C.2 Wave Model Questions | IB Physics HL 2025

1C-1A-12

Intensity and amplitude·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A small loudspeaker radiates sound uniformly in all directions. At a distance of 2.0 m the amplitude of the pressure variation is A. The intensity of a wave is proportional to the square of its amplitude.

What is the amplitude at a distance of 6.0 m?

Show mark scheme

Marking point	Mark	Notes
Step 1For a point source the intensity obeys an inverse-square law: I ∝ 1/r², so tripling the distance divides the intensity by 9.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Intensity is proportional to amplitude squared: I ∝ A².	—
Step 3So the amplitude falls by √9 = 3: new amplitude = A/3.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis applies the inverse-square law to the amplitude. It is the intensity that falls by 9; the amplitude falls by √9.
BCorrect: the power spreads over a sphere, so the intensity falls by (6.0/2.0)² = 9, and the amplitude by √9 = 3.
CThis takes the intensity to fall in proportion to 1/r (by 3), as if the sound spread over a circle rather than a sphere.
DThis assumes the amplitude is conserved as the wave travels. The energy is conserved, but it spreads over a larger area, so the intensity and the amplitude both fall.

Syllabus understandingC.2 — the nature of sound waves (Linking question: why does the intensity of a wave decrease with distance according to the inverse square law?); B.1 — b = L/4πd² Command term: Determine

2C-1A-14

The wave equation·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A transverse wave travels along a rope. The distance between a crest and the adjacent trough is d. A point on the rope takes a time t to move from its maximum displacement to its equilibrium position.

What is the speed of the wave?

Show mark scheme

Marking point	Mark	Notes
Step 1A crest and the adjacent trough are half a wavelength apart: λ = 2d.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Maximum displacement to equilibrium is a quarter of a cycle: T = 4t.	—
Step 3v = λ/T = 2d/(4t) = d/(2t).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis takes d as the whole wavelength (λ = d) with T = 4t.
BCorrect: λ = 2d and T = 4t, so v = d/(2t).
CThis takes the time from maximum displacement to equilibrium as half a period (T = 2t), with λ = 2d.
DThis takes t as the whole period, with λ = 2d.

Syllabus understandingC.2 — the wave equation v = fλ = λ/T; the motion of the particles of a medium as a wave passes, in terms of displacement with position and with time Command term: Determine

3C-1A-23

Electromagnetic and mechanical waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

Which statement is correct for both sound waves in air and electromagnetic waves in a vacuum?

Show mark scheme

Marking point	Mark	Notes
Step 1Sound in air is longitudinal (oscillations of air molecules parallel to the direction of travel); electromagnetic waves are transverse oscillations of electric and magnetic fields.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Sound is a mechanical wave and needs a medium; electromagnetic waves travel through a vacuum, at c = 3.0 × 10⁸ m s⁻¹, far faster than sound (≈ 340 m s⁻¹).	—
Step 3What all travelling waves share is that they carry energy from one place to another while the medium (if any) has no resultant displacement.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

ASound in air is longitudinal, not transverse.
BElectromagnetic waves need no medium — light reaches us from the Sun through the vacuum of space.
COnly electromagnetic waves travel at c; sound in air travels at about 340 m s⁻¹.
DCorrect: every travelling wave transfers energy without transferring matter.

Syllabus understandingC.2 — transverse and longitudinal travelling waves; the nature of sound waves; the nature of electromagnetic waves; the differences between mechanical and electromagnetic waves; travelling waves transfer energy without a resultant displacement of the medium Command term: Identify

4C-1A-29

Wave graphs·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Graph 1 shows the displacement of the particles of a medium against position at one instant, and graph 2 shows the displacement of one particle against time, for the same wave.

What is the speed of the wave?

Two graphs describing the same transverse wave.

Show mark scheme

Marking point	Mark	Notes
Step 1From graph 1 (against position) the wavelength is the repeat distance: λ = 4.0 cm = 0.040 m.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From graph 2 (against time) the period is the repeat time: T = 0.20 s, so f = 5.0 Hz.	—
Step 3v = fλ = 5.0 × 0.040 = 0.20 m s⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is T/λ with the wavelength left in cm (0.20/4.0) — the ratio is inverted and the units are inconsistent.
BCorrect: v = λ/T = 0.040/0.20 = 0.20 m s⁻¹.
CThis uses 8.0 cm, the full length of the position axis (two wavelengths), as the wavelength: 0.080/0.20 = 0.40 m s⁻¹.
DThis is the frequency in Hz, not the speed.

Syllabus understandingC.2 — graphs of displacement against position and against time; v = fλ = λ/T Command term: Determine

5C-1A-37

Electromagnetic and mechanical waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A microwave signal of frequency 10 GHz travels through a vacuum. An ultrasonic wave travels through water, in which the speed of sound is 1500 m s⁻¹. The ultrasonic wave has the same wavelength as the microwave signal.

What is the frequency of the ultrasonic wave?

Show mark scheme

Marking point	Mark	Notes
Step 1For the microwave: λ = c/f = 3.00 × 10⁸/1.0 × 10¹⁰ = 0.030 m.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The ultrasound is a mechanical wave and travels at only 1500 m s⁻¹, but has the same wavelength.	—
Step 3f = v/λ = 1500/0.030 = 5.0 × 10⁴ Hz.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses the wavelength in millimetres (30) without converting it to metres: 1500/30 = 50 Hz.
BCorrect: λ = 0.030 m, so f = 1500/0.030 = 5.0 × 10⁴ Hz.
CThis assumes the two waves have the same frequency; only their wavelengths are equal, and their speeds differ by a factor of 2 × 10⁵.
DThis inverts the ratio of the speeds: 1.0 × 10¹⁰ × (3.00 × 10⁸/1500) = 2.0 × 10¹⁵ Hz.

