Beats and the Doppler effect

Two loudspeakers driven at f₁ and f₂ put y₁ = A cos(2πf₁t) and y₂ = A cos(2πf₂t) at the same point, and the medium is linear, so the ear receives their sum. The identity cos A + cos B = 2 cos((A−B)/2) cos((A+B)/2) rewrites that sum as y = 2A cos(2π(Δf/2)t) cos(2π f̄ t), with f̄ = (f₁ + f₂)/2 and Δf = f₁ − f₂. Read it as a fast carrier at f̄ multiplied by a slowly varying amplitude. Take 440.0 Hz and 443.0 Hz: the carrier is 441.5 Hz, which the ear hears as a single pitch, and the envelope oscillates at 1.5 Hz. But loudness tracks the square of the amplitude, and cos² runs at twice the frequency of cos, so the sound swells to full volume at both the positive and the negative peak of the envelope. The audible rate is therefore 2 × 1.5 Hz = 3.0 Hz, one loud moment every 0.333 s, and in general f_beat = |f₁ − f₂|.

Draw each tone as a phasor of length A turning at ω₁ = 2πf₁ and ω₂ = 2πf₂. Step into a frame turning at the mean rate and the two phasors counter-rotate at ±Δω/2, so their resultant has length 2A|cos(Δω t/2)| — the envelope again, now as geometry. The phase difference between the tones advances by 2π in a time 1/Δf: in phase means reinforcement, antiphase means cancellation, and the beat period is simply how long that drift takes. Two consequences follow. A spectrum analyser fed this signal shows exactly two lines, at 440.0 and 443.0 Hz, and nothing at 3 Hz — the beat belongs to the pair, it is not a frequency present in the air. And equal amplitudes are what make the nulls silent: with A₁ = 1.0 and A₂ = 0.6 the resultant swings between 1.6 and 0.4, a factor of 16 in intensity but never zero. Tuning uses the same picture — tighten the string, and if the beats slow you were flat.

A source at rest emits a crest every Tₛ = 1/fₛ, and each crest spreads at v, so crests sit λ₀ = v/fₛ apart in every direction. Let the source now move at vₛ toward you. Between emitting one crest and the next it travels vₛ Tₛ, so the gap ahead of it closes to λ₊ = v Tₛ − vₛ Tₛ = (v − vₛ)/fₛ, while behind it opens to λ₋ = (v + vₛ)/fₛ. This is a statement about the air, not about hearing: with fₛ = 500 Hz and v = 343 m s⁻¹ the rest wavelength is λ₀ = 0.686 m, and a source approaching at 30.0 m s⁻¹ leaves crests only 0.626 m apart ahead of it — a microphone anywhere in front measures that spacing while standing perfectly still. A stationary listener still meets crests at speed v, so f = v/λ₊ = fₛ v/(v − vₛ) = 343/0.626 = 547.9 Hz. Set vₛ = 0 and λ₊ returns to λ₀, as it must.

Now leave the source at rest, so the air carries crests λ₀ = 0.686 m apart in every direction, and walk into them at vₒ. Nothing about the wave has changed; what changes is how fast you meet it. Relative to you the crests approach at v + vₒ, so you cross one every λ₀/(v + vₒ) seconds and hear f = (v + vₒ)/λ₀ = fₛ (v + vₒ)/v. At vₒ = 30.0 m s⁻¹ this gives f = 500 × 373/343 = 543.7 Hz. Two features separate this from the moving-source case. It is strictly linear: every extra 1 m s⁻¹ of your own speed adds fₛ/v = 1.46 Hz, however fast you are already travelling. And its extreme case is finite rather than singular — at vₒ = −v you travel with the wave, no crest ever overtakes you, and f = 0. The moving-source formula has no such well-behaved limit.

Write x = u/v for a closing speed u. The moving-source form is f/fₛ = 1/(1 − x) = 1 + x + x² + …, while the moving-observer form is f/fₛ = 1 + x exactly. To first order they agree, which is why a carelessly chosen formula usually lands within a per cent, and the ratio of the two is exactly f(source moving)/f(observer moving) = 1/(1 − x²). At u = 30.0 m s⁻¹ in air, x = 0.0875 and the ratio is 1.0077: 547.9 Hz against 543.7 Hz, 4.2 Hz apart and easily counted as a beat. No change of frame turns one case into the other, because the air is in a different state of motion in each. The correspondence check is light: it needs no medium, so its shift can depend only on relative velocity, and the relativistic result f = fₛ √((1 + β)/(1 − β)) expands as 1 + β + β²/2 — the same first order, and a second order sitting exactly halfway between the two acoustic forms.

Four assumptions hold these formulae up. Only the component of velocity along the line of sight enters: at angle θ the source form becomes f = fₛ v/(v − vₛ cos θ), so a car passing you sweeps its pitch continuously and is momentarily unshifted at closest approach, where cos θ = 0. The air must be still, or the relations must be applied in the air's own frame — a wind of 9.0 m s⁻¹ blowing from listener to source opposes the source's motion, raising its speed relative to the air by 9.0 m s⁻¹, and leaves the ground-fixed listener drifting upwind through the air at 9.0 m s⁻¹, away from the source. What you hear left the source at a retarded time, so the velocity that matters is the one it had when it emitted, not the one it has now. And the source form demands vₛ < v: as vₛ → v the denominator vanishes, crests arrive on top of one another, amplitudes stop being small, and linear acoustics gives way to a shock cone of half-angle μ with sin μ = v/vₛ.