C.5 The Doppler effect | IB DP Physics

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Note the change: the Doppler effect used to be additional higher level material. It is now studied by everyone, so SL students meet it for the first time.

🎯What you need to be able to do

Explain the Doppler effect in terms of wavefronts bunching ahead of a source and stretching behind it.
State that the source keeps emitting at its own unchanged frequency, and say what actually changes.
Use \( f' = f\,v/(v \pm v_s) \) for a moving source and \( f' = f\,(v \pm v_o)/v \) for a moving observer, and choose the sign by checking the direction of the shift.
Explain why the two cases are physically different, and say when the difference becomes large.
Describe what happens when a source reaches the wave speed, and name the result.
Use \( \Delta f/f = \Delta\lambda/\lambda = v/c \) for light when \( v \ll c \), and interpret redshift and blueshift.
Explain applications: galaxy recession, radar speed measurement and Doppler ultrasound, including why a reflected wave is shifted twice.

📚Where the effect actually comes from

The mechanism first, because the formulas only make sense afterwards. A stationary source emits wavefronts as a set of concentric circles, evenly spaced in every direction. Now let the source move. It emits each new front from a point slightly further along than the last, so the fronts crowd together ahead of it and spread apart behind it. Crowded fronts mean a shorter wavelength and so a higher frequency; spread fronts mean the opposite.

The circles themselves are never squashed — each one still expands at the wave speed. It is their centres that march along, and that alone produces the whole effect.

That single picture earns more marks than a recalled equation, and it is the thing to draw first in any Doppler question that asks you to explain.

The siren does not change its note. The source is still emitting at its own unchanged frequency throughout — nothing on the ambulance is behaving differently as it passes you. What changes is the rate at which wavefronts arrive at your ear. Say it that way round in an explanation question; the intuitive story, that the siren somehow drops in pitch, is simply wrong, and examiners look for exactly this distinction.

🔄A moving source changes the wavelength itself

In one period \( T = 1/f \) the source emits one whole wavefront and moves a distance \( v_sT \) in the process. Ahead of it, the next front is emitted from that much closer to the previous one, so the wavelength in the medium is genuinely shortened:

\[ \lambda' = \frac{v - v_s}{f} \quad\text{ahead},\qquad \lambda' = \frac{v + v_s}{f} \quad\text{behind} \]

The wave still travels at \( v \), the speed the medium fixes, so the frequency an observer receives is \( f' = v/\lambda' \), which rearranges to the equation you will use:

A source moves to the right at speed v-sub-s while emitting waves. The distance between successive wavefronts ahead of it is labelled as the ordinary wavelength v over f reduced by v-sub-s over f, giving a compressed wavelength of v minus v-sub-s all over f. Behind the source the same construction gives v plus v-sub-s all over f, a stretched wavelength. A note states that the wave speed in the medium is unchanged, so the received frequency is the wave speed divided by this new wavelength. — The source travels \( v_sT \) between one front and the next, and that distance is subtracted from the wavelength ahead and added to the wavelength behind.

Moving source\( f' = f\dfrac{v}{v \pm v_s} \)

Moving observer\( f' = f\dfrac{v \pm v_o}{v} \)

🕶️A moving observer changes how often the fronts arrive

Now hold the source still and move the observer instead. The wavefronts in the medium are completely undisturbed — evenly spaced, wavelength \( \lambda = v/f \) in every direction. An observer running towards them simply meets them more often, because their own motion adds to the closing speed. The relative speed of approach becomes \( v + v_o \) rather than \( v \), and

\[ f' = \frac{v + v_o}{\lambda} = f\,\frac{v + v_o}{v} \]

A stationary source at the left emits evenly spaced straight wavefronts travelling right at speed v. An observer at the right moves left, towards the source, at speed v-sub-o. The wavefronts are drawn with identical spacing throughout, and the annotation states that the wavelength in the medium is unchanged; only the closing speed is different, being v plus v-sub-o rather than v, so fronts are met more often. A second annotation notes that an observer moving away meets them less often, at a closing speed of v minus v-sub-o. — Nothing has happened to the wave here. The spacing is untouched — the observer is simply running into the fronts, so more of them arrive each second.