Syllabus understandingC.2 — the nature of sound waves; the nature of electromagnetic waves; the differences between mechanical and electromagnetic waves; the wave equation v = fλ Command term: Determine

6C-1A-38

Longitudinal waves·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A longitudinal wave travels along a long spring. The graph shows, at one instant, the displacement s of the coils against their equilibrium position x. A positive displacement is to the right.

Which point is at the centre of a rarefaction?

Displacement of the particles against their equilibrium position at one instant (positive = displaced to the right). Drawn to scale.

Show mark scheme

Marking point	Mark	Notes
Step 1At a rarefaction the coils are furthest apart: the coils on its left are displaced to the left (s < 0) and those on its right are displaced to the right (s > 0).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2This happens where s = 0 and the graph has a positive gradient — point Q (x = 20 cm).	—
Step 3At S the coils on either side are displaced towards S (s > 0 on the left, s < 0 on the right): S is the centre of a compression.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AP has the largest displacement to the left, but its neighbours are displaced by almost the same amount, so the spacing there is normal.
BCorrect: the coils on either side of Q are displaced away from it, so the coils are spread out.
CR has the largest displacement to the right; a maximum displacement does not mean a compression or rarefaction.
DAt S the neighbouring coils move towards S: this is a compression, not a rarefaction.

Syllabus understandingC.2 — transverse and longitudinal travelling waves (Guidance: the motion of particles of a medium in terms of displacement with respect to position along the wave) Command term: Identify

7C-1A-39

Wave graphs·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A transverse wave travels to the right along a rope. The graph shows the displacement of the rope against position at one instant. X and Y are points on the rope.

Which row describes the motion of X and of Y at this instant?

A transverse wave on a rope at one instant, travelling to the right (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1A short time later the whole waveform has shifted slightly to the right. The part of the rope just to the left of X is below the axis, so X will move down.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2X is at zero displacement, where the speed of an oscillating point is greatest.	—
Step 3Y is at a crest (maximum displacement), where it is momentarily at rest before moving down.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: X moves down at maximum speed; Y is instantaneously at rest at maximum displacement.
BY is at maximum displacement, where the speed is zero; zero displacement (not maximum displacement) is where the speed is greatest.
CThis is the direction for a wave travelling to the left; for a wave travelling to the right X takes on the displacement of the rope to its left, which is below it.
DIn a transverse wave the rope moves perpendicular to the direction of travel; the rope itself does not travel with the wave.

Syllabus understandingC.2 — transverse and longitudinal travelling waves (Guidance: the motion of particles of a medium when a wave passes through it, in terms of displacement with respect to position along the wave and with time) Command term: Deduce

8C-1A-40

Longitudinal waves·C.2 Wave model

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A sound wave travels through air. The graph shows how the displacement s of one air molecule M varies with time. A positive displacement is in the direction in which the sound travels, and T is the period.

At which time is M at the centre of a compression?

Displacement of one air molecule against time; time is shown as a multiple of the period T. Positive displacement is in the direction in which the sound travels.

Show mark scheme

Marking point	Mark	Notes
Step 1At the centre of a compression (or rarefaction) the molecule itself has zero displacement, so only T/2 and T are possible.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2As the wave moves forward, M takes on the displacement of the molecules just behind it. At t = T M is moving forwards (displacement increasing), so the molecules just behind it are displaced forwards, towards M; the molecules just ahead of M were displaced backwards, also towards M.	—
Step 3Molecules on both sides are displaced towards M: M is at the centre of a compression at t = T. (At T/2 M moves backwards and is at the centre of a rarefaction.)	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AAt T/4 M has its maximum forward displacement; its neighbours are displaced by nearly the same amount, so the air there is at normal density.
BAt T/2 M has zero displacement but is moving backwards, against the direction of travel: this is the centre of a rarefaction.
CAt 3T/4 M has its maximum backward displacement; this is not a compression or rarefaction.
DCorrect: zero displacement while moving in the direction of travel of the wave marks the centre of a compression.

Syllabus understandingC.2 — the nature of sound waves; transverse and longitudinal travelling waves (Guidance: displacement with respect to position along the wave and with time) Command term: Deduce

9C-1A-64

Longitudinal waves·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A longitudinal wave travels in the positive x direction along a long spring. The graph shows, at one instant, the displacement s of the coils against their equilibrium position x. A positive displacement is in the positive x direction. P is a coil at x = 10 cm.

Which statements about P at this instant are correct?

I. P is momentarily at rest.

II. P is at the centre of a compression.

III. The acceleration of P has its maximum magnitude and is in the negative x direction.

Displacement s of the coils against their equilibrium position x at one instant (drawn to scale). The wave travels in the positive x direction.

Show mark scheme

Marking point	Mark	Notes
Step 1P has its maximum displacement, so it is at an extreme of its oscillation: its velocity is zero (I correct).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2In SHM a = −ω²s, so at the maximum positive displacement the acceleration has its maximum magnitude and points in the negative x direction (III correct).	—
Step 3The coils on either side of P (at 8 cm and 12 cm, say) have almost the same displacement as P, so the spacing around P is the normal spacing. The centre of a compression is where neighbouring coils are displaced towards each other, at x = 20 cm, where s = 0 and the graph has its steepest negative gradient (II incorrect).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AII is incorrect: at the maximum displacement the coils near P are displaced by almost the same amount, so the spacing there is normal, not compressed.
BCorrect: P is at rest at an extreme of its oscillation and its acceleration −ω²s is greatest in magnitude and in the negative x direction.
CI is correct: a coil at maximum displacement is momentarily at rest. II is incorrect.
DThis treats the maximum of the displacement graph as a compression, as if it were a picture of a transverse wave. II is incorrect.

