1 higher-level section hidden.
Educerie · IB Diploma · Physics
Theme C Wave behaviour · C.5 Doppler effect
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe the Doppler effect for sound and for electromagnetic waves | SL, HL | "Outline what is meant by the Doppler effect" (2 marks) |
| Explain the effect with wavefront diagrams for a moving source and for a moving observer | SL, HL | "Using a diagram, explain why the frequency heard is higher" (3 marks) |
| Use Δf/f = Δλ/λ ≈ v/c for light when v ≪ c | SL, HL | Calculate the speed of a galaxy from a shifted line (2–3 marks) |
| Explain how shifted spectral lines reveal the motion of stars and galaxies | SL, HL | Redshift or blueshift; a rotating star; an extended body (2–4 marks) |
| Describe Doppler radar and Doppler ultrasound as applications | SL, HL | "Outline how the speed of blood flow is measured" (2–3 marks) |
| Use f′ = f v/(v ± us) for a moving source of sound | HL only | Calculate the frequency heard as a siren approaches and recedes (2–3 marks) |
| Use f′ = f (v ± uo)/v for a moving observer | HL only | Calculate the frequency heard by a moving cyclist (2 marks) |
| Determine the speed of a source or observer from observed frequencies | HL only | From the two pitches of a passing car, find its speed (3–4 marks) |
Before you start
You need v = fλ and the idea of wavefronts from C.2 and C.3, and one fact you must hold on to all the way through: the speed of a wave is set by the medium, not by the source. A siren on a speeding ambulance does not throw its sound forward faster. For light you also need spectral lines, each at a fixed wavelength, which E.1 develops in full. Have the data booklet open at C.5.
1The idea in one paragraph
When a source of waves and an observer move towards each other, the observer receives a higher frequency than the source emits; when they move apart, a lower one. That change is the Doppler effect. For sound, it matters who is moving: a moving source squeezes its wavefronts together in front of it, shortening the wavelength, while a moving observer simply runs into the wavefronts more often. For light there is no medium, so only the relative speed matters, and when that speed is much less than c the fractional shift is Δf/f = Δλ/λ ≈ v/c. Astronomers read the shift from spectral lines: a redshift means the body is moving away, a blueshift means it is approaching. Radar speed guns and medical ultrasound measure speeds the same way, from waves reflected off a moving target. At HL, you calculate the sound shifts exactly.
2What changes, and what does not
Two statements settle almost every Doppler question.
- The source sets how many waves are emitted per second, its frequency f.
- The medium sets how fast the waves travel, the wave speed v. For sound in air this is about 340 m s⁻¹ whatever the source is doing.
So when something moves, v stays the same. What changes is either the wavelength in the medium (if the source moves) or the rate at which the observer meets the waves (if the observer moves). Either way, the observed frequency f′ differs from f.
This is why sound and light differ. Sound moves through air, so a source moving through the air and an observer moving through it are physically different situations, with slightly different formulas at HL. Light needs no medium and its speed is c for every observer (A.5), so only the relative speed matters, and one formula covers both.
3A moving source: the wavefronts bunch up
Figure 1 compares a source at rest with a source moving to the right at half the speed of the waves. Each circle is a wavefront, drawn one period apart.
In panel (a) each wavefront is a circle centred on the source, and the spacing, the wavelength, is the same in every direction.
In panel (b) the source has moved on between emitting one wavefront and the next. Each circle is centred on the point where it was emitted and grows at speed v. So:
- Ahead of the source, towards observer A, the wavefronts are squeezed together. The wavelength is shorter. The waves still travel at v, so more of them reach A each second: A hears a higher frequency.
- Behind the source, at B, the wavefronts are spread out. The wavelength is longer, fewer arrive each second, and B hears a lower frequency.
That is the pitch drop you hear as an ambulance passes. The siren itself vibrates at its usual frequency, and the driver hears it unchanged; what changes is the wavelength in the air, and so the frequency received.
4A moving observer: meeting the waves more often
Now keep the source still and move the observer. Figure 2 shows the observer O moving towards the source through the wavefronts.
The wavefronts are spaced exactly as they would be anyway: the wavelength is unchanged. But the observer is running into them, so relative to the observer the waves go past at v + uo, where uo is the observer's speed. More wavefronts pass each second, and the frequency received is higher. Move away from the source and the waves pass at v − uo, and the frequency is lower.
Moving source: the wavelength in the medium changes. Moving observer: the wavelength is unchanged, but the rate of meeting wavefronts changes. Either way, approaching raises the frequency and receding lowers it.
5The Doppler effect for light
For light and all electromagnetic waves, the guide gives the approximate equation, valid when the relative speed v is much less than the speed of light c:
Here λ (or f) is the value emitted, measured in a laboratory at rest; Δλ (or Δf) is the size of the shift; and v is the relative speed of source and observer along the line joining them.