✏️Worked example 1 — the ambulance

An ambulance siren emits 780 Hz. It drives past a stationary observer at 22 m s\(^{-1}\). Take the speed of sound as 340 m s\(^{-1}\).

(a) Frequency heard as it approaches. Moving source, approaching, so the wavelength ahead is shortened and the denominator is \( v - v_s \):

\[ f' = 780 \times \frac{340}{340 - 22} = 780 \times \frac{340}{318} = 834\ \text{Hz} \]

(b) Frequency heard after it passes. Now the denominator is \( v + v_s \): \( f' = 780 \times 340/362 = 733 \) Hz. Lower, as receding must be.

(c) The drop the observer hears. \( 834 - 733 = 101 \) Hz, and it happens over the second or so the ambulance takes to pass. That abruptness is the whole reason the effect is so obvious in everyday life: the shift itself is only about 13%, but it arrives all at once.

(d) The same observer now cycles towards a stationary siren at 22 m s\(^{-1}\). Moving observer, approaching: \( f' = 780 \times (340 + 22)/340 = 830 \) Hz.

Close to (a) but not identical — 834 Hz against 830 Hz. That 4 Hz gap is not rounding. The two situations are genuinely different physics, and the next section is about why.

⚖️The two cases are not the same

At everyday speeds the two equations agree closely, which is why it is tempting to treat them as one. They are not. A moving source changes the wavelength in the medium; a moving observer does not. The wave that a stationary microphone would record is a different wave in the two cases, and if you froze the medium and measured the crest spacing you could tell them apart immediately.

The disagreement grows as the speed approaches the wave speed. Ratio the two expressions for an approaching motion at speed \( u \) and the moving-source result runs away to infinity as \( u \to v \), while the moving-observer result only ever doubles.

A graph of received frequency divided by emitted frequency against speed as a fraction of the wave speed, from zero to nearly one, for motion of approach. The moving-observer line is straight, rising steadily from one to two. The moving-source curve starts on top of it but bends upward ever more steeply, passing two at a half of the wave speed and rising towards infinity as the speed approaches the wave speed. A shaded band at the far left marks everyday speeds, where the two are indistinguishable. Annotations state that the moving observer at most doubles the frequency while the moving source has no upper limit. — Below about a tenth of the wave speed the two are indistinguishable in practice — which is every ambulance you will ever hear. The difference only becomes dramatic near the wave speed.

✏️Worked example 2 — where they part company

A 400 Hz source and a listener approach each other. Compare the frequency received when the source moves at 170 m s\(^{-1}\) with the listener still, and when the listener moves at 170 m s\(^{-1}\) with the source still. Take \( v = 340 \) m s\(^{-1}\).

Both speeds are exactly half the speed of sound.

Moving source: \( f' = 400 \times 340/(340 - 170) = 400 \times 2 = 800 \) Hz. The wavelength ahead has been halved, so the frequency doubles.

Moving observer: \( f' = 400 \times (340 + 170)/340 = 400 \times 1.5 = 600 \) Hz. The wavelength is untouched; the listener meets the fronts one and a half times as often.

800 Hz against 600 Hz — a difference of a third, from the same relative speed. At 22 m s\(^{-1}\) the same comparison gave 834 against 830. The two equations are not interchangeable, and a question that specifies which body is moving is specifying which equation you must use.

➰Getting the sign right, every time

For the moving source the minus sign goes with approaching and the plus with receding; for the moving observer it is the other way round. That is easy to state and easy to misremember under exam pressure, so do not rely on it.

A decision panel in two halves. The left half, for a moving source, shows that approaching takes v minus v-sub-s in the denominator, giving a higher frequency, and receding takes v plus v-sub-s, giving a lower one. The right half, for a moving observer, shows that approaching takes v plus v-sub-o in the numerator and receding takes v minus v-sub-o. Beneath both, a wide band gives the check that works without memorisation: approaching must raise the frequency and receding must lower it, so work out the answer, compare it with the emitted frequency, and flip the sign if it moved the wrong way. — The physical check costs five seconds and cannot be misremembered. Use it on every Doppler answer you write, including the ones you are sure about.