Syllabus understandingC.2 — transverse and longitudinal travelling waves; the motion of the particles of a medium in terms of displacement with position along the wave; C.1 — the defining equation of SHM a = −ω²x Command term: Deduce

10C-1A-72

Wave graphs·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A transverse wave travels in the positive x direction along a rope. The graph shows the displacement y of the rope against position x at time t = 0 (solid line) and at t = 0.010 s (dashed line). In this time the wave moves less than one wavelength.

What is the frequency of the wave?

Displacement y against position x at t = 0 (solid) and t = 0.010 s (dashed), drawn to scale.

Show mark scheme

Marking point	Mark	Notes
Step 1From the graph λ = 12 cm, and each crest moves 3.0 cm in the positive x direction in 0.010 s.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2v = 0.030/0.010 = 3.0 m s⁻¹.	—
Step 3f = v/λ = 3.0/0.12 = 25 Hz.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: v = 3.0 m s⁻¹ and λ = 0.12 m, so f = 25 Hz.
BThis takes the distance from a crest to the adjacent trough (6 cm) as the wavelength.
CThis reads the shift in the wrong direction (9.0 cm, the distance to the next dashed crest to the left): 9.0 m s⁻¹/0.12 m = 75 Hz.
DThis treats 0.010 s as the period of the wave.

Syllabus understandingC.2 — wavelength, frequency, time period and wave speed applied to wave motion as given by v = fλ = λ/T; displacement with respect to position along the wave Command term: Determine

11C-1A-86

Electromagnetic waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A radio wave of wavelength 1.5 km and a gamma ray of wavelength 1.5 pm both travel through a vacuum.

Which quantities are the same for both waves?

I. the speed of the wave

II. the angle between the direction in which the electric field oscillates and the direction in which the wave travels

III. the number of oscillations per second

Show mark scheme

Marking point	Mark	Notes
Step 1All electromagnetic waves travel through a vacuum at the same speed c = 3.00 × 10⁸ m s⁻¹, whatever their wavelength (I is the same).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2All electromagnetic waves are transverse: the electric (and magnetic) field oscillates at 90° to the direction of travel (II is the same).	—
Step 3The frequency is c/λ: 2 × 10⁵ Hz for the radio wave and 2 × 10²⁰ Hz for the gamma ray, a factor of 10¹⁵ apart (III differs).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the speed and the transverse nature are shared by all electromagnetic waves in a vacuum; the frequencies differ by the same factor as the wavelengths.
BThis assumes the frequency is the same because the speed is the same. With c fixed, f = c/λ, so a wavelength 10¹⁵ times shorter means a frequency 10¹⁵ times higher.
CThis assumes that the more energetic gamma ray travels faster than the radio wave. In a vacuum every electromagnetic wave travels at c.
DThis treats all electromagnetic waves as identical apart from their wavelength, forgetting that c = fλ links a different wavelength to a different frequency.

Syllabus understandingC.2 — the nature of electromagnetic waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; Guidance: the approximate orders of magnitude of the wavelengths of radio, microwave, infrared, visible, ultraviolet, X-rays and gamma rays (data booklet) Command term: Deduce

12C-1A-87

Particle motion and wave speed·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A transverse wave of amplitude A and wavelength λ travels along a rope. In a time t, a point on the rope moves through a total distance D, measured along its path. D is much greater than A.

What is the speed of the wave?

Show mark scheme

Marking point	Mark	Notes
Step 1In one period the point moves from equilibrium to +A, back to equilibrium, to −A and back again: a total distance of 4A.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The number of oscillations in time t is D/(4A), so the period is T = 4At/D.	—
Step 3v = λ/T = λD/(4At).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis inverts the count of oscillations, taking it as 4A/D, so that T = tD/(4A) and v = 4Aλ/(Dt).
BCorrect: the point travels 4A in each period, so T = 4At/D and v = λD/(4At).
CThis takes the distance moved in one period as 2A (crest to trough only), so the period found is half the true period.
DThis takes the distance moved in one period as A, the amplitude, so the period found is a quarter of the true period.

Syllabus understandingC.2 — wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; Guidance: the motion of particles of a medium when a wave passes through it, in terms of displacement with respect to the position along the wave and with time Command term: Determine

13C-1A-88

The wave equation·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The left-hand end of a long rope, initially at rest, is made to oscillate with frequency f. By the time the front of the wave has travelled a distance L along the rope, the end has completed n oscillations.

What is the wavelength of the wave?

Show mark scheme

Marking point	Mark	Notes
Step 1During each complete oscillation of the end, the wave advances by one wavelength.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2So n oscillations correspond to n wavelengths: nλ = L.	—
Step 3λ = L/n; the frequency is not needed.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the relationship, as if each wavelength contained n oscillations.
BThis is the speed of the wave, not its wavelength: the time taken is n/f, so v = Lf/n.
CThis finds L/n correctly and then divides by f again, applying λ = v/f to a quantity that is already a wavelength.
DCorrect: the wave moves forward one wavelength for every oscillation of the source, so λ = L/n.

Syllabus understandingC.2 — wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; transverse and longitudinal travelling waves Command term: Determine

14C-1A-89

Wavelength, frequency and medium·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A vibrator in a ripple tank produces straight water waves of wavelength λ. The frequency of the vibrator is doubled, and the waves then pass into a region of deeper water where they travel 1.5 times faster than in the original region.

What is the wavelength of the waves in the deeper region?

