IB DP Physics: Travelling waves and the Doppler effect (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topics C.2 (Wave model) and C.5 (Doppler effect) from the IB Physics guide (first assessment 2025). Questions 1 to 9 are SL and HL. Questions 10 and 11 use the formulas for sound that are additional higher level, so they are marked HL only. Where a value is needed it is given in the question. Teaching is in the study guide and the revision notes. The course hub and printable checklist show where this unit sits. Not sure where to start? Try the free 10-minute diagnostics.

Questions

1. A wave travels along a rope. Its frequency is 3.2 Hz and its wavelength is 0.75 m. Calculate the period and the speed of the wave. [3]

2. A loudspeaker emits a note of frequency 2.5 kHz. The speed of sound in air is 340 m s⁻¹ and in water 1480 m s⁻¹. The note travels from air into water. Calculate the wavelength in air and in water, and state what happens to the frequency. [4]

3. A longitudinal wave travels in the positive x direction through a gas. At one instant the displacement of the particles, taken as positive in the direction of travel, is 0, +2.0 mm, 0, −2.0 mm and 0 at x = 0, 0.1, 0.2, 0.3 and 0.4 m. The wavelength is 0.4 m.

(a) Identify the position of the centre of a compression and the centre of a rarefaction. [2] (b) Explain why there is no overall movement of the gas along the direction of travel although energy is transferred. [2]

4. A radio station transmits at 101.3 MHz. Use c = 3.00 × 10⁸ m s⁻¹.

(a) Calculate the wavelength of the radio wave. [2] (b) Give three ways in which this wave differs from a sound wave. [3]

5. A transverse wave travels in the positive x direction along a rope at 6.0 m s⁻¹. The wavelength is 1.5 m and the amplitude is 2.0 cm. At t = 0 the displacement is y = 2.0 sin(2πx/1.5) cm.

(a) Calculate the period of the wave. [2] (b) State the displacement of the rope at x = 0.375 m at t = 0. [1] (c) Find the displacement at x = 0.375 m after one quarter of a period, and state whether the rope at that point is moving up or down. [2]

6. A police car with its siren on drives past you at constant speed. Use wavefronts to explain why you hear a higher pitch as it approaches and a lower pitch after it passes. [4]

7. The sodium line of rest wavelength 589.0 nm is observed at 593.1 nm in the light from a galaxy. Use c = 3.00 × 10⁸ m s⁻¹. Calculate the speed of the galaxy along the line of sight and state whether it is moving towards or away from Earth. [4]

8. In the spectrum of a star, a line of rest wavelength 434.0 nm is observed to move repeatedly between 433.80 nm and 434.20 nm.

(a) Suggest an explanation for the repeated shift. [2] (b) Calculate the maximum speed of the star along the line of sight. [3]

9. Doppler ultrasound is used to examine blood flow and radar is used to check vehicle speeds.

(a) Describe how Doppler ultrasound gives information about the flow of blood. [2] (b) State what a radar speed gun compares to find the speed of a vehicle. [2]

10. (HL only) A lorry sounds a horn of frequency 900 Hz as it drives at 18 m s⁻¹ along a straight road past a stationary observer. The speed of sound is 340 m s⁻¹.

(a) Calculate the frequency heard as the lorry approaches. [2] (b) Calculate the frequency heard as it moves away. [2] (c) Calculate the wavelength of the sound reaching the observer ahead of the lorry. [2]

11. (HL only) A stationary source emits sound of frequency 660 Hz. The speed of sound is 340 m s⁻¹.

(a) A cyclist rides towards the source at 12 m s⁻¹. Calculate the frequency heard. [2] (b) A second cyclist riding towards the source hears 700 Hz. Calculate the speed of this cyclist. [3]

Answers

1. T = 1/f = 1/3.2 [1] = 0.313 s [1]. v = fλ = 3.2 × 0.75 = 2.4 m s⁻¹ [1] [3] Examiner insight: A unit and a sensible number of significant figures are expected on both values; showing v = fλ with the numbers substituted lets a slip in arithmetic be seen for what it is.

2. In air: λ = v/f = 340/2500 [1] = 0.136 m [1]. The frequency stays the same, 2.5 kHz, because it is set by the source [1]. In water: λ = 1480/2500 = 0.592 m [1] [4] Examiner insight: Convert kHz to Hz before substituting. Saying that frequency is unchanged, with the reason, is the point that is easiest to leave out.

3. (a) Compression centre at x = 0.2 m, because particles on either side are displaced towards it [1]. Rarefaction centre at x = 0 (or 0.4 m) [1] (b) Each particle vibrates about its own rest position [1]. The disturbance passes along as energy, so the particles return to where they began and there is no overall displacement of the gas [1] [4] Examiner insight: A compression sits where the displacement is zero, not where it is largest. State the neighbours’ displacement directions to justify the position.