Work out the number, then ask a single question: is the answer on the right side of the emitted frequency? Approaching must give a higher frequency and receding a lower one, in every case, for sound and for light alike. If your answer went the wrong way, you have the sign upside down; flip it and move on.

The shift is zero at closest approach, not maximum. The equations use the component of velocity along the line joining source and observer. When a car passes you on a straight road, the instant it is level with you it is moving entirely sideways, that component is zero, and so is the shift — the observed frequency passes exactly through the emitted frequency at that moment. The shift is largest when the source is far away and heading straight at you. This is why a passing siren glides from high to low through the true pitch rather than jumping.

✈️When the source catches its own waves

Push the source speed up towards \( v \) and the wavefronts ahead have less and less of a head start. At \( v_s = v \) they cannot get away at all: every front emitted since the source reached that speed lies on top of the source, piled into one front of enormous amplitude. The moving-source equation says \( f' \to \infty \), which is the mathematics announcing that the model has broken down.

Three panels showing a source moving faster in each. In the first, at a source speed well below the wave speed, the circular wavefronts are bunched ahead but still nested inside one another. In the second, at exactly the wave speed, all the fronts touch at a single point at the source and pile into one flat front. In the third, at twice the wave speed, the source has outrun its own wavefronts and their common tangent forms a V-shaped cone trailing behind, with the half-angle theta marked and the relation sine theta equals the wave speed divided by the source speed, equal to one over the Mach number. — Past the wave speed the source outruns its own sound and the fronts have a common tangent: a cone. The boom is not a one-off bang at the moment of breaking the barrier — the cone trails along continuously, and you hear it when its edge sweeps over you.

The geometry is worth a line. In a time \( t \) the source travels \( v_st \) while the front it emitted at the start has grown to radius \( vt \). The cone's half-angle therefore satisfies

\[ \sin\theta = \frac{vt}{v_st} = \frac{v}{v_s} = \frac{1}{M} \]

where \( M = v_s/v \) is the Mach number. A faster aircraft makes a narrower cone.

💡Light: only the relative velocity matters

For light there is no medium. There is nothing for a moving source to compress and nothing for a moving observer to run into, and there is no experiment that could tell you which of the two is "really" moving. So the two cases collapse into one, and the shift depends on the relative velocity alone. When \( v \ll c \) it is well approximated by

\[ \frac{\Delta f}{f} = \frac{\Delta\lambda}{\lambda} = \frac{v}{c} \]

Light from a receding object shifts to longer wavelengths — redshift — and from an approaching object to shorter, blueshift. The names are about the direction of the shift, not about the colour you end up with: a redshifted blue star is still blue.

Three horizontal spectra one above the other, each showing the same pattern of four dark absorption lines. The middle spectrum is the laboratory reference, with the lines at their rest wavelengths. The top spectrum, from an approaching source, has the whole pattern shifted bodily towards the blue end at shorter wavelengths. The bottom spectrum, from a receding source, has it shifted towards the red end at longer wavelengths. Vertical guides connect corresponding lines to show that the pattern is displaced as a whole and its spacing is preserved, which is what makes the shift measurable. — It is the *pattern* that identifies the element, and the pattern shifts as a whole. That is what makes the measurement possible: you know exactly which line you are looking at, so you know exactly how far it has moved.

✏️Worked example 3 — a galaxy's speed from one line

In the laboratory, hydrogen emits a red line at 656.3 nm. In the spectrum of a distant galaxy the same line is measured at 662.9 nm. Find the galaxy's speed relative to us and state its direction. Take \( c = 3.00\times10^{8} \) m s\(^{-1}\).

The shift is \( \Delta\lambda = 662.9 - 656.3 = 6.6 \) nm, towards longer wavelengths.

\[ \frac{v}{c} = \frac{\Delta\lambda}{\lambda} = \frac{6.6}{656.3} = 1.01\times10^{-2} \]

\( v = 1.01\times10^{-2} \times 3.00\times10^{8} = 3.0\times10^{6} \) m s\(^{-1}\), and because the shift is to longer wavelengths it is a redshift, so the galaxy is receding.