Show mark scheme

Marking point	Mark	Notes
Step 1The frequency is set by the vibrator and does not change when the waves enter deeper water: it is 2f everywhere.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2λ = v/f, so λ ∝ v/f.	—
Step 3New wavelength = (1.5/2)λ = 0.75λ.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis multiplies by 1/(2 × 1.5), treating the wavelength as inversely proportional to the speed as well as to the frequency.
BCorrect: doubling f halves λ, and the 1.5 times greater speed then increases it by 1.5: 0.75λ.
CThis uses λ ∝ f/v, the inverse of the wave equation (2/1.5 = 1.3).
DThis multiplies by both factors (2 × 1.5), treating the wavelength as proportional to the frequency.

Syllabus understandingC.2 — wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; the guiding question: the effect of a change in the frequency of oscillation or in the medium on the wavelength of a travelling wave Command term: Determine

15C-1A-90

Longitudinal waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A vibrator at the left-hand end of a long spring produces a longitudinal wave of frequency 3.0 Hz. The diagram shows the spring at one instant. P and Q are at the centres of compressions, 0.80 m apart.

What is the speed of the wave?

A long spring carrying a longitudinal wave, at one instant. P and Q are at the centres of compressions (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1Between P and Q there is a third compression: P and Q are two wavelengths apart, so λ = 0.80/2 = 0.40 m.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2v = fλ = 3.0 × 0.40.	—
Step 3v = 1.2 m s⁻¹.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis divides the wavelength by the frequency (λ/f = 0.40/3.0) instead of multiplying.
BThis takes the distance from a compression to the next rarefaction (0.20 m, half a wavelength) as the wavelength.
CCorrect: PQ contains two whole wavelengths, so λ = 0.40 m and v = 3.0 × 0.40 = 1.2 m s⁻¹.
DThis takes PQ = 0.80 m as one wavelength, overlooking the compression between P and Q.

Syllabus understandingC.2 — transverse and longitudinal travelling waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T Command term: Determine

16C-1A-91

Particle velocity and wave speed·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A transverse wave of wavelength 0.60 m travels along a rope. The graph shows how the velocity of one point on the rope varies with time.

What is the speed of the wave?

Velocity of one point on the rope against time (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The velocity of the point repeats every 0.20 s, so the period of the wave is T = 0.20 s (two cycles are shown).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The wave speed is the speed of the waveform along the rope: v = λ/T.	—
Step 3v = 0.60/0.20 = 3.0 m s⁻¹.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is the maximum speed of the point as it oscillates across the rope, not the speed at which the wave travels along it.
BThis takes the whole time axis, 0.40 s, which contains two cycles, as the period.
CCorrect: T = 0.20 s from the graph, so v = 0.60/0.20 = 3.0 m s⁻¹.
DThis takes the time between successive zeros of velocity, 0.10 s, as the period; that is half a period.

17C-1A-92

Mechanical and electromagnetic waves·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A sound wave and a light wave both pass from air into a block of glass.

Which statement describes the change in wavelength of each wave as it enters the glass?

Show mark scheme

Marking point	Mark	Notes
Step 1For both waves the frequency is set by the source and does not change at the boundary, so λ ∝ v.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Sound is a mechanical wave passed on by the particles of the medium; it travels much faster in a stiff solid such as glass (thousands of m s⁻¹) than in air (about 340 m s⁻¹), so its wavelength increases.	—
Step 3Light is an electromagnetic wave; it travels more slowly in glass than in air (c/n), so its wavelength decreases.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: sound speeds up in the solid and light slows down, and at constant frequency the wavelength changes in the same way as the speed.
BThis assumes that every wave slows down in a denser medium. That is true for light but not for sound, which travels faster in solids.
CThis assumes that light, like sound, travels faster in a solid than in air.
DThis reverses both changes: it treats sound as slowing down and light as speeding up in glass.

Syllabus understandingC.2 — the differences between mechanical waves and electromagnetic waves; the nature of sound waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; C.3 — n₁/n₂ = v₂/v₁ Command term: Deduce

18C-1A-93

Wave graphs·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A transverse wave travels along a rope that lies along the x-axis. Graph 1 shows the displacement y of the rope against position x at time t = 0. Graph 2 shows the displacement of point P on the rope against time, starting at t = 0.

Which row gives the direction of travel of the wave and its speed?

Graph 1 (top): displacement against position at t = 0. Graph 2 (bottom): displacement of P against time. Both drawn to scale.

Direction of travelSpeed

Show mark scheme

Marking point	Mark	Notes
Step 1Graph 2: at t = 0 P is at zero displacement and then moves upwards; the period is T = 0.25 s.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Graph 1: the rope just to the left of P is above the axis and the rope just to the right is below. P moves up only if the profile moves to the right, bringing the raised part of the rope to P, so the wave travels in the positive x direction.	—
Step 3λ = 0.80 m from graph 1, so v = λ/T = 0.80/0.25 = 3.2 m s⁻¹.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis reverses the direction rule and inverts the speed (T/λ = 0.25/0.80).
BThe direction is right but the speed is inverted: 0.25/0.80 = 0.31 is T/λ, in s m⁻¹.
CThe speed is right but the direction is reversed: a wave travelling in the negative x direction would bring the lowered rope on the right to P, so P would move down first.
DCorrect: P moves up first, so the raised rope on its left is moving towards it (positive x), and v = 0.80/0.25 = 3.2 m s⁻¹.

Syllabus understandingC.2 — transverse and longitudinal travelling waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; Guidance: the motion of particles of a medium when a wave passes through it, in terms of displacement with respect to the position along the wave and with time Command term: Deduce

19C-1A-94

Particle motion and wave speed·C.2 Wave model

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The left-hand end of a long horizontal rope, initially at rest, is attached to a vibrator. At t = 0 the vibrator starts to oscillate vertically with frequency 2.5 Hz, its first movement being upwards from the equilibrium position. The waves on the rope have a wavelength of 0.80 m. Point P on the rope is 0.60 m from the vibrator. There are no reflections.

