The Doppler Effect in Sound

The pitch-drop of a passing siren is the everyday face of it, but the Doppler effect quietly runs whole technologies. Medical Doppler ultrasound bounces high-frequency sound off flowing blood; the shift in the reflected frequency reveals the speed and direction of the flow, letting a cardiologist see a leaking valve without a scalpel — and because it involves a moving reflector, it is a double Doppler shift, the hard archetype you will drill next. Weather radar and police speed guns use the same idea with radio waves and light, obeying the relativistic cousin of this formula. Bats and dolphins echolocate with ultrasonic chirps and read the Doppler shift of the echo to judge how fast their prey is closing. And on the grandest scale, the redshift of galaxies — light, not sound, stretched by cosmic recession — is how we know the universe expands. Every one of these is the same core insight from a train platform: motion between a wave's maker, its medium, and its reader changes the frequency you measure.

Fix three things in your mind before any algebra, because getting them straight is 90% of the marks. First, sound is a disturbance of a medium — the air — and it travels at a fixed speed $v$ relative to that air, whatever the source is doing. A jet can fly at twice the speed of sound; the sound it makes still crawls outward through the air at the same $340\ \mathrm{m/s}$. Second, frequency is a rate of arrival: how many wave crests pass your ear each second. Third, the pitch you hear is that arrival rate. So anything that changes how often crests reach you changes the pitch — and there are two independent ways to do that.

Way one: move the source. A stationary source sits at one point and pumps out spherical wavefronts, one every period $T = 1/f$, each an expanding shell centred on that point — a set of neat concentric circles. Now let the source drift while it emits. Each new shell is launched from a point a little further along the source's path than the last. Ahead of the source the shells pile up, their spacing — the wavelength — squeezed short; behind, they thin out, the wavelength stretched long. The source has physically altered the wavelength of the sound in the air. A listener ahead meets short waves at the normal sound speed and reads a higher frequency; a listener behind meets long waves and reads lower.

Way two: move the listener. Leave the source still, so the wavefronts in the air are perfect concentric circles of the ordinary wavelength $\lambda = v/f$. Now walk toward the source. You crash into the crests sooner than if you stood still, because you and the next crest are closing on each other at $v + v_o$ instead of $v$. The wavelength in the air is untouched; what changed is your closing speed with it. You collect crests faster, so you hear a higher frequency. Walk away and the crests have to chase you: closing speed $v - v_o$, lower frequency. These two mechanisms — reshaping the wave versus running into an unchanged wave — are genuinely different, and that difference is not a technicality. It is the reason sound's Doppler effect is asymmetric, the single most examined and most misunderstood point on this topic, and we will make it quantitative below.

Put the two mechanisms together and you get one equation that covers every case:

where: $f$ — the frequency actually emitted by the source (its true, at-rest note); $f'$ — the frequency the observer hears; $v$ — the speed of sound in the medium; $v_o$ — the observer's speed through the medium; $v_s$ — the source's speed through the medium. The sign rule is the whole game, and it is chosen so that approach always raises the pitch:

• Numerator, the observer term. Use $(v + v_o)$ when the observer moves toward the source, $(v - v_o)$ when away. A moving observer scales the closing speed, so it multiplies the top.
• Denominator, the source term. Use $(v - v_s)$ when the source moves toward the observer, $(v + v_s)$ when away. A moving source shrinks the wavelength, and a smaller wavelength sits in the bottom.

There is a foolproof memory hook that never lies: toward is up, away is down. Whichever combination of signs makes $f' > f$ for a net approach is the correct one — and if your signs give a lower pitch for an approaching object, you have made a mistake, full stop. We build a whole technique around this check in a later article; for now, just know that the physics guarantees it. One warning that separates the students who lose marks from those who don't: $v$ is the speed of sound in the air, measured in the air's frame — it is not the relative speed of source and observer. Sound is not like two cars closing on a road, where only the closing speed matters. The air is a referee. This is exactly why the two motions do not cancel and why the effect is asymmetric.

The four elementary cases

Everything else is a combination of these four, computed with $v = 340\ \mathrm{m/s}$ and a source of $f = 1000\ \mathrm{Hz}$:

Look hard at (2) and (4). Same medium, same $1000\ \mathrm{Hz}$ note, the same $34\ \mathrm{m/s}$ of approach — and the answers differ: $1111.11\ \mathrm{Hz}$ when the source moves, only $1100.00\ \mathrm{Hz}$ when the observer moves. This is not rounding; it is real, measurable physics.

