Doppler Effect Simulator

See how relative motion between source and observer changes the perceived frequency.

Starting values — Doppler Effect Simulator

Source frequency (Hz)	700
Source speed (m/s)	30
Observer speed (m/s)	0
Wave speed (m/s)	343

Change any control above and the results recompute from these.

Loading interactive simulation...

Approaching and receding are not mirror images 🖖

Nothing about the sound itself changes when the source moves. Every crest still travels at 343 m/s through the same air; what moves is the point each one was launched from, so they crowd together ahead of the source and spread out behind it. That is why the panel prints two formulas rather than one — f₀(v + v_obs)/(v − v_src) for the observer in front, f₀(v − v_obs)/(v + v_src) for the one behind — and why the two can never agree. It is bookkeeping on air, and it does not carry over to light.

Why the siren drops as it passes 🖖

A common surprise: an ambulance approaching at a steady speed does not rise in pitch — it holds one high, constant tone, then snaps to a lower constant tone the instant it passes you. The pitch changes only because the direction of motion relative to you flips from "toward" to "away". Set the source speed and press play: the front-observer readout stays fixed while approaching, then jumps down at the moment of passing.

Sound's Doppler betrays who is really moving 🖖

For sound, moving the source and moving the observer are not equivalent, even at an identical closing speed. A source approaching a still listener at 34.3 m/s raises 440 Hz to f·343/308.7 ≈ 489 Hz; a listener approaching a still source at the same 34.3 m/s hears only f·377.3/343 ≈ 484 Hz. The air is a preferred frame, so sound's Doppler shift secretly reveals who moves relative to the medium — something light's Doppler effect, with no medium, can never do.

THE DOPPLER EFFECT — WHO IS MOVING DECIDES WHICH FORMULA YOU NEED

Which Doppler Case Are You In?

Pitch changes when the distance between source and listener is changing — but the formula depends on which of them is doing the moving, because sound travels through air and air is a preferred frame. Answer three questions in order: is the source moving, is the observer moving, and is the source faster than the wave it is making? Light needs its own answer, since it has no medium at all.

Source moving, observer still — the classic siren f′ = f₀ v / (v − v_s)

Both moving — the two speeds are not interchangeable f′ = f₀ (v + v_o) / (v − v_s)

Source faster than sound — no Doppler, a shock cone M = v_s/v ≥ 1; sinθ = 1/M

Light — no medium, so only the relative speed counts f′ = f₀√((1 − β)/(1 + β))

Source moving, observer still — the classic siren

What you know: The source travels at vₛ through the air; you stand still. Only the denominator changes: the source chases its own wavefronts forward and stretches them behind.

Formula: f′ = f₀ v / (v − v_s)

Worked example: f₀ = 700 Hz, vₛ = 30 m/s, v = 343 m/s → approaching 700 × 343/313 = 767 Hz, receding 700 × 343/373 = 644 Hz

Open this case: Ambulance

Source moving, observer still — the classic siren. Wavefronts bunch up ahead of the moving source and spread out behind it. The source travels at vₛ through the air; you stand still. Only the denominator changes: the source chases its own wavefronts forward and stretches them behind. — Wavefronts bunch up ahead of the moving source and spread out behind it.

Both moving — the two speeds are not interchangeable

What you know: Source and observer are both moving through the air. Observer speed goes in the numerator, source speed in the denominator — and swapping them does not give the same answer.

Formula: f′ = f₀ (v + v_o) / (v − v_s)

Worked example: f₀ = 500 Hz, vₛ = 50 m/s, vₒ = 10 m/s, v = 343 m/s → approaching 500 × 353/293 = 602 Hz, receding 500 × 333/393 = 424 Hz

Open this case: Train

Both moving — the two speeds are not interchangeable. Source speed and observer speed enter the formula on opposite sides of the fraction. Source and observer are both moving through the air. Observer speed goes in the numerator, source speed in the denominator — and swapping them does not give the same answer. — Source speed and observer speed enter the formula on opposite sides of the fraction.

Source faster than sound — no Doppler, a shock cone

What you know: vₛ ≥ v, so the Mach number M = vₛ/v is at least 1. The denominator v − vₛ goes to zero and then negative: the approaching-frequency formula stops meaning anything.

Formula: M = v_s/v ≥ 1; sinθ = 1/M

Worked example: vₛ = 400 m/s, v = 343 m/s → M = 1.17, and the wavefronts pile into a cone of half-angle sin θ = 1/M → θ = 59°

Open this case: Supersonic

Source faster than sound — no Doppler, a shock cone. At M > 1 the wavefronts have an envelope: a cone with half-angle sin θ = 1/M. vₛ ≥ v, so the Mach number M = vₛ/v is at least 1. The denominator v − vₛ goes to zero and then negative: the approaching-frequency formula stops meaning anything. — At M > 1 the wavefronts have an envelope: a cone with half-angle sin θ = 1/M.