Which describes P at t = 0.50 s?

Show mark scheme

Marking point	Mark	Notes
Step 1v = fλ = 2.5 × 0.80 = 2.0 m s⁻¹, so the wave front reaches P at t = 0.60/2.0 = 0.30 s; P is at rest until then.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P then copies the motion of the vibrator: it starts by moving upwards. T = 1/2.5 = 0.40 s.	—
Step 3By t = 0.50 s P has oscillated for 0.20 s = T/2: it has gone up and come back to equilibrium, and is now moving downwards.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis ignores the time taken for the wave to reach P: 0.50 s is 1.25 periods of the vibrator itself, which is then at its highest point.
BCorrect: P starts to move at 0.30 s and has completed half an oscillation by 0.50 s, so it is passing through equilibrium moving down.
CThis uses the travel time, 0.30 s (three-quarters of a period), as the time for which P has been oscillating.
DThis inverts the wave equation, v = λ/f = 0.32 m s⁻¹, so the wave would not reach P until 1.9 s.

20C-1A-95

Energy transfer by waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A transverse wave of wavelength λ travels along a rope. Consider one point on the rope over one complete period of the wave.

Which statements are correct?

I. The point moves a distance λ along the rope.

II. The resultant displacement of the point is zero.

III. Energy is transferred along the rope past the point.

Show mark scheme

Marking point	Mark	Notes
Step 1Each point of the rope oscillates at right angles to the rope about its equilibrium position; it does not travel along the rope (I is incorrect).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2After one complete period the point is back where it started: its resultant displacement is zero (II is correct).	—
Step 3The wave carries energy along the rope, each part doing work on the next, even though no material is carried along (III is correct).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis treats the wave as carrying the rope with it. It is the waveform that moves λ in one period, not the material.
BThis combines the misconception that the point travels λ with the correct energy statement, and misses that the point returns to its starting position.
CCorrect: the point only oscillates about its equilibrium position, so its resultant displacement is zero, while energy travels along the rope.
DI is incorrect: the rope does not move along the direction in which the wave travels.

Syllabus understandingC.2 — transverse and longitudinal travelling waves; Guidance: travelling waves transfer energy, even if there is no resultant displacement of the medium Command term: Deduce

21C-1A-96

Orders of magnitude of EM wavelengths·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksEstimate

A mobile phone transmits electromagnetic waves of frequency 1.8 GHz.

What is the order of magnitude of (wavelength of the phone signal) / (wavelength of visible light)?

Show mark scheme

Marking point	Mark	Notes
Step 1λ = c/f = 3.00 × 10⁸/1.8 × 10⁹ = 0.17 m (a microwave).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Visible light has wavelengths from about 400 nm to 700 nm, of order 5 × 10⁻⁷ m.	—
Step 3Ratio = 0.17/(5 × 10⁻⁷) ≈ 3 × 10⁵; for any visible wavelength the ratio lies between 2 × 10⁵ and 4 × 10⁵, so the order of magnitude is 10⁵.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio (visible light / phone signal).
BThis takes the wavelength of visible light as about 5 × 10⁻⁴ m (reading 500 nm as 500 μm), giving 0.17/(5 × 10⁻⁴) ≈ 3 × 10².
CCorrect: 0.17 m divided by about 5 × 10⁻⁷ m is about 3 × 10⁵.
DThis takes the wavelength of visible light as about 10⁻⁹ m, which is the X-ray region: 0.17/10⁻⁹ ≈ 2 × 10⁸.

22C-1A-97

Sound waves·C.2 Wave model

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A loudspeaker emits sound of frequency 500 Hz into still air.

Which statement describes the motion of an air molecule in front of the loudspeaker while the sound passes?

Show mark scheme

Marking point	Mark	Notes
Step 1Sound in air is a longitudinal wave: the molecules oscillate about their equilibrium positions along the direction of travel.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Every molecule is driven, through its neighbours, at the frequency of the source: 500 oscillations per second.	—
Step 3The molecules do not travel with the wave; the energy (the disturbance) travels at the speed of sound.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis treats the sound as a flow of air. The disturbance travels at the speed of sound; the molecules only oscillate about fixed positions.
BThis describes a transverse wave. Sound in air is longitudinal.
CAs the sound spreads out its intensity, and so its amplitude, decreases; its frequency does not change.
DCorrect: a longitudinal oscillation, parallel to the direction of travel, at the source frequency of 500 Hz.

Syllabus understandingC.2 — the nature of sound waves; transverse and longitudinal travelling waves; Guidance: the motion of particles of a medium when a wave passes through it, in terms of displacement with respect to the position along the wave and with time Command term: Identify

23C-1A-98

Pulses: time and position·C.2 Wave model

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A single transverse pulse travels at 3.0 m s⁻¹ in the positive x direction along a long rope. The graph shows the displacement y of a point P on the rope against time t as the pulse passes P.

Which row describes the shape of the pulse on the rope at an instant?

Displacement of point P against time as the pulse passes (drawn to scale).

Steeper side of the pulseLength of the pulse

Show mark scheme

Marking point	Mark	Notes
Step 1The part of the pulse that reaches P first is the part furthest ahead on the rope. P rises steeply first, so the steep side of the pulse is at the front.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The pulse takes 0.30 − 0.10 = 0.20 s to pass P.	—
Step 3Length on the rope = speed × time = 3.0 × 0.20 = 0.60 m.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the steep rise arrives first, so it is the leading edge, and the pulse is 3.0 × 0.20 = 0.60 m long.
BThis copies the time graph as if it were a picture of the rope: the steep side is on the left of the graph, but on the rope the earliest part to arrive is at the front.
CThe front is correct but the length is the time divided by the speed (0.20/3.0) instead of multiplied by it.
DThis copies the time graph as a picture of the rope and also divides the time by the speed.