Why is the source shift bigger? Because the source-motion shrinks the wavelength, and $f' = v/\lambda'$ has that shrunken $\lambda'$ in the denominator, where a small change is amplified. The observer-motion only adds to the closing speed linearly. Algebraically, source-toward gives a factor $\dfrac{v}{v - v_s} = \dfrac{1}{1 - v_s/v}$, which grows without bound as $v_s \to v$; observer-toward gives $\dfrac{v + v_o}{v} = 1 + v_o/v$, which only grows in proportion. For small speeds the two nearly agree — both are about $1 + v_s/v$ — but the source term always wins, and it runs away completely near the speed of sound.

Contrast this with light. Light needs no medium; there is no "air" for a light wave to ride, and no frame in which the wave speed is special — it is $c$ for everyone. So a light source and a light observer cannot be distinguished by any experiment: only their relative velocity can matter, and the relativistic Doppler formula depends on that relative velocity alone. Sound's asymmetry is therefore a direct fingerprint of the medium. When you see a problem that asks about a moving source and a moving observer and quietly expects the same answer for each, it is testing whether you know that sound is not light. (The light formula is its own separate article; do not import equation (1) into a relativity problem.)

Now the piece the syllabus demands you can produce yourself — the derivation, done with algebra and a clear picture, no calculus.

The second row is worth dwelling on. As a source approaches the speed of sound, equation (1) blows up because $\lambda'$ collapses to nothing: every crest is emitted essentially on top of the last, and they superpose into a single high-pressure shock front — the sonic boom. The word Mach attached to supersonic speeds comes from exactly this limit, and from the physicist whose portrait sits below.

Board level — a train horn approaching, then receding (CBSE/ICSE pattern)

A train sounds a horn of frequency $400\ \mathrm{Hz}$ and moves at $20\ \mathrm{m/s}$ along a straight track. A person stands beside the track. Take the speed of sound as $340\ \mathrm{m/s}$. Find the frequency heard (a) as the train approaches and (b) after it has passed and is receding. (c) What is the sudden drop in pitch the person hears at the instant of passing?

Step 1. Identify who moves. The source moves; the observer is stationary ($v_o = 0$). So only the denominator changes, and $v = 340$, $v_s = 20$, $f = 400$.

Step 2. Approaching, the source moves toward the observer, so take $(v - v_s)$ in the denominator: $f' = 400\times\dfrac{340}{340 - 20} = 400\times\dfrac{340}{320} = \boxed{425.00\ \mathrm{Hz}}$. Higher than $400$ — correct for approach.

Step 3. Receding, the source moves away, so take $(v + v_s)$: $f' = 400\times\dfrac{340}{340 + 20} = 400\times\dfrac{340}{360} = \boxed{377.78\ \mathrm{Hz}}$. Lower than $400$ — correct for recession.

Step 4. The drop the ear hears as the train passes is the difference: $425.00 - 377.78 = \boxed{47.22\ \mathrm{Hz}}$. That step of about a semitone-and-a-half is exactly the "eeeeee-ooooo" you hear at a level crossing. Notice the note does not pass smoothly through $400\ \mathrm{Hz}$ — it jumps from $425$ straight to $378$ as the geometry flips from approach to recession.

Competitive level — both moving, with a wind (JEE-Main / NEET pattern)

A car sounding a $1000\ \mathrm{Hz}$ horn drives toward a cyclist at $34\ \mathrm{m/s}$; the cyclist rides toward the car at $17\ \mathrm{m/s}$. A steady wind of $10\ \mathrm{m/s}$ blows from the car toward the cyclist. Take the still-air speed of sound as $340\ \mathrm{m/s}$. What frequency does the cyclist hear? Compare with the no-wind case.

Step 1. Handle the wind first. A wind blowing from source to observer is carried by the whole medium, so it adds to the speed of sound along that path: the effective sound speed becomes $v_{\text{eff}} = v + w = 340 + 10 = 350\ \mathrm{m/s}$. This single substitution is the entire wind trick — replace $v$ by $v_{\text{eff}}$ everywhere, because all velocities are now measured against the ground and the sound moves at $350$ over the ground.

Step 2. Assign signs. Both approach, so numerator $(v_{\text{eff}} + v_o)$ and denominator $(v_{\text{eff}} - v_s)$: $f' = 1000\times\dfrac{350 + 17}{350 - 34} = 1000\times\dfrac{367}{316}$.

