Doppler effect in a medium

Sound in air at 20 °C travels at about 343 m s⁻¹ because of the air's density and stiffness, and nothing a whistle does changes that. Once a crest has left the source it propagates at v relative to the air, indifferent to where the source went next. That single fact fixes the frame for everything below: v, the source speed vₛ and the observer speed vₒ must all be quoted relative to the medium at rest, never relative to each other, and not to the ground unless the air happens to be still. Set the problem on the straight line joining source and observer and adopt one sense throughout: vₛ is positive when the source moves toward the observer, vₒ is positive when the observer moves toward the source. Write βₛ = vₛ/v and βₒ = vₒ/v; both are dimensionless, and for road vehicles they are about 0.05 or smaller.

Let the source emit a crest every T = 1/fₛ. A crest leaves, and before the next one leaves it has travelled vT while the source has advanced vₛT toward the observer. The next crest therefore starts from a point vₛT nearer, and the gap between the two, ahead of the source, is λ = vT − vₛT = (v − vₛ)/fₛ. Behind the source the two motions add instead and the gap is (v + vₛ)/fₛ. The observer, at rest in the air, still meets crests travelling at v, so f_obs = v/λ = fₛ v/(v − vₛ). Numbers: a 520 Hz whistle in air at 340 m s⁻¹ lays down λ₀ = 0.6538 m at rest; move it at 32 m s⁻¹ and the wavelength ahead falls to 308/520 = 0.5923 m, so the arriving frequency is 340/0.5923 = 574 Hz. Notice what did not change — the speed of the sound, and the number of crests the source makes each second. Only their spacing in the air changed.

Now hold the source still and let the observer move. Nothing the observer does can alter the pattern already in the air: the wavelength is λ = v/fₛ everywhere, ahead and behind. What changes is the rate at which the observer crosses it. Moving toward the source at vₒ, the observer sees crests approach at v + vₒ, so f_obs = (v + vₒ)/λ = fₛ(1 + vₒ/v). The same 520 Hz source and the same 32 m s⁻¹ now give 520 × (1 + 32/340) = 569 Hz, five hertz below the moving-source answer. Two limiting checks confirm the form. At vₒ = 0 it returns fₛ. At vₒ = −v the observer retreats exactly as fast as the wave advances, never meets another crest, and receives f_obs = 0 — a finite, sensible answer, where the moving-source expression instead blows up. That contrast is the first clue that these are not one case in two disguises.

Put both on the same footing: let one or the other carry the same speed u, and write β = u/v. The moving observer gives f/fₛ = 1 + β exactly. The moving source gives f/fₛ = 1/(1 − β) = 1 + β + β² + β³ + …. The two share their constant and first-order terms and differ from β² onward, so f_source − f_observer = fₛ β²/(1 − β). At β = 0 both return fₛ, the correspondence any Doppler formula has to satisfy. At β = 0.0904, a car at 31.0 m s⁻¹ in still air, the gap is 0.90% — 7.2 Hz on an 800 Hz horn, trivial for a phone and near the limit of what an ear can separate. Recession mirrors it: a receding observer gives fₛ(1 − β), a receding source gives fₛ/(1 + β) = fₛ(1 − β + β² − …), and the receding source is always the higher of the two. Push β to 1 and the running observer hears silence while the receding source still delivers fₛ/2.

When both move, the two mechanisms act on different things — one on the wavelength in the medium, one on the crossing rate — so their factors simply multiply: f_obs = fₛ (v + vₒ)/(v − vₛ), with vₒ positive toward the source and vₛ positive toward the observer. The rest is bookkeeping, and four checks catch nearly every slip. Set vₛ = 0 and the moving-observer result must reappear; set vₒ = 0 and the moving-source result must. Any net approach has to raise the frequency and any net recession has to lower it, so if the arithmetic says otherwise a sign is wrong. The numerator may legitimately vanish, at vₒ = −v, but the denominator may not: vₛ → v is the edge of this model, not a prediction of infinite pitch. And a few per cent is what ordinary speeds buy — a factor of two means a speed comparable with v has crept into the sum.

Every line above assumes the medium is at rest, and air rarely is. Work in the air's frame and nothing changes; convert back to the ground and, for a uniform wind along the line of sight, there is one clean shortcut: replace v by v + w, where w is the wind component pointing from source to observer, and keep the source and observer speeds as measured over the ground, giving f_obs = fₛ (v + w + vₒ)/(v + w − vₛ). Test it where the answer is known: with both parties at rest the wind cancels top and bottom and f_obs = fₛ. A wind never shifts pitch on its own — downwind it stretches the wavelength and speeds the arrivals by exactly compensating factors. What it does change is the yardstick. A 960 Hz siren closing at a net 17.0 m s⁻¹ is heard at 1011 Hz with a 6.00 m s⁻¹ tailwind and 1013 Hz with that wind reversed, because a smaller effective speed makes both vₒ/v and vₛ/v larger. The shortcut covers a uniform wind only.