4. (a) λ = c/f = 3.00 × 10⁸ / 101.3 × 10⁶ [1] = 2.96 m [1] (b) Any three of these: the radio wave needs no medium while sound does [1]; the radio wave is transverse while sound is longitudinal [1]; the radio wave travels at c in a vacuum while the speed of sound depends on the medium [1] [5] Examiner insight: Each difference has to be stated for both waves. Writing only “radio is faster” does not make a contrast that can be credited.

5. (a) T = λ/v = 1.5/6.0 [1] = 0.25 s [1] (b) y = 2.0 sin(2π × 0.375/1.5) = 2.0 sin(π/2) = +2.0 cm [1] (c) A quarter period is 0.0625 s, in which the pattern moves 6.0 × 0.0625 = 0.375 m to the right, so the displacement at x = 0.375 m is the old value at x = 0, which is 0 [1]. The particle has gone from +2.0 cm to zero, so it is moving down [1] [5] Examiner insight: Show how the pattern shifts by comparing with the earlier profile. A direction with no supporting reasoning is a weak answer.

6. Each wavefront is emitted from a different point along the car’s path, so the wavefronts are not concentric [1]. Ahead of the car they are closer together, so the wavelength is shorter [1]. The speed of sound is unchanged, so f = v/λ is higher as it approaches [1]. After it passes you, the wavefronts reaching you are further apart, so the wavelength is longer and the frequency lower [1] [4] Examiner insight: A good answer links spacing to wavelength and then to frequency through the unchanged wave speed. Saying the sound speed changes is a common error that loses credit.

7. Δλ = 593.1 − 589.0 = 4.1 nm [1]. v = cΔλ/λ [1] = 3.00 × 10⁸ × 4.1/589.0 = 2.09 × 10⁶ m s⁻¹ [1]. The observed wavelength is longer (redshift), so the galaxy is moving away [1] [4] Examiner insight: Use the rest wavelength in the denominator. The direction is a separate point and needs the redshift reason, not only the word “away”.

8. (a) The star is moving in orbit (for example around another star) [1], so it alternately moves towards and away from us, shifting the line to shorter and then longer wavelength [1] (b) Maximum Δλ = 0.20 nm [1]. v = cΔλ/λ = 3.00 × 10⁸ × 0.20/434.0 [1] = 1.4 × 10⁵ m s⁻¹ [1] [5] Examiner insight: The greatest shift on one side of the rest wavelength gives the greatest speed. Quoting the full range of 0.40 nm would double the answer.

9. (a) High-frequency sound is reflected from moving blood [1]. The frequency of the echo differs from that emitted, and the size and sign of the shift give the speed and direction of the flow [1] (b) It compares the frequency of the microwaves it sends out with the frequency of the reflection [1]; a larger shift means a greater speed [1] [4] Examiner insight: Mention what is emitted, what is reflected and what is compared. Descriptions that only say “the Doppler effect is used” gain little.

10. (a) f′ = f v/(v − us) [1] = 900 × 340/(340 − 18) = 900 × 340/322 = 950 Hz [1] (b) f′ = 900 × 340/(340 + 18) [1] = 900 × 340/358 = 855 Hz [1] (c) λ′ = (v − us)/f = 322/900 [1] = 0.358 m [1] [6] Examiner insight: Choose the sign by checking the outcome: approaching must give a frequency above 900 Hz. A frequency below the source value for approach is a signal to recheck the sign.

11. (a) f′ = f(v + uo)/v [1] = 660 × (340 + 12)/340 = 683 Hz [1] (b) 700 = 660 × (340 + uo)/340 [1]. 340 + uo = 700 × 340/660 = 360.6 [1]. uo = 20.6 m s⁻¹ [1] [5] Examiner insight: Rearrange before substituting and show the intermediate value. Using the source formula for a moving observer gives a different and incorrect answer.

Where marks are usually lost

• Leaving kHz or MHz unconverted in v = fλ.
• Saying the medium is carried forward by the wave rather than vibrating about its rest position.
• Putting a compression at the largest displacement instead of at zero displacement with neighbours displaced towards it.
• Writing that frequency changes at a boundary.
• Explaining the Doppler effect by a change in the speed of sound rather than a change in wavelength or arrival rate.
• Using the observed wavelength instead of the rest wavelength in Δλ/λ.
• Stating “redshift” without saying the source is moving away.
• Choosing the wrong sign in v ± us or v ± uo.
• Leaving out units or giving a speed to too many significant figures.

Next steps

• Recall the formulas in the revision notes.
• Re-read the explanations in the study guide.
• Open the IB DP Physics course hub and tick off the printable checklist.
• Take the free 10-minute diagnostics to find gaps.
• Or book a free trial class.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topics C.2 Wave model and C.5 Doppler effect.