The Doppler Effect, With Both the Source and the Observer Moving

Q: What sign convention does this page use?

Both speeds are positive when they CLOSE the gap: vo positive when the observer moves towards the source, vs positive when the source moves towards the observer, giving f′ = f(v + vo)/(v − vs). Several texts instead measure every velocity positive in one fixed direction, usually from observer to source, and then write f′ = f(v + vo)/(v + vs). That is the same physics with the source term's sign flipped. Check your book before substituting; this is the commonest source of lost marks on the topic.

Move the source at 30 m/s and you get one answer. Move the observer at 30 m/s instead and you get a different one. The formula is not symmetric and this page prints both so you can see by how much — along with the whole drop as a horn passes you, the wavelength in front and behind, and what happens as the source approaches the speed of sound.

Last updated 11 Nov 2026

Observed frequency, with the asymmetry shown

Both speeds, and what each one is worth

Frequency emitted by the sourceIn hertz, as measured by somebody riding with the source. This never changes: the siren does not know you are moving. Everything on this page is about what arrives, not about what was sent.

Speed of the wave in the mediumIn metres per second, and measured relative to the MEDIUM rather than to anybody. 343.24 m/s is dry air at 20 °C, derived from √(γRT/M) rather than quoted; at 0 °C it is 331.32. This is the number that makes the acoustic Doppler effect different from the relativistic one for light: sound has a medium that defines a frame, so the source’s speed and the observer’s speed enter the formula DIFFERENTLY. For light there is no medium, only the relative velocity, and the formula on this page does not apply.

Observer’s speed towards the sourcePositive when the observer moves TOWARDS the source, negative when moving away. The convention matters more than the number: this page implements f′ = f(v + v_o)/(v − v_s) with both speeds positive when they close the gap. Some texts write the same physics as f′ = f(v + v_o)/(v + v_s) with every velocity measured positive in one fixed direction, which flips the sign of the source term. Check which form your book uses before you substitute.

Source’s speed towards the observerPositive when the source moves TOWARDS the observer. 30 m/s is about 108 km/h, a car on a motorway. Note the asymmetry the formula builds in: this speed sits in the DENOMINATOR, so it has a stronger effect than the same speed given to the observer, and the effect runs away as the source approaches the wave speed — at exactly the wave speed the denominator is zero and the page refuses to answer, which is the correct response rather than a failure.

547.8872HzExample

A 500 Hz horn on a car doing 30 m/s towards a stationary listener, air at 20 °C

Two speeds, two different mechanisms, and a formula that is not symmetric

f′ = f · (v + v_o) / (v − v_s) · both positive when closing · λ_ahead = (v − v_s)/f · λ_behind = (v + v_s)/f

v_o, in the numerator: the observer’s speed towards the source. Changes the rate at which waves are met, not the waves themselves. Walk into a line of waves and you cross them faster; the spacing in the air is untouched. The effect is therefore linear and bounded: at v_o = v you meet them twice as fast and the frequency doubles, and that is all it ever does
v_s, in the denominator: the source’s speed towards the observer. Changes the waves. Each crest is emitted from further forward than the last, so the wavelength in the air really is shortened to (v − v_s)/f. That is why it sits in the denominator and why it runs away: it has no upper bound, and as v_s approaches v the spacing goes to zero
the asymmetry: the single most useful thing on this page. 30 m/s given to the observer turns 500 Hz into 543.71 Hz. The same 30 m/s given to the source turns it into 547.89 Hz. Same closing speed, different answer, because the medium defines a frame and the two motions do physically different things. A relative-velocity-only formula would give one answer and would be wrong for sound
v_s = v: the denominator is zero and there is no observed frequency. Every wave the source has ever emitted arrives at the same instant, piled into one front. That is the sonic boom, and it is a statement about the formula having reached the edge of its domain rather than about an infinite pitch. Beyond it the waves trail behind the source in a cone and a different description is needed
light: this formula does not apply to light. There is no medium, so there is no way for the source’s motion and the observer’s motion to differ, and the relativistic result depends only on their relative velocity — with a time-dilation factor √(1 − β²) that has no acoustic counterpart. For slow speeds the two agree to first order, which is why the acoustic picture is a useful intuition and a bad formula

