What the HL part adds: At SL the Doppler effect for sound is described in words, and light is handled with Δλ/λ ≈ v/c. The HL job is to put a number on the sound shift, for a moving source or a moving observer.
SL: qualitative
- A source moving towards you sounds higher
- A source moving away sounds lower
HL: a number
- Calculate the observed frequency f' (sound)
- Find the speed of the moving source or observer
When the source moves, each wavefront is emitted from a point closer to (or further from) the observer than the last. This squashes the wavelength ahead of the source and stretches it behind. The wave still travels at the medium speed v, so a shorter wavelength is detected as a higher frequency, and a longer one as a lower frequency.
Which sign?: Decide the sign physically first, then check it against your answer:
- Approaching ⇒ frequency goes up ⇒ f' > f. - Receding ⇒ frequency goes down ⇒ f' < f.
If your arithmetic disagrees with that, you have the sign the wrong way round.
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For a source moving through the medium at speed us, with the wave speed v (e.g. the speed of sound), the observed frequency is given in the data booklet.
Which sign?
- The top sign (−) is for the source approaching.
- The bottom sign (+) is for it receding.
- observed (detected) frequency (Hz)
- frequency emitted at source (Hz)
- speed of the wave in the medium, e.g. sound (m s⁻¹)
- speed of the source through the medium (m s⁻¹)
Worked example — an approaching siren
An ambulance siren emits f = 512 Hz and approaches a stationary observer at us = 30 m s⁻¹. The speed of sound is v = 340 m s⁻¹. Find the frequency the observer hears.
Solution
- Write the given equation; approaching ⇒ use the minus sign in the denominator:
- Substitute the values:
- Work it out — keep the unit:
Final answer
f' = 561 Hz — higher than 512 Hz, exactly as expected for an approaching source.
Sanity check the sign: Approaching must give f' bigger than f. Here 561 Hz > 512 Hz. If you had used + by mistake you would get 470 Hz (lower) — a red flag that the sign is wrong.
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Now the source is still and the observer moves through the medium at speed uo. The wavelength in the air does not change; the observer just meets the wavefronts faster (towards) or slower (away).
- observed (detected) frequency (Hz)
- frequency emitted at the source (Hz)
- speed of the wave in the medium (m s⁻¹)
- speed of the observer through the medium (m s⁻¹)
Which sign?: Towards the source: + (meets more wavefronts, f' > f).
Away from the source: − (f' < f).
The sign is in the numerator here, not the denominator as for a moving source.
Worked example — how fast is the detector moving?
A loudspeaker emits 2000 Hz. A detector moving along the line joining them records 1900 Hz. The speed of sound is 340 m s⁻¹. Find the detector's speed and its direction.
Solution
- The frequency is lower, so the detector moves away: use the minus sign.
- Substitute:
- Rearrange: 340 − uo = 340 × 0.95 = 323, so
Final answer
uo = 17 m s⁻¹, moving away from the loudspeaker.
Only one of them moves: Exam problems never have the source and the observer both moving. Decide which one moves, then pick its equation.
Where it is used: Doppler ultrasound: ultrasound reflected from moving blood cells comes back at a shifted frequency, which gives the speed of the blood.
Radar speed guns: microwaves reflected from a moving car return with a frequency shift that gives the car's speed.
Light is different: Light needs no medium, so the sound equation does not apply. For a source moving at speed v much smaller than c, the fractional change in frequency (or wavelength) is simply v/c.
- change in frequency / wavelength
- emitted frequency / wavelength
- speed of source towards/away from observer (m s⁻¹)
- speed of light, 3.0 × 10⁸ m s⁻¹
Worked example — a receding galaxy
A known spectral line from a distant galaxy is shifted so that Δλ/λ = 0.020. The light is shifted to longer wavelengths (a redshift). Estimate the galaxy's recession speed. Take c = 3.0 × 10⁸ m s⁻¹.
Solution
- Write the given relation:
- Make v the subject and substitute:
- Work it out — keep the unit:
Final answer
v ≈ 6.0 × 10⁶ m s⁻¹ (about 0.02c). The redshift means the galaxy is moving away from us.
Redshift vs blueshift: A redshift (λ longer, towards the red end) means the source is receding. A blueshift (λ shorter) means it is approaching. The redshift of distant galaxies is the key evidence that the Universe is expanding.
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Sound — f' = f·v/(v ∓ u_s)
- Needs a medium; v is the speed of sound
- Gives the actual observed frequency f'
- − sign (smaller denominator) ⇒ approaching ⇒ higher f'
- + sign (larger denominator) ⇒ receding ⇒ lower f'
- Source and observer speeds matter separately
Light — Δf/f = Δλ/λ ≈ v/c
- No medium; valid only for v ≪ c
- Gives the fractional shift, not f' directly
- Shorter λ (blueshift) ⇒ approaching
- Longer λ (redshift) ⇒ receding
- Only the relative speed v matters
| Situation | Sign / shift | Effect on detected wave |
|---|---|---|
| Sound source approaching | use − in (v ∓ us) | f' > f (higher pitch) |
| Sound source receding | use + in (v ∓ us) | f' < f (lower pitch) |
| Light source approaching | blueshift | λ shorter, f higher |
| Light source receding | redshift | λ longer, f lower |
| Observer moving towards a sound source | use + in (v ± uo) | f' > f (higher pitch) |
| Observer moving away from a sound source | use − in (v ± uo) | f' < f (lower pitch) |
Common slip: The minus/plus is in the denominator, not the numerator. A smaller denominator (the − case) makes the fraction bigger, so the approaching case correctly gives the higher frequency.
Where it shows up: Quantitative Doppler is HL only (C.5). It comes up in Paper 1A and Paper 2.
| Paper | What they ask |
|---|---|
| Paper 1A | A one-step 'is this a red- or blue-shift?' or 'which sign gives the higher frequency?' |
| Paper 2 | Calculate the observed frequency of a moving sound source, or estimate a recession speed from a measured Δλ/λ |
Three easy marks: (1) Choose the sign physically (approaching ⇒ higher) before substituting. (2) For light, work with the fraction Δλ/λ — don't reach for the sound formula. (3) Always quote the unit (Hz or m s⁻¹) and check the direction matches a red/blueshift.
IB-style question — a passing train whistle
A train whistle emits a steady 660 Hz tone. The train moves away from a stationary observer on the platform at 25 m s⁻¹. The speed of sound is 340 m s⁻¹. Determine the frequency the observer hears.
Solution
- Receding source ⇒ use the plus sign in the denominator:
- Substitute:
- Evaluate — keep the unit:
Final answer
f' = 615 Hz — lower than 660 Hz, as expected for a receding source.