Is the approximation allowed? \( v/c \approx 0.01 \), so \( v \) is one per cent of the speed of light — comfortably in the range where \( \Delta\lambda/\lambda = v/c \) holds. Note also that \( \lambda \) in the denominator is the rest wavelength, 656.3 nm, not the measured one. At this size of shift it makes no visible difference, but it is the correct quantity and it matters for large redshifts.

🌌Redshift and the expanding universe

Measure the spectra of many galaxies and two things stand out. Almost every one is redshifted, and the more distant the galaxy the larger its redshift. Nothing about the Doppler effect on its own predicts that pattern — a random collection of galaxies should show as many blueshifts as redshifts — so the pattern itself is the evidence. It says that everything is receding from everything else, which is what an expanding universe looks like from the inside.

Speed proportional to distance is the signature of uniform expansion. The handful of blueshifted galaxies are the nearby ones, where ordinary orbital motion inside a group is larger than the expansion between us.

📡Reflected waves shift twice

Bounce a wave off a moving object and the Doppler effect happens twice, because the object plays both roles in turn. On the way in it is a moving observer, receiving the transmitted wave at a shifted frequency. It then re-radiates what it received, so on the way back it is a moving source. Both steps shift in the same direction, so the returning signal carries roughly double the single shift.

A radar transmitter on the left sends a wave at frequency f to a car approaching from the right. Step one is labelled: the car acts as a moving observer and receives a frequency raised once. Step two is labelled: the car re-radiates that already-raised frequency, now acting as a moving source, so it is raised a second time by the same factor. The wave returning to the transmitter is labelled with a total shift of twice f times u over c for the case of light. A note states that the same double shift is what a medical Doppler ultrasound scan measures when it is reflected from moving blood cells. — Two shifts in the same direction, one on each leg of the journey. Forgetting the factor of two halves every answer in this application, and it is the single most common mistake here.

For light and small speeds this gives a beautifully simple result. The single shift is \( fu/c \), so the total shift measured on return is

\[ \Delta f = \frac{2fu}{c} \]

✏️Worked example 4 — a speed camera

A police radar gun transmits at 24.0 GHz and detects a beat frequency of 4.80 kHz between the transmitted and returning signals from an approaching car. Find the car's speed. Take \( c = 3.00\times10^{8} \) m s\(^{-1}\).

The beat frequency is the total Doppler shift, so \( \Delta f = 4.80\times10^{3} \) Hz. Rearranging \( \Delta f = 2fu/c \):

\[ u = \frac{c\,\Delta f}{2f} = \frac{3.00\times10^{8} \times 4.80\times10^{3}}{2 \times 24.0\times10^{9}} = 30.0\ \text{m s}^{-1} \]

That is 108 km h\(^{-1}\).

Check the fraction. \( \Delta f/f = 4800/(24.0\times10^{9}) = 2.0\times10^{-7} \), and \( 2u/c = 60/(3.00\times10^{8}) = 2.0\times10^{-7} \). They match, so the factor of two is in the right place. Had the two been omitted, the answer would have come out as 60 m s\(^{-1}\) — a driver reported at double their actual speed.

🔭See it happen

PhET, Sound Waves, has a Doppler mode where you can drag the source and watch the wavefronts crowd together in front of it. Push the source speed up to the wave speed and the shock front forms on screen — the sonic boom drawn rather than described. Then set the source moving in a straight line past the listener and watch the received frequency pass smoothly through the emitted value at the moment of closest approach.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A train horn sounds at 500 Hz as the train approaches a stationary observer at 30 m s\(^{-1}\). Take the speed of sound as 340 m s\(^{-1}\). Find the frequency heard.

Moving source, approaching, so the denominator is \( v - v_s \): \( f' = 500 \times \dfrac{340}{310} = 548 \) Hz. Higher, as approaching must be.

2. Find the frequency heard by the same observer after the train has passed, and the size of the drop.

\( f' = 500 \times \dfrac{340}{370} = 459 \) Hz. Lower, as receding must be. The drop is \( 548 - 459 = 89 \) Hz, heard over the second or two the train takes to pass.

3. Now the source is stationary at 500 Hz and the observer cycles towards it at 30 m s\(^{-1}\). Find the frequency heard, and explain why it differs from your answer to question 1.

