The Doppler Effect for Sound Waves

The Doppler effect is the change in the observed frequency of a wave when the source and the observer move relative to each other: the frequency is higher when they approach and lower when they recede. In brief: for sound with a moving source and a stationary observer, f’ = f × v ÷ (v ∓ v_s), using the minus sign when the source approaches; for a moving observer and stationary source, f’ = f × (v ± v_o) ÷ v, with plus when the observer approaches. The pitch of an ambulance siren drops as it passes.

The change in a siren’s pitch as an ambulance passes is the best-known example, but the effect is also used in radar speed guns, medical blood-flow measurements and astronomy. This guide explains why the effect occurs, gives the formulas for sound, compares the cases in a table, works through calculations and corrects common mistakes.

Why does the frequency change?

A source emits waves at a steady frequency f. If the source moves toward the observer, each successive wave is emitted from a point closer to the observer. The wavefronts are bunched together in front of the source (shorter wavelength) and spread out behind it (longer wavelength). Since the wave speed is set by the medium, a shorter wavelength reaches the observer at a higher frequency (v = fλ). When the observer moves toward a stationary source, the observer meets the wavefronts more frequently, so the frequency is higher. Only the relative motion along the line between source and observer matters. The pitch the source emits does not actually change; the observer receives a different frequency.

What are the formulas?

Let v be the speed of sound, v_s the speed of the source, and v_o the speed of the observer, all relative to the air.

• Moving source, stationary observer: f’ = f × v ÷ (v − v_s) if the source approaches, and f’ = f × v ÷ (v + v_s) if it recedes.
• Moving observer, stationary source: f’ = f × (v + v_o) ÷ v if the observer approaches, and f’ = f × (v − v_o) ÷ v if the observer recedes.
• Both moving: f’ = f × (v ± v_o) ÷ (v ∓ v_s), taking the upper signs for motion toward each other.

If the source moves faster than sound, v_s > v, the waves pile up into a shock wave (a sonic boom), and the formula fails.

Doppler cases compared

Worked example: a passing siren

A siren emits a 440 Hz tone. The speed of sound is 343 m/s. A stationary listener hears an ambulance moving at 30 m/s.

1. Approaching: f’ = 440 × 343 ÷ (343 − 30) = 440 × 343 ÷ 313 = 440 × 1.0958 = 482 Hz.
2. Receding: f’ = 440 × 343 ÷ (343 + 30) = 440 × 343 ÷ 373 = 440 × 0.9196 = 405 Hz.
3. Change in pitch: the frequency drops from 482 Hz to 405 Hz as the ambulance passes, a difference of 77 Hz, which is noticeable (about 3 semitones).
4. Moving observer: a cyclist travels at 10 m/s toward a stationary 440 Hz source. f’ = 440 × (343 + 10) ÷ 343 = 440 × 1.0292 = 453 Hz.
5. Speed from the shift: if a source of 1,000 Hz is heard at 1,060 Hz approaching, then 1,060 = 1,000 × 343 ÷ (343 − v_s), so 343 − v_s = 343 × 1,000 ÷ 1,060 = 323.6, and v_s = 19.4 m/s.

Who discovered the effect?

Christian Doppler predicted the effect for light and sound in 1842. In 1845, Christophorus Buys Ballot tested it for sound by having trumpeters play a steady note on a moving train, while musicians with perfect pitch on the platform noted the change. Hippolyte Fizeau discovered the equivalent effect for light in 1848, and Edwin Hubble’s measurements of redshifts of galaxies in 1929 led to the discovery that the universe is expanding.

Where is it used?

Police radar guns measure the speed of vehicles from the shift in frequency of the reflected radio waves. Doppler ultrasound measures the speed of blood flow in arteries and the heart, and weather radar detects the motion of rain and wind in storms. Astronomers use it to find the speeds of stars and galaxies, and to detect exoplanets from the small periodic shifts in the frequency of the light of their stars.

Which mistakes are common?

Mistake 1: Thinking that the source changes its frequency

The source emits the same frequency. The observer receives a different one because of relative motion.

Mistake 2: Using the wrong sign

Approaching raises the frequency (the denominator v − v_s becomes smaller). Receding lowers it.

Mistake 3: Believing that the shift occurs only at the moment of passing

An observer on the line of motion hears a steady higher frequency while it approaches and a steady lower one after it passes, and the transition appears rapid at the closest point. For a source not on the line, only the component of velocity along the line of sight matters.

Mistake 4: Applying the sound formulas to light

Light needs a relativistic formula, and does not depend on a medium.