Worked example

A 500 Hz horn on a car doing 30 m/s towards a stationary listener, air at 20 °C
THE SUBSTITUTION. f′ = f(v + v_o)/(v − v_s) = 500 × (343.24 + 0)/(343.24 − 30) = 500 × 343.24/313.24 = 547.89 Hz. A rise of 47.89 Hz, or 9.58 per cent — about a semitone and a half, which is why a passing horn is so obviously out of tune with itself.
WHAT ACTUALLY CHANGED IN THE AIR. The wavelength ahead of the car is (v − v_s)/f = 313.24/500 = 0.6265 m, against 0.6865 m if the car were still. Behind it the wavelength is (v + v_s)/f = 373.24/500 = 0.7465 m. The waves in front are genuinely closer together: this is a change to the sound in the air, not an illusion of the listener, and anyone standing anywhere ahead of the car measures it.
NOW GIVE THE SPEED TO THE LISTENER INSTEAD. A stationary horn with the listener running at 30 m/s towards it: f′ = 500 × (343.24 + 30)/343.24 = 543.71 Hz. The same 30 m/s of closing speed gives 4.18 Hz less than the moving source did. The mechanisms are different: the runner meets unchanged waves faster, the car lays down shortened waves. This is the point of the page — a formula written in terms of relative velocity alone cannot produce both numbers.
THE WHOLE PASS, which is what you actually hear. Approaching, 547.89 Hz. Receding, f′ = 500 × 343.24/373.24 = 459.81 Hz. The total drop as the car goes by is 88.08 Hz, from 9.58 per cent sharp to 8.04 per cent flat. The drop is not symmetric about 500 Hz either, and the asymmetry is the denominator again. Note also that the change is not instantaneous in practice: what matters is the component of velocity along the line of sight, so the glide happens over the moment of closest approach rather than at a point.
AND THE LIMIT. At 30 m/s the car is at Mach 0.0874 and the effect is a tenth of the emitted frequency. At Mach 0.5 the approaching frequency is double. At Mach 0.9 it is ten times. At Mach 1 the denominator is zero, every wave arrives together, and this page refuses to print a number — correctly, because there is no observed frequency to report, only a shock front.

Who is moving, at 30 m/s, with a 500 Hz source in air at 20 °C

The first two rows are the whole argument of this page: the same 30 m/s of closing speed, 4.18 Hz apart. A relative-velocity formula would have to give them the same answer, and it would be wrong, because the medium distinguishes them — the moving source changes the wavelength in the air and the moving observer does not. Notice too that the moving observer’s shift is exactly symmetric, +43.71 and −43.71, while the moving source’s is not: +47.89 approaching against −40.19 receding. The observer’s term is linear in the numerator and the source’s is in the denominator, and only linear terms are symmetric. The last two rows are what a listener in one car hears from the horn of another; combining the two effects multiplies the factors rather than adding the shifts, which is why the combined 19.15 per cent is slightly more than 9.58 plus 8.74 minus their product.
Who moves	Towards or away	Observed frequency, Hz	Shift, Hz	Shift, %
Source	towards	547.89	+47.89	+9.58
Observer	towards	543.71	+43.71	+8.74
Both, each at 30	towards	595.77	+95.77	+19.15
Nobody	—	500.00	0.00	0.00
Observer	away	456.30	−43.70	−8.74
Source	away	459.81	−40.19	−8.04
Both, each at 30	away	419.62	−80.38	−16.08

Why the source’s 30 m/s and the observer’s 30 m/s are not the same 30 m/s

The acoustic Doppler formula is not symmetric, and that is not an accident of the algebra. f′ = f(v + v_o)/(v − v_s) puts the observer’s speed in the numerator and the source’s in the denominator because the two motions do physically different things. A moving observer runs into waves that are already in the air and encounters them faster; the spacing between them is untouched. A moving source emits each crest from a point further forward than the last, so the spacing in the air genuinely shrinks. One changes the measurement, the other changes the thing being measured.

The consequence is a number you can check. With a 500 Hz source in air at 20 °C, 30 m/s of closing speed given to the observer yields 543.71 Hz; the same 30 m/s given to the source yields 547.89 Hz. Four hertz of difference from the same relative motion. It matters because the intuitive picture — “what counts is whether they are getting closer” — is exactly the wrong intuition for sound. The medium is a frame, the two speeds are measured against it, and they are not interchangeable. For light there IS no medium, the relativistic formula depends on the relative velocity alone, and the symmetry everyone expects is restored — which is a much deeper statement than it looks and is worth noticing where it fails.

The second thing the asymmetry buys you is a runaway. The observer’s term is linear and bounded: at v_o = v the frequency doubles and that is the end of it. The source’s term has no bound at all. At Mach 0.5 the approaching frequency is twice the emitted one; at Mach 0.9, ten times; at Mach 1 the denominator is zero. That last case is not an infinite pitch, it is the collapse of the model: every wave the source has emitted arrives at the same instant, which is the pressure front of a sonic boom, and this page refuses to print a frequency because there is no frequency to print. Beyond the wave speed the algebra returns a negative number, and a negative frequency is a sign that you have left the domain rather than a result.

What you actually hear as a horn passes is the whole glide, not either endpoint. Approaching at 30 m/s, 547.89 Hz; receding, 459.81 Hz; a drop of 88.08 Hz, which is well over a tone. The page prints that total drop as a row because it is the measurable thing: you do not get to hear the emitted 500 Hz at all unless the car stops. One refinement worth knowing for a real pass: only the component of velocity ALONG THE LINE OF SIGHT counts, so a car that passes ten metres away from you is at full approach speed when far off, at zero radial speed exactly abeam, and at full recession afterwards. The transition is a smooth glide over a second or two rather than a jump, and the closer the pass the faster the glide.