- If source and observer are moving apart, the received wavelength is longer and the frequency lower. For visible light this moves every line towards the red end of the spectrum: a redshift.
- If they are moving together, the wavelength is shorter: a blueshift.
Figure 3 shows the same set of absorption lines in a laboratory, from a receding body and from an approaching one. The shifts are exaggerated so you can see them.
Each line moves by Δλ = λ × v/c, a fixed fraction of its own wavelength, so the line at 656 nm moves further than the one at 410 nm, and the pattern stays recognisable.
Worked example: a receding galaxy. A hydrogen line with laboratory wavelength 656.3 nm is observed in the light of a galaxy at 662.9 nm. Determine the galaxy's speed and direction relative to Earth.
Divide by the laboratory wavelength (nanometres over nanometres is fine), and state the direction. If v were not much less than c, the approximation would fail and a relativistic formula would be needed; that is beyond this course, but checking v/c is not.
6What spectral lines tell us about stars and galaxies
A star's light carries absorption lines from the elements in its outer layers, at wavelengths fixed by atomic physics and measurable on Earth. When those lines are found shifted, the shift gives the star's speed along the line of sight, towards or away from us. Motion across the sky produces no first-order shift at all.
Three uses follow.
Stars and galaxies moving through space. A single shift gives a single speed along the line of sight. The light from almost all distant galaxies is redshifted, and more distant galaxies show larger redshifts: the evidence that the universe is expanding.
Rotation of an extended body. A star, a planet or a galaxy is not a point. If it rotates, one edge moves towards us and the opposite edge moves away. Figure 4 shows what that does to a single spectral line.
The approaching edge is blueshifted, the receding edge redshifted. If the body is also moving as a whole, both edges share that shift too; so take half the difference between the two edge shifts to get the rotation speed, and the average of the two to get the speed of the whole body.
Binary stars. Two stars orbiting each other alternately approach and recede, so their lines shift back and forth with the orbital period. Even when the two stars cannot be seen separately, the periodic shift reveals the pair and its orbital speed.
7Radar and ultrasound: speed from a reflected wave
The guide asks for the Doppler effect in radar and in medical physics as examples. Both send a wave of known frequency at a moving target and measure the frequency of the reflection.
Figure 5 shows why the shift from a reflection is twice the shift for one-way travel.
- The target moves towards the emitter. It receives the wave as a moving observer, at a raised frequency.
- It reflects that raised frequency, and in doing so it acts as a moving source, so the frequency is raised again on the way back.
For target speeds u much less than the wave speed, each step contributes about u/(wave speed), so the total fractional shift is about twice that. For radar, which uses microwaves travelling at c:
Worked example: a speed gun. A police radar gun emits microwaves at 24.0 GHz. The reflection from an approaching car is 4.80 kHz higher. Find the car's speed.
Doppler ultrasound. A probe on the skin sends ultrasound, a few megahertz, into the body. Red blood cells moving in an artery reflect it with a small shift, and the machine turns the shift into a blood speed. A narrowed artery shows up as a region of unusually fast flow, and the same method monitors a baby's heartbeat before birth. The wave speed here is the speed of sound in soft tissue, about 1540 m s⁻¹, and blood moves at well under 1 m s⁻¹, so the small-speed rule still applies with the speed of sound in place of c:
A 2 kHz shift is audible, so many machines let the operator listen to the flow.
8HLThe Doppler equations for sound
For sound and other mechanical waves, with v the wave speed in the medium, the data booklet gives two equations.
Moving source: f′ = f v / (v ± us). Moving observer: f′ = f (v ± uo) / v.
us is the speed of the source and uo the speed of the observer, both measured relative to the medium. The guide promises that only one of them moves in any problem.
Where they come from. Both come straight from sections 3 and 4.
Choosing the sign. Do not memorise a sign table. Decide first whether the frequency must go up (approaching) or down (receding), then pick the sign that makes it do so. Approaching source: minus in the denominator. Receding source: plus. Approaching observer: plus in the numerator. Receding observer: minus.
Worked example 1: an ambulance. A siren emits 700 Hz and the ambulance travels at 25 m s⁻¹. The speed of sound is 340 m s⁻¹. Calculate the frequencies heard by a pedestrian as it approaches and as it recedes.
Notice the shifts are not equal: +56 Hz on the way in, −48 Hz on the way out. The moving-source formula is not symmetric.
Worked example 2: a moving listener. A cyclist rides at 8.0 m s⁻¹ towards a stationary horn sounding at 500 Hz.
Worked example 3: finding the speed. A train's horn emits 400 Hz. A person on the platform hears 430 Hz as it approaches. Determine the train's speed.