24C-1B-06

Intensity and amplitude·C.2 Wave model

Paper 1BHard7 marks

Data-based question6 steps to full marksDeduce

A small filament lamp is placed in a darkened room. A radiation sensor, which responds to all wavelengths, measures the intensity I of the radiation at distance r from the filament. With the lamp switched off, the sensor reads 0.40 W m⁻² because of radiation from the warm surroundings; this reading does not change during the experiment.

The student wants to test whether the radiation from the filament obeys an inverse-square law.

r / m	0.200	0.300	0.400	0.500	0.600
Sensor reading / W m⁻²	24.47	10.91	6.42	4.25	3.03

(a)

(i)

Determine the intensity due to the lamp alone at r = 0.600 m.

(1)

(b)

(i)

Deduce whether the data support the inverse-square law.

(2)

(c)

(i)

Show that the total power radiated by the filament is about 12 W.

(1)

(d)

(i)

The filament has a surface area of 6.0 × 10⁻⁶ m² and an emissivity of 0.40. Determine the temperature of the filament.

(2)

(ii)

The power is uncertain by ±8 %. Determine the absolute uncertainty in the temperature.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
3.03 − 0.40 = 2.63 W m⁻²	✓ 1	Subtraction of the background must be seen.
Part (b)(i)
Ir² calculated, after subtracting the background, for at least three distances, e.g. 0.963, 0.963, 0.947 W	✓ 1	Allow ECF from (a). Using uncorrected readings gives 0.979, 1.027, 1.091 (rising): [1 max] for a correct conclusion from these.
The values are constant within ≈ 2 %, so I ∝ 1/r²: the law is supported	✓ 1	The conclusion must refer to the constancy of their product (or ratio).
Part (c)(i)
P = 4πr²I = 4π × 0.958 = 12.0 W	✓ 1	Must use a background-corrected value; allow ECF from (b). A 2-s.f. value or the full substitution is required.
Part (d)(i)
T = (P/(eσA))^1/4 = (12.0/(0.40 × 5.67 × 10⁻⁸ × 6.0 × 10⁻⁶))^1/4	✓ 1	Allow ECF from (c).
T = 3.06 × 10³ K	✓ 1	Accept 3.0–3.1 × 10³ K.
Part (d)(ii)
Percentage uncertainty in T = ¼ × 8 % = 2 %; ΔT = 0.02 × 3060 ≈ ±60 K	✓ 1	Absolute uncertainty to 1 s.f. Using 4 × 8 % = 32 % is a named error: [0]. Allow ECF from (d)(i).

Answers: (a)(i) 2.63 W m⁻² · (b)(i) Ir² ≈ 0.96 W (constant) · (d)(i) 3.06 × 10³ K · (d)(ii) ±60 K (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — the Stefan–Boltzmann law as given by L = σAT⁴; B.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature; C.2 (Linking question) — the inverse-square decrease of intensity with distance; Tools — taking background into account, testing a hypothesis, propagating uncertainty through a power Command term: Deduce

25C-1B-14

The wave equation·C.2 Wave model

Paper 1BEasy6 marks

Data-based question5 steps to full marksDetermine

A ripple tank is lit from above so that the wavefronts form an image on a screen below the tank. The image is 1.5 times larger than the real wave pattern. For each frequency f the student freezes the image with a stroboscope and measures the distance across five wavelengths, 5λ, on the screen with a ruler; each reading has an uncertainty of ±0.2 cm.

A graph of λ against 1/f is a straight line through the origin.

f / Hz	10	14	18	22	26	30
5λ / cm	11.9	8.6	6.7	5.4	4.6	3.9
λ / cm	2.38	1.72	1.34	1.08	0.92	0.78
f⁻¹ / s	0.100	0.0714	0.0556	0.0455	0.0385

(a)

(i)

Complete the table.

(1)

(b)

(i)

Using the full range of the data, determine the speed of the image of the waves on the screen.

(2)

(c)

(i)

Determine the speed of the water waves in the tank.

(1)

(d)

(i)

Calculate the percentage uncertainty in λ at 30 Hz. Suggest how the student could halve this uncertainty using the same ruler.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/30 = 0.0333 s	✓ 1	Accept 0.033.
Part (b)(i)
Gradient of λ against f⁻¹ is v: (2.38 − 0.78) × 10⁻²/(0.100 − 0.0333)	✓ 1	The 10 Hz and 30 Hz rows must be used (full range); use of the value from (a): allow ECF. Adjacent rows only: [1 max].
v = 0.24 m s⁻¹	✓ 1	Accept 0.23–0.25 m s⁻¹.
Part (c)(i)
Every wavelength on the screen is 1.5 times the real one while the frequency is unchanged, so v_tank = 0.24/1.5 = 0.16 m s⁻¹	✓ 1	Multiplying by 1.5: [0]. Allow ECF from (b).
Part (d)(i)
0.2/3.9 × 100 = 5 %	✓ 1	Accept 5.1 %.
Measure across 10 wavelengths (about 7.8 cm) instead of 5, so the same ±0.2 cm is a fraction half as large: ≈ 2.6 %	✓ 1	A change of instrument or "repeat readings" alone: [0] for this mark.