Where this goes wrong, in order. Signs: more marks are lost to the sign convention here than to anything else, which is why both fields on this page are labelled “towards” and why the hint under each says what the alternative convention looks like. Adding the two shifts instead of multiplying the two factors: the effects compound, so both parties closing at 30 m/s gives 19.15 per cent rather than a plain sum. Using a relative speed for sound, which collapses the asymmetry the page exists to show. And using this formula for light, where it is wrong at second order in v/c and conceptually wrong at first order.

Frequently asked questions

Why does it matter whether the source or the observer is moving?

Because sound travels in a medium, and both speeds are measured against that medium rather than against each other. A moving source changes the wavelength that is physically present in the air — each crest leaves from further forward than the last — so its speed enters the denominator. A moving observer leaves the air alone and simply meets unchanged waves at a different rate, so its speed enters the numerator. With a 500 Hz source in air, 30 m/s of closing speed gives 547.89 Hz if the source moves and 543.71 Hz if the observer does.

What sign convention does this page use?

Both speeds are positive when they CLOSE the gap: v_o positive when the observer moves towards the source, v_s positive when the source moves towards the observer, giving f′ = f(v + v_o)/(v − v_s). Several texts instead measure every velocity positive in one fixed direction, usually from observer to source, and then write f′ = f(v + v_o)/(v + v_s). That is the same physics with the source term’s sign flipped. Check your book before substituting; this is the commonest source of lost marks on the topic.

What happens when the source reaches the speed of sound?

The denominator becomes zero and there is no observed frequency, so this page refuses to answer rather than printing infinity. Physically, every wave the source has emitted arrives at the listener at the same instant, piled into a single pressure front — the leading edge of a sonic boom. Above the wave speed the waves trail behind the source inside a cone and the formula returns a negative frequency, which signals that the model no longer applies rather than describing anything.

Can the observer move so fast that they hear nothing?

Yes, for sound: an observer receding at or above the wave speed is outrunning the waves, (v + v_o) goes to zero or below, and nothing arrives. This page refuses rather than printing a zero or negative frequency. It cannot happen for light, because nothing outruns light — which is one of the sharpest places where the acoustic and relativistic Doppler effects are not merely different formulae but different situations.

Does this formula work for light and for redshift?

No. It depends on a medium, and light has none, so there is no way for a moving source and a moving observer to give different answers — the relativistic result depends only on the relative velocity, and it carries a time-dilation factor of √(1 − v²/c²) with no acoustic counterpart. For speeds far below c the two agree to first order in v/c, so the acoustic picture is a serviceable intuition and a wrong calculation. Do not use this page for astronomical redshifts.

Why is the rise as something approaches bigger than the fall as it recedes?

Because the source’s speed is in the denominator. At 30 m/s with a 500 Hz source, approaching gives +47.89 Hz and receding gives −40.19 Hz — not equal and opposite. Dividing by a smaller denominator moves the answer further than dividing by a larger one moves it back. A MOVING OBSERVER, whose speed is in the numerator, does give an exactly symmetric pair: +43.71 and −43.71. That contrast is a neat test of whether you have the formula the right way round.

Related calculators

References

NCERT, Physics Textbook for Class XI, Part II, Chapter 14, Waves. The chapter number was checked against NCERT’s own answers appendix for Part II (keph2an.pdf, read 11 November 2026), which runs Chapters 8 to 14 and puts Oscillations at 13 and Waves at 14; older printings numbered Waves as 15 and much of the web still says 15. Cited for the syllabus treatment of wave speed, standing waves, beats and the Doppler effect. Citing the chapter is not a claim that any board has approved this page.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre is defined by fixing the speed of light in vacuum at c = 299 792 458 m/s exactly, so c carries no uncertainty at all: it is a definition, not a measurement, and every refractive index on these pages divides that exact number. The same brochure fixes the Boltzmann and Avogadro constants exactly, which makes the molar gas constant R = 8.314 462 618 153 24 J mol⁻¹ K⁻¹ exact as well — and that is what the speed-of-sound figures here are derived from rather than copied out of a table.
A convention this batch could not source and therefore does not attribute. Two forms of the acoustic Doppler equation are in wide circulation: the one implemented here, f′ = f(v + v_o)/(v − v_s) with both speeds positive when closing, and f′ = f(v + v_o)/(v + v_s) with every velocity measured positive in one fixed direction. They are the same physics with the source term’s sign flipped. Which form a given board’s textbook prints was not established for this batch: the NCERT Class XI waves chapter could not be read directly, because ncert.nic.in refused the chapter file to an automated request, and every secondary source found for it was a coaching site or an answer mill of the kind this project’s brief rules out. So the page states its own convention explicitly under each input, prints both forms side by side, and makes no claim about which book uses which.
The Mach-number figures quoted on this page (Mach 0.5 doubles the approaching frequency, Mach 0.9 multiplies it by ten) are arithmetic on the formula above and not measurements: 1/(1 − 0.5) = 2 and 1/(1 − 0.9) = 10. They are included because the runaway in the denominator is the feature of the acoustic Doppler effect that a worked example at 30 m/s cannot show — at 30 m/s everything looks comfortably linear.