Source against observer. Figure 6 plots f′/f against speed for all four cases. For a moving observer the graph is a straight line, since uo appears only on top. For a moving source it curves, because us is in the denominator; an approaching source's frequency grows without limit as us approaches v, when the wavefronts pile on top of each other (the sonic boom of a supersonic aircraft). At low speeds the four lines crowd together: when u ≪ v, both formulas reduce to Δf/f ≈ u/v, the same form as the light equation.
A passing source. A listener near a road hears a steady high pitch as a car approaches, a rapid fall as it passes, and a steady low pitch as it moves away, as in Figure 7. The fall is rapid because only the component of the car's velocity along the line to the listener counts, and that component swings from +u to −u as the car goes by. The two flat levels give you two equations in two unknowns, the car's speed and its horn's true frequency; Q5 asks you to solve them.
9Where marks are lost
Saying the wave speed changes. The speed of sound is set by the air. A moving source changes the wavelength; a moving observer changes the rate of meeting wavefronts. Neither changes v in the medium.
Saying the source's frequency changes. The source vibrates at its usual frequency; the driver hears the normal siren. It is the frequency received that changes.
Dividing by the observed wavelength. In Δλ/λ, λ is the laboratory (emitted) value.
Giving a speed with no direction. A redshift means moving apart, a blueshift moving together. State which.
Forgetting the factor of 2 for a reflection. Radar and ultrasound shift the frequency twice: Δf ≈ 2uf/(wave speed).
Thinking sideways motion causes a shift. Only the velocity component along the line of sight counts. Sound a passing car emits at its point of closest approach reaches you unshifted.
HL: choosing signs by memory. Decide up or down first, then choose the sign that gives it. An approaching source with a plus sign in the denominator gives a lower frequency, which is impossible.
HL: using the observer formula for a moving source, or vice versa. Ask who is moving through the air. The formulas give different answers for the same speed.
10Draw it right
- Moving source: circles not concentric, each centred on the point where it was emitted, all expanding at the same speed; closer together ahead, further apart behind. Mark the source's direction with an arrow.
- Stationary source for comparison: concentric, evenly spaced circles.
- Moving observer: unchanged, evenly spaced wavefronts, with the observer's velocity arrow; the diagram shows no change in λ.
- Shifted spectra: the same pattern of lines, every line moved the same way, longer-wavelength lines moved further; label red and violet ends.
- Rotating body: the approaching edge blueshifted, the receding edge redshifted, arrows showing the direction of rotation relative to the observer.
- HL passing source: frequency against time, flat high, falling fastest at closest approach, flat low; the high level further above f than the low level is below it.
11Try it
Marks in brackets. Answers and marker's notes are at the end. Take c = 3.00 × 10⁸ m s⁻¹ and the speed of sound in air as 340 m s⁻¹.
Q1. A source of sound moves at constant speed directly towards a stationary observer. Compared with when the source is at rest, what does the observer measure? 1 mark
A. same wavelength, higher frequency, higher wave speed
B. shorter wavelength, higher frequency, same wave speed
C. shorter wavelength, same frequency, lower wave speed
D. same wavelength, same frequency, higher wave speed
Q2. A drone carrying a buzzer flies in a straight line past a stationary microphone. Using a sketch of the wavefronts, explain why the frequency recorded is higher while the drone approaches than while it moves away. 3 marks
Q3. A line in the laboratory spectrum of hydrogen has a wavelength of 434.0 nm. In the spectrum of a distant galaxy, the same line is at 441.8 nm.
(a) State whether the galaxy is moving towards or away from Earth. 1 mark
(b) Calculate the speed of the galaxy relative to Earth. 2 marks
(c) Explain why the equation you used is valid here. 1 mark
Q4. A line with laboratory wavelength 589.00 nm is observed at 589.04 nm from one edge of a star's equator and at 588.96 nm from the opposite edge.
(a) Explain why the two edges give different wavelengths. 2 marks
(b) Determine the speed of the star's equator. 2 marks
(c) The star's radius is 1.2 × 10⁹ m. Estimate its rotation period in days. 2 marks
Q5 (HL). Figure 7 shows the frequency heard as a car sounding its horn passes a listener: 510 Hz while it approaches and 453 Hz while it recedes. Determine the speed of the car and the frequency emitted by the horn. 4 marks
Q6 (HL). A bell rings at 800 Hz.
