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 and stretching.
Use the Doppler equations for sound, distinguishing a moving source from a moving observer.
Use the approximate relation for light when the speed is much less than \(c\).
Describe applications: astronomy, radar speed measurement and medical ultrasound.

📚The physics

The mechanism first, because the formulas make sense only afterwards. A stationary source emits wavefronts as concentric circles. Now let it move. Each new wavefront is emitted from a point slightly further along, so the fronts bunch up ahead of the source and stretch out behind it. Bunched fronts mean a shorter wavelength and therefore a higher frequency; stretched fronts mean the opposite. That is the entire effect, and a sketch of it earns more marks than a recalled equation.

Crucially, the source is still emitting at its own unchanged frequency. The siren does not change pitch as it passes you; what changes is the rate at which wavefronts arrive at your ear. This is worth being explicit about, because the intuitive story — that the siren somehow drops in pitch — is wrong.

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

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

For the moving source, the minus sign is for approaching and plus for receding; for the moving observer, plus for approaching and minus for receding. Rather than memorise which sign goes where, apply a physical check every time: approaching must give a higher frequency, receding a lower one. Work out the answer, then ask whether it moved in the right direction. If it did not, flip the sign.

The two cases are not physically identical, even though they give similar numbers at low speeds. A moving source genuinely changes the wavelength in the medium; a moving observer simply meets the unchanged wavefronts more or less often. At speeds approaching the wave speed the two predictions diverge noticeably. If the source reaches the wave speed the fronts pile up into a shock front — the sonic boom.

For light there is no medium and no distinction between source and observer motion; only the relative velocity matters. When \( v \ll c \) the shift 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.

Applications worth knowing. In astronomy, redshift in the spectral lines of distant galaxies is the evidence that they are receding, and the fact that almost everything is redshifted is the observational foundation of an expanding universe. In a speed camera or weather radar, a wave is reflected from a moving object, so the shift happens twice — the vehicle first acts as a moving observer and then as a moving source — which doubles the measured shift. Medical Doppler ultrasound uses the same double shift to measure the speed of blood flow.

✏️Worked example

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: \( f' = 780 \times 340/(340 - 22) = 780 \times 340/318 = 834 \) Hz. Higher, as it must be.

(b) Frequency heard after it passes. \( f' = 780 \times 340/(340 + 22) = 780 \times 340/362 = 733 \) Hz. Lower, as it must be.

(c) The drop the observer hears. \( 834 - 733 = 101 \) Hz, and it happens over the second or so it takes the ambulance to pass — which is why the change sounds so abrupt.

(d) The same observer now cycles towards a stationary siren at 22 m s\(^{-1}\). Moving observer: \( f' = 780 \times (340 + 22)/340 = 831 \) Hz.

Close to (a) but not identical — 834 against 831. The two situations are genuinely different physics, and at these speeds the difference is small but real.

🔭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.

📝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: \( f' = f\dfrac{v}{v - v_s} = 500 \times \dfrac{340}{310} = 548 \) Hz. Higher, as approaching must be.

2. Find the frequency heard by the same observer after the train has passed.

\( f' = 500 \times \dfrac{340}{370} = 459 \) Hz. Lower, as receding must be. The observer hears a drop of about 89 Hz as it passes.

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 compare it with the first answer.

Moving observer, approaching: \( f' = f\dfrac{v + v_o}{v} = 500 \times \dfrac{370}{340} = 544 \) Hz. Close to the 548 Hz of the moving source but not identical — a moving source genuinely changes the wavelength in the medium, while a moving observer just meets unchanged wavefronts more often.

4. 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.

5. 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, then re-radiates that shifted wave as a moving source. Both steps shift in the same direction, so the returning signal carries double the single shift. Medical Doppler ultrasound works the same way.

6. State what happens to the wavefronts when a source reaches the speed of the wave in the medium, and name the result.

The wavefronts can no longer outrun the source, so they pile up on top of one another into a single high-amplitude shock front. For sound this is the sonic boom.

🔗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