Answers: (a)(i) 0.0333 s · (b)(i) 0.24 m s⁻¹ · (c)(i) 0.16 m s⁻¹ · (d)(i) 5 %; measure 10λ (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — wave speed v = fλ = λ/T applied to mechanical waves; Tools — using the full data range, correcting a systematic (scale) error, reducing a percentage uncertainty with the same instrument Command term: Determine

26C-1B-15

Sound waves·C.2 Wave model

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A loudspeaker emits sound of frequency 2500 Hz. Two small microphones lie on a straight line in front of it. One is fixed and the other is moved away from the loudspeaker so that their separation d increases. Both signals are displayed on an oscilloscope, and the phase difference Δφ between them is measured for each separation. The separation is read from a scale fixed to the bench.

The graph shows Δφ against d with the line of best fit.

d / cm	2.0	4.0	6.0	8.0	10.0	12.0
Δφ / rad	1.16	2.04	3.00	3.89	4.75	5.76

Phase difference Δφ against microphone separation d with the line of best fit (drawn to scale).

(a)

(i)

Determine the gradient of the line in rad m⁻¹ and hence the wavelength of the sound.

(2)

(b)

(i)

Hence determine the speed of sound.

(1)

(c)

(i)

The frequency is known to ±1 % and the wavelength to ±3 %. Determine the absolute uncertainty in the speed of sound.

(2)

(d)

(i)

The model predicts a line through the origin. Use the graph to determine the zero error in the separation scale and state its effect on your answer to (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Large triangle, e.g. (0, 0.23) to (13.5, 6.41): gradient = 0.457 rad cm⁻¹ = 45.7 rad m⁻¹	✓ 1	Accept 44.5–47.0 rad m⁻¹.
A separation of one wavelength gives 2π, so λ = 2π/gradient = 0.137 m	✓ 1	Accept 0.134–0.141 m. Allow ECF.
Part (b)(i)
v = fλ = 2500 × 0.137 = 343 m s⁻¹	✓ 1	Accept 334–353 m s⁻¹. Allow ECF from (a).
Part (c)(i)
Percentage uncertainty = 1 % + 3 % = 4 %	✓ 1
Δv = 0.04 × 343 = ±10 m s⁻¹	✓ 1	Absolute uncertainty to 1 s.f. (±14 m s⁻¹ also accepted). Allow ECF from (b).
Part (d)(i)
Intercept 0.23 rad corresponds to 0.23/gradient = 0.5 cm: the true separation is ≈ 0.5 cm more than the scale reading	✓ 1	Accept 0.4–0.6 cm. Allow ECF from (a).
A constant offset changes only the intercept; the gradient, λ and v are unaffected	✓ 1

Answers: (a)(i) 45.7 rad m⁻¹; 0.137 m · (b)(i) 343 m s⁻¹ · (c)(i) ±10 m s⁻¹ · (d)(i) ≈ 0.5 cm; no effect (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of sound waves; v = fλ; C.1/C.2 — phase difference in radians; Tools — gradient with its unit, propagation of uncertainties, a zero error read from an intercept Command term: Determine

27C-1B-23

The wave equation·C.2 Wave model

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A rope passes over a pulley at one end of a bench, and a load hanging from the end of the rope sets the tension F. A pulse is made at the other end, and a video camera recording 250 frames per second shows the pulse passing two markers 3.00 m apart. The time t taken by the pulse to travel between the markers is found from the frame count to ±0.004 s.

A student suggests that the speed of the pulse is v = KFⁿ, where K and n are constants, and plots lg v against lg F. Theory predicts v = √(F/μ), where μ is the mass per unit length of the rope.

F / N	t / s	lg(F / N)	lg(v / m s⁻¹)
1.5	0.536	0.176	0.748
3.0	0.380	0.477	0.897
5.0	0.296	0.699	1.006
8.0	0.232	0.903	1.112
12.0	0.190	1.079
18.0	0.155	1.255	1.287

lg v against lg F with the line of best fit (drawn to scale).

(a)

(i)

Calculate the missing value of lg(v / m s⁻¹) for F = 12.0 N.

(1)

(b)

(i)

Determine the gradient of the line and deduce whether the data support the theoretical relationship.

(2)

(c)

(i)

The rope breaks if the tension exceeds 40 N. Use the line of best fit to predict the greatest speed at which a pulse can travel along this rope.

(2)

(d)

(i)

Determine the percentage uncertainty in v for F = 18.0 N. Suggest how the student could reduce this uncertainty using the same camera.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 3.00/0.190 = 15.79 m s⁻¹, so lg v = 1.198	✓ 1	Accept 1.20. Answer to 3 d.p. not required.
Part (b)(i)
Gradient from a large triangle = 0.50	✓ 1	Accept 0.48–0.52. Award [0] for a gradient from two adjacent table values only.
The theory gives v = F^½/√μ, so lg v = ½ lg F − ½ lg μ: the gradient n ≈ 0.5 supports the theory	✓ 1	Conclusion must be consistent with their gradient. Allow ECF from their gradient.
Part (c)(i)
Equation of the line: lg v = 0.50 lg F + 0.66; with lg 40 = 1.602, lg v = 0.50 × 1.602 + 0.66 = 1.46	✓ 1	Intercept read from the graph: accept 0.65–0.67. Allow ECF from (b). The prediction lies beyond the data, so the equation of the line (not a reading from the graph) is expected.
v = 10^1.46 = 29 m s⁻¹	✓ 1	Accept 28–30 m s⁻¹. Using 1.46 as the speed scores [0] for this mark.
Part (d)(i)
% uncertainty in v = % uncertainty in t = 0.004/0.155 × 100 = 2.6 %	✓ 1	Accept 3 %. The uncertainty in the 3.00 m distance is negligible.
Place the markers further apart (use a longer rope), or time the pulse over several reflections, so that the measured time is longer while the ±0.004 s stays the same	✓ 1	The suggestion must increase the measured time. "Repeat the readings" alone is not enough here.