(a) A cyclist rides at 6.0 m s⁻¹ directly towards the bell, which is stationary. Calculate the frequency she hears. 2 marks
(b) Instead, the bell moves at 6.0 m s⁻¹ towards the stationary cyclist. Calculate the frequency she hears. 1 mark
(c) Explain why the answers differ although the relative speed is the same. 2 marks
12In one breath
The source sets the frequency, the medium sets the wave speed. A moving source bunches its wavefronts ahead (shorter λ, higher f) and spreads them behind (longer λ, lower f); a moving observer leaves λ alone but meets the wavefronts more or less often. Approaching raises the frequency, receding lowers it. For sound it matters which one moves through the air; for light only the relative speed counts, and when v ≪ c, Δf/f = Δλ/λ ≈ v/c, with λ the laboratory value. Moving apart gives a redshift, moving together a blueshift. Shifted spectral lines give the line-of-sight speed of stars and galaxies; opposite edges of a rotating body shift opposite ways, and half the difference gives the rotation speed. Radar guns and Doppler ultrasound measure speed from a reflection, which is shifted twice: Δf ≈ 2uf/(wave speed). HL: moving source f′ = f v/(v ± us), moving observer f′ = f (v ± uo)/v; decide up or down first, then choose the sign; the two formulas give different answers for the same speed.
Answers
Q1. B. A moving source shortens the wavelength ahead of it; the wave speed is fixed by the air, so v = fλ gives a higher frequency. A and D change the wave speed, which the source cannot do; C keeps f the same, which contradicts v = fλ with v fixed.
Q2. Sketch: circular wavefronts, each centred on the point where the drone emitted it, closer together on the side it is moving towards and further apart behind. Explanation: the drone moves on between emitting successive wavefronts, so ahead of it the wavelength is shorter and behind it longer. The wave speed in air is unchanged, so while approaching more wavefronts reach the microphone each second (higher frequency), and while receding fewer do (lower frequency). 1 for a correct sketch with non-concentric wavefronts bunched ahead, 1 for shorter wavelength ahead and longer behind, 1 for same wave speed so f = v/λ is higher when approaching. "The sound is compressed" with no wavelength argument scores 0 for the explanation.
Q3. (a) Away from Earth: the observed wavelength is longer, a redshift. (b) Δλ = 441.8 − 434.0 = 7.8 nm. v = cΔλ/λ = 3.00 × 10⁸ × 7.8 ÷ 434.0 = 5.4 × 10⁶ m s⁻¹. (c) v/c = 0.018, so the galaxy's speed is much less than the speed of light, which is the condition for Δλ/λ ≈ v/c. (a) 1 for away, with redshift or longer wavelength as the reason. (b) M1 for Δλ/λ with the laboratory λ, A1 for 5.4 × 10⁶ m s⁻¹. (c) 1 for v ≪ c with the ratio or a comparison. Dividing by the observed 441.8 nm uses the wrong λ and loses the M1, even though the answer differs only slightly.
Q4. (a) The star rotates, so one edge moves towards Earth and the opposite edge moves away. Light from the approaching edge is blueshifted to a shorter wavelength and light from the receding edge is redshifted to a longer one. (b) The average of the two is 589.00 nm, so there is no overall line-of-sight motion; half the difference is 0.04 nm. v = cΔλ/λ = 3.00 × 10⁸ × 0.04 ÷ 589.00 = 2.0 × 10⁴ m s⁻¹. (c) T = 2πr/v = 2π × 1.2 × 10⁹ ÷ 2.04 × 10⁴ = 3.7 × 10⁵ s ≈ 4.3 days. (a) 1 for rotation, one edge towards and one away, 1 for blueshift and redshift. (b) M1 for Δλ = 0.04 nm (half the difference), A1 for 2.0 × 10⁴ m s⁻¹. (c) M1 for 2πr/v, A1 for about 4 days. Using the full difference 0.08 nm doubles the speed and scores A0 in (b); ECF applies in (c).
Q5 (HL). Approaching: 510 = f × 340/(340 − u). Receding: 453 = f × 340/(340 + u). Dividing the first by the second: 510/453 = (340 + u)/(340 − u), so 1.126 × (340 − u) = 340 + u, giving u = 340 × 0.126 ÷ 2.126 = 20 m s⁻¹. Then f = 510 × (340 − 20) ÷ 340 = 480 Hz. M1 for both equations with correct signs, M1 for eliminating f by dividing, A1 for 20 m s⁻¹, A1 for 480 Hz. Taking f as the average of the two readings (481.5 Hz) is a guess, not a method, and scores 0 for that part.
Q6 (HL). (a) Moving observer: f′ = 800 × (340 + 6.0) ÷ 340 = 814.1 Hz. (b) Moving source: f′ = 800 × 340 ÷ (340 − 6.0) = 814.4 Hz. (c) Sound travels through air, so it matters which one moves relative to the medium. A moving source changes the wavelength in the air, putting us in the denominator; a moving observer leaves the wavelength unchanged and changes only the speed of the waves relative to it, putting uo in the numerator. The two give different results. (a) M1 for the observer formula with plus, A1 for 814 Hz. (b) A1 for 814.4 Hz, to 4 s.f. to show the difference. (c) 1 for motion relative to the medium, 1 for wavelength changed (source) versus rate of meeting wavefronts changed (observer). "Light only depends on relative speed" is true but does not answer this question.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section C.5 Doppler effect. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.