Free C.5 Doppler effect questions for IB Physics (SL & HL), part of C Wave behaviour. Each question has the answer and the marking points an examiner would look for.
Generate a full C.5 Doppler Effect worksheet →All C Wave behaviour questionsThe hydrogen line of wavelength 656.3 nm in the laboratory is observed at 662.9 nm in the spectrum of a galaxy.
What is the speed of the galaxy relative to the Earth?
Answer: C
An ambulance siren emits sound of frequency 800 Hz. The speed of sound is 340 m s−1.
The ambulance moves at 30 m s−1 towards a stationary observer. Calculate the frequency heard.
Calculate the frequency heard after the ambulance has passed.
The ambulance is now stationary and the observer moves towards it at 30 m s−1. Calculate the frequency heard.
Explain why the answers to (a) and (c) differ.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 880 Hz f′ = 800 × 340/(340 − 30) ✓ = 877 Hz ✓ | 2 |
| (b) | 740 Hz f′ = 800 × 340/370 ✓ = 735 Hz ✓ | 2 |
| (c) | 870 Hz f′ = 800 × 370/340 ✓ = 871 Hz ✓ | 2 |
| (d) | source changes λ; observer changes rate a moving source changes the wavelength in the air ✓ a moving observer changes the rate at which wavefronts are received but not the wavelength ✓ | 2 |
A police car siren emits 900 Hz. The car moves at 30 m s−1. Speed of sound = 340 m s−1.
Calculate the frequency heard by a stationary observer as the car approaches.
Calculate the frequency heard as the car moves away.
Explain why the observed frequency changes suddenly as the car passes.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 987 Hz f′ = f v/(v − u) ✓ 987 Hz ✓ | 2 |
| (b) | 827 Hz 827 Hz ✓ | 1 |
| (c) | source velocity component reverses the velocity of the source relative to the observer changes from towards to away ✓ | 1 |
A hydrogen line of wavelength 656.3 nm in the laboratory is observed at 662.9 nm in the light from a distant galaxy.
Calculate the speed of the galaxy relative to the Earth.
State whether the galaxy is moving towards or away from the Earth.
State why this equation is valid only for speeds much less than c.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 3.0 × 106 m s−1 v = cΔλ/λ ✓ = 3.0 × 106 m s−1 ✓ | 2 |
| (b) | away away (red-shift) ✓ | 1 |
| (c) | non-relativistic it is the non-relativistic approximation of the Doppler shift ✓ | 1 |
An ambulance siren emits sound of frequency 700 Hz. The ambulance moves at 25 m s−1 past a stationary observer. The speed of sound is 340 m s−1.
Explain, with a diagram of wavefronts, why the observer hears a higher frequency as the ambulance approaches.
Calculate the frequency heard as the ambulance approaches and as it recedes.
A spectral line of wavelength 656.3 nm in the laboratory is observed at 676.0 nm in the light from a galaxy. Calculate the speed of the galaxy and state its direction of motion.
State why the sound equation cannot be used for light, and the condition for the light equation to be valid.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | wavefronts compressed wavefronts are closer together in front of the moving source ✓ shorter wavelength, so higher frequency at the observer (speed unchanged) ✓ | 2 |
| (b) | 756 Hz; 652 Hz approaching: f′ = fvv − us = 756 Hz ✓ receding: f′ = fvv + us ✓ = 652 Hz ✓ | 3 |
| (c) | 9.0 × 106 m s−1, away Δλ = 19.7 nm ✓ v = Δλλc = 9.0 × 106 m s−1 ✓ moving away from the Earth (redshift) ✓ | 3 |
| (d) | no medium; v ≪ c light needs no medium; only the relative speed matters ✓ valid only when v ≪ c ✓ | 2 |
A sound detector moves along a straight line joining it to a stationary loudspeaker that emits 1700 Hz. The detector records 1580 Hz. The speed of sound is 340 m s−1.
State the direction of motion of the detector.
Calculate the speed of the detector.
A spectral line of wavelength 656.3 nm in the laboratory is observed at 660.2 nm in the light from a galaxy. Calculate the speed of the galaxy.
Outline why the Doppler equation for light is the same for a moving source and a moving observer, but not for sound.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | away from the loudspeaker away from the loudspeaker ✓ | 1 |
| (b) | 25 m s−1 f′ = f(v − uo)/v ✓ uo = 25 m s−1 ✓ | 2 |
| (c) | 1.8 × 106 m s−1 Δλ/λ = v/c ✓ v = 1.8 × 106 m s−1 away from Earth ✓ | 2 |
| (d) | light needs no medium sound travels through a medium, so the motion relative to the medium matters ✓ light needs no medium, only the relative speed matters ✓ | 2 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in