiver again, so the source has $u_s=-v$: $f_o=f_{s}c/(c+v)<f_s$. For the deliberately chosen values $c=343\mathrm{m}{\mathrm{s}}^{-1}$, $v=20\mathrm{m}{\mathrm{s}}^{-1}$ and $f_s=500\mathrm{Hz}$, the two predictions areThe corresponding wavelengths are $0.646\mathrm{m}$ and $0.726\mathrm{m}$, compared with $c/f_s=0.686\mathrm{m}$ for a stationary source. These are calculations for chosen parameters, not reported measurements.

A receiver approaching a stationary source at $20\mathrm{m}{\mathrm{s}}^{-1}$ instead has $u_s=0$, $u_o=-20\mathrm{m}{\mathrm{s}}^{-1}$, giving $f_o=500(363/343)\mathrm{Hz}\simeq 529.15\mathrm{Hz}$. Moving away gives $f_o=500(323/343)\mathrm{Hz}\simeq 470.85\mathrm{Hz}$. The equal closing speeds do not make these equal to the moving-source values. If both source and receiver move together with $u_s=u_o$, however, the numerator and denominator cancel and $f_o=f_s$. For speeds small compared with $c$, the first-order shift is $(f_o-f_s)/f_s\simeq (u_s-u_o)/c$; the difference between moving-source and moving-receiver experiments appears beyond this first order.