Light — no medium, so only the relative speed counts

What you know: The wave is light, not sound. There is no air to define a frame, so "who is moving" is not a meaningful question and the classical formula is only an approximation.

Formula: f′ = f₀√((1 − β)/(1 + β))

Worked example: f₀ = 600 THz receding at 5 × 10⁷ m/s (β = 0.167) → the classical formula gives 514 THz, the relativistic one 507 THz — a 1.4% difference already

Open this case: Redshift

Light — no medium, so only the relative speed counts. No medium: the shift depends on the relative speed alone, and needs the relativistic formula. The wave is light, not sound. There is no air to define a frame, so "who is moving" is not a meaningful question and the classical formula is only an approximation. — No medium: the shift depends on the relative speed alone, and needs the relativistic formula.

References (2)

The effect as first proposed: C. Doppler, "Über das farbige Licht der Doppelsterne und einiger anderer Gestirne des Himmels." Abhandlungen der königlich böhmischen Gesellschaft der Wissenschaften 2, 465–482, 1842.
Insight block 3 — why source motion and observer motion are not equivalent for sound: L. E. Kinsler, A. R. Frey, A. B. Coppens and J. V. Sanders, Fundamentals of Acoustics, 4th edition, ch. 5. Wiley, 2000. ISBN 978-0-471-84789-2 — the medium as a preferred frame, and the asymmetry it produces.

Problem solved in full

Frequency ahead and behind a 700 Hz siren moving at 30 m/s 5 steps
A 700 Hz siren moving at 30 m/s through air. Find the frequency ahead and behind, then decide whether it matters that the source is moving rather than you.
1. A source moving forward emits each crest closer to the last, so the wavelength ahead is shortened by the distance it travelled in one period. Divide the fixed wave speed by that shortened wavelength.
  
  $f_{front} = f_{0} \frac{v}{v - v _{src}} = 700 \cdot \frac{343}{313} = 767.1 Hz$
2. Behind, the same reasoning with the opposite sign. Note the denominator, not the numerator, is what changes — the wave speed is set by the air and is completely indifferent to the source.
  
  $f_{rear} = f_{0} \frac{v}{v + v _{src}} = 700 \cdot \frac{343}{373} = 643.7 Hz$
3. Compare the two shifts. They differ because 1/(1−x) and 1/(1+x) are not mirror images, and the gap is second order in the Mach number.
  
  $Δ f_{front} = + 67.1 Hz, Δ f_{rear} = - 56.3 Hz, \frac{1}{2} (767.1 + 643.7) = 705.4 \neq = 700$
4. Now move the observer instead and hold the source still. The speeds enter the numerator, the two shifts are exactly symmetric, and their average is the emitted frequency.
  
  $observer moving instead: f = f_{0} \frac{v \pm v _{obs}}{v} = 761.2 and 638.8, \frac{1}{2} (761.2 + 638.8) = 700$
5. The Mach number is what governs how far apart those two descriptions drift. At 0.087 they agree to within a percent; the disagreement is quadratic, so it is invisible here and total near Mach 1.
  
  $M = \frac{v _{src}}{v} = \frac{30}{343} = 0.087; f_{front} \to 2 f_{0} at v_{src} = \frac{v}{2} = 171.5 m/s$
Answer

767.1 Hz and 643.7 Hz — and yes, it matters. The two shifts are not equal: +67.1 Hz ahead against −56.3 Hz behind, averaging 705.4 rather than 700. Run the same 30 m/s as observer motion instead and you get 761.2 and 638.8, which average to exactly 700. Same relative speed, different answer, because sound has a medium and the air knows which of you is moving through it. This is the point where acoustics and relativity part company: light has no medium, only the relative velocity survives, and the relativistic Doppler formula is symmetric by construction. The asymmetry here runs away with speed — at Mach 0.5 the forward frequency doubles, and at Mach 1 the denominator hits zero, which is the shock front rather than an infinite pitch.

$767.1 Hz and 643.7 Hz; a moving source is not a moving observer$

By the 42Math team · Updated 2026-07-27

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Physics lessons →

Example problems

Ambulance - Ambulance siren at 30 m/s
Train - Train horn with moving observer
Supersonic - Supersonic: Mach > 1, shock cone
Redshift - Cosmological redshift analogy