Step 3. Evaluate: $f' = 1000\times\dfrac{367}{316} = \boxed{1161.39\ \mathrm{Hz}}$.

Step 4. Compare. With no wind, $v_{\text{eff}} = 340$ and $f' = 1000\times\dfrac{357}{306} = 1166.67\ \mathrm{Hz}$. The tailwind actually lowers the observed shift slightly, from $1166.67$ to $1161.39\ \mathrm{Hz}$, because adding the same $10\ \mathrm{m/s}$ to both the top and the bottom of the fraction dilutes the ratio toward $1$. A common exam distractor adds the wind only to one side — don't. Timing note: about two minutes. The wind step is the whole difficulty; once you have replaced $v$ by $v + w$, it is a routine sign assignment. If you find yourself adding the wind to the source but not the observer, stop — the medium moves as one.

Using relative speed instead of the medium-frame $v$. The number-one killer. For sound you may never collapse source-motion and observer-motion into a single closing speed; identify the medium, then ask about each mover separately. The asymmetry ($1111.11$ vs $1100.00$) is the whole reason.

Putting the observer term in the denominator (or the source term in the numerator). The observer's speed scales the closing speed and belongs on top; the source's speed changes the wavelength and belongs on the bottom. Swap them and an examiner's distractor is waiting. The physical picture protects you: reshaping a wavelength (source) is not the same act as running into a wave (observer).

Losing the sign. Sanity-check every answer against "toward is up." An approaching object that comes out lower in pitch means a sign is flipped. This one check catches most slips and is the seed of the sign-discipline technique in a later article.

Speeds in km/h. A car doing $72\ \mathrm{km/h}$ is $20\ \mathrm{m/s}$; a horn's $v_s$ must be in $\mathrm{m/s}$ to sit beside $v = 340\ \mathrm{m/s}$. Mixing units silently wrecks the ratio.

Using this formula for light. Radar guns, police speed cameras and galaxy redshifts are light (electromagnetic) Doppler and obey the relativistic formula that depends only on relative velocity — a different equation in a different article. Equation (1) is for waves in a medium: sound, ripples on water, ultrasound.

Board scope — say it plainly. IB (both SL and HL) keeps the wave model and the Doppler effect, so this page is on-syllabus for IB. IGCSE 0625 does not teach the Doppler effect at all — if you are an IGCSE candidate this is enrichment, not examinable, though the ultrasound and echo pages are yours. AP Physics 2 treats Doppler qualitatively — you must know that approach raises pitch and recession lowers it, and why, but AP will not ask you to grind equation (1) for a number. CBSE, ICSE, JEE-Main, JEE-Advanced and NEET want the full quantitative formula, the derivation, and the asymmetry.

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), §15.8 on the Doppler effect · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves", the Doppler section · Lord Rayleigh, The Theory of Sound, Vol. 2, for the classical wave treatment · Cambridge A-Level 9702 and the IB Physics guide (wave phenomena) for the reduced-algebra board versions.

Provenance. Standard results. A verifier should recompute: the four elementary cases $1000\times340/306 = 1111.11$, $1000\times340/374 = 909.09$, $1000\times374/340 = 1100.00$, $1000\times306/340 = 900.00$; the both-toward case $1000\times357/306 = 1166.67$; the train example $400\times340/320 = 425.00$, $400\times340/360 = 377.78$, drop $47.22$; the wind example $1000\times367/316 = 1161.39$ against the no-wind $1166.67$; and the temperature note that sound rises from about $332\ \mathrm{m/s}$ at $0\ \mathrm{^\circ C}$ to $343\ \mathrm{m/s}$ at $20\ \mathrm{^\circ C}$.

Prerequisites. The Speed of Sound — where the $v$ in every formula on this page comes from, and why it depends on the medium and its temperature · Characteristics of Sound — pitch, frequency and how the ear reads them. Leads to. Doppler-Effect Numericals — the problem gym that drills every variant including the reflected-wall double shift · Doppler Sign Discipline — the one procedure that kills every sign error · Ultrasound and Its Applications — where a moving reflector turns this effect into a diagnostic tool.

Go deeper. The relativistic Doppler effect for light — where there is no medium and only relative velocity survives — is the natural sequel once you meet special relativity; it is a different formula for a genuinely different physical situation, and confusing the two is the classic trap this page has been guarding against.