Moving observer, approaching: \( f' = f\dfrac{v + v_o}{v} = 500 \times \dfrac{370}{340} = 544 \) Hz. It is close to the 548 Hz of the moving source but not equal, because the physics is different: a moving source genuinely shortens the wavelength in the air, whereas a moving observer meets an undisturbed set of wavefronts more often. At 30 m s\(^{-1}\), less than a tenth of the speed of sound, the two predictions have barely separated.

4. A car drives past you at constant speed along a straight road, sounding its horn continuously. Sketch how the frequency you hear varies with time, and state its value at the moment the car is level with you.

The frequency starts above the emitted value while the car is distant and heading towards you, falls smoothly — most steeply as the car passes — and settles below the emitted value once it is far away and receding. At the instant it is level with you the car's velocity is entirely across your line of sight, so the component along that line is zero and you hear the true emitted frequency. The curve therefore crosses the emitted frequency exactly at closest approach; it does not jump.

5. Light from a distant galaxy shows \( \Delta\lambda/\lambda = 0.004 \). Find its recession speed.

For \( v \ll c \), \( v = c\,\Delta\lambda/\lambda = 3.0 \times 10^{8} \times 0.004 = 1.2 \times 10^{6} \) m s\(^{-1}\). The shift is to longer wavelengths, so it is a redshift and the galaxy is receding.

6. A sodium line with a rest wavelength of 589.0 nm is observed in a star's spectrum at 588.6 nm. Find the star's speed along the line of sight and state whether it is approaching or receding.

\( \Delta\lambda = 0.4 \) nm, towards shorter wavelengths, so it is a blueshift and the star is approaching. \( v = c\,\Delta\lambda/\lambda = 3.00\times10^{8} \times 0.4/589.0 = 2.0\times10^{5} \) m s\(^{-1}\). Note this is only the component along the line of sight; any motion across the sky produces no shift at all.

7. Explain why the Doppler shift measured by a police speed camera is twice what you might first expect.

The shift happens twice. The moving vehicle first acts as a moving observer, receiving the transmitted wave at a shifted frequency; it then re-radiates that already-shifted wave, now acting as a moving source, which shifts it again in the same direction. The returning signal therefore carries double the single shift, \( \Delta f = 2fu/c \). Medical Doppler ultrasound and weather radar work the same way.

8. A bat emits 45.0 kHz and receives the echo from a moth flying directly away from it at 5.0 m s\(^{-1}\). Take the speed of sound as 340 m s\(^{-1}\). Find the frequency of the echo the bat receives.

Two stages. The moth is first a receding observer: it receives \( 45.0\ \text{kHz} \times (340 - 5)/340 \). It then re-radiates that as a receding source, so the bat receives that frequency multiplied by \( 340/(340 + 5) \). Combining: \[ f' = 45.0\ \text{kHz} \times \frac{340 - 5}{340 + 5} = 45.0 \times \frac{335}{345} = 43.7\ \text{kHz} \] Lower than 45.0 kHz, as receding must be. The bat's nervous system reads that 1.3 kHz deficit as "moving away".

9. An aircraft flies at Mach 1.5. Find the half-angle of its shock cone, and state what happens to that angle as the aircraft goes faster.

\( \sin\theta = 1/M = 1/1.5 = 0.667 \), so \( \theta = 41.8^\circ \). As \( M \) increases, \( \sin\theta \) decreases and the cone gets narrower — at Mach 2 it is \( 30.0^\circ \). The cone trails continuously behind the aircraft; the boom is heard on the ground when the edge of the cone sweeps past, not at the moment the aircraft passed overhead.

10. State what happens to the wavefronts when a source reaches the speed of the wave in the medium, name the result, and say what the moving-source equation does at that point.

The wavefronts can no longer outrun the source, so every one of them piles up on top of the source into a single front of very large amplitude. For sound this is the shock front heard as a sonic boom. In the equation \( f' = fv/(v - v_s) \) the denominator goes to zero and \( f' \to \infty \) — not a real infinity, but the model telling you it no longer applies, because it assumed the fronts get away from the source.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The Physics Classroom — the Doppler effect and shock waves
The Physics Hypertextbook — the Doppler effect