Answers: (a)(i) 1.198 · (b)(i) 0.50; supports · (c)(i) 29 m s⁻¹ · (d)(i) 2.6 % (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the wave equation v = fλ = λ/T, speed of a travelling pulse; A.2 — tension; Tools — linearising a power law with logarithms, gradient and intercept of a graph, extrapolation, percentage uncertainty Command term: Determine

28C-1B-29

Speed of light in an optical fibre·C.2 Wave model

Paper 1BMedium7 marks

Data-based question5 steps to full marksDetermine

A pulse generator makes an infrared LED emit a short pulse of light and, at the same instant, starts the trace of an oscilloscope. The pulse travels through a coil of optical fibre of length L to a photodiode connected to the oscilloscope, which displays the time t at which the pulse arrives. The circuits add the same unknown delay to every reading.

Coils of six different lengths are used. Each value of t has an uncertainty of ±20 ns; the uncertainty in L is negligible. The graph shows t against L with error bars and the line of best fit. c = 3.00 × 10⁸ m s⁻¹.

L / m	25	50	75	100	125	150
t / ns	155	282	404	525	645	772

Arrival time t against fibre length L, with error bars and the line of best fit (drawn to scale).

(a)

(i)

Determine the gradient of the line of best fit. State its unit.

(2)

(ii)

Hence determine the speed of the light in the fibre and the refractive index of the fibre.

(2)

(b)

(i)

Draw the steepest and shallowest lines that pass through all the error bars. Use them to determine the absolute uncertainty in the refractive index.

(2)

(c)

(i)

Use the graph to determine the delay added by the circuits.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (771 − 34)/(150 − 0) = 4.91	✓ 1	Accept 4.75–5.05. Two adjacent data points only: [0] for this mark.
Unit: ns m⁻¹	✓ 1	Accept s m⁻¹ with the value converted (4.9 × 10⁻⁹ s m⁻¹).
Part (a)(ii)
v = 1/gradient = 1/(4.91 × 10⁻⁹) = 2.04 × 10⁸ m s⁻¹	✓ 1	Allow ECF from (a)(i).
n = c/v = 3.00 × 10⁸/2.04 × 10⁸ = 1.47	✓ 1	Accept 1.43–1.53. Allow ECF.
Part (b)(i)
Steepest line ≈ 5.3 ns m⁻¹ and shallowest ≈ 4.6 ns m⁻¹ (e.g. through the ends of the first and last error bars)	✓ 1	Accept 5.1–5.4 and 4.5–4.75.
n ranges from 1.38 to 1.58, so Δn = ±0.10	✓ 1	Accept ±0.07 to ±0.12; half the range of the gradients times c. Allow ECF from (a)(ii).
Part (c)(i)
Intercept on the t axis = 34 ns	✓ 1	Accept 25–45 ns. The delay does not affect the gradient, so the answers to (a) are unchanged.

Answers: (a)(i) 4.91 ns m⁻¹ · (a)(ii) 2.04 × 10⁸ m s⁻¹; 1.47 · (b)(i) ±0.10 · (c)(i) 34 ns (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of electromagnetic waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; C.3 — Snell's law as given by n₁/n₂ = v₂/v₁; Tools — gradient and intercept of a graph with units, steepest and shallowest lines through error bars, systematic error Command term: Determine

29C-1B-30

Longitudinal pulses on a spring·C.2 Wave model

Paper 1BHard7 marks

Data-based question5 steps to full marksDeduce

A long spring of mass m = 0.240 kg is stretched along a smooth floor between two fixed posts; the stretched length L can be changed. A sharp push on one end starts a longitudinal pulse, which reflects back and forth between the posts. A student uses a stopwatch to measure the time t for the pulse to make three complete round trips (a total distance of 6L) and calculates the speed v of the pulse.

A model predicts that, for a spring stretched to many times its unstretched length, v = L√(k/m), where k is the spring constant of the whole spring.

L / m	t / s	v / m s⁻¹
2.0	2.38	5.04
2.5	2.36	6.36
3.0	2.42	7.44
3.5	2.35
4.0	2.40	10.00
4.5	2.38	11.34
5.0	2.35	12.77

Speed v of the pulse against stretched length L, with the line of best fit (drawn to scale).

(a)

(i)

Calculate the missing value of v.

(1)

(ii)

The student expected the time t to increase with L. Deduce, using at least three rows of the table, whether the data support the model.

(2)

(b)

(i)

Determine the gradient of the line of best fit, and hence the spring constant k of the spring.

(2)

(ii)

Each time has an uncertainty of ±0.10 s and the mass has an uncertainty of ±0.005 kg. Taking the percentage uncertainty in the gradient to be equal to the percentage uncertainty in a typical time, determine the absolute uncertainty in k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 6 × L/t = 6 × 3.5/2.35 = 8.94 m s⁻¹	✓ 1	Accept 8.9 m s⁻¹. Using 2L or L instead of 6L: [0].
Part (a)(ii)
v/L calculated for at least three rows, e.g. 2.52, 2.55, 2.55 s⁻¹ (or t nearly constant at about 2.4 s for every L)	✓ 1	Allow ECF from (a)(i).
v/L is constant within about 3 %, i.e. v ∝ L (and t = 6√(m/k) does not depend on L), so the data support the model	✓ 1	The conclusion must be consistent with the candidate's values; "the times are close" without reference to the model: [0].
Part (b)(i)
Gradient from a large triangle = 2.55 s⁻¹	✓ 1	Accept 2.45–2.62 s⁻¹.
Gradient = √(k/m), so k = m × gradient² = 0.240 × 2.55² = 1.56 N m⁻¹	✓ 1	Accept 1.44–1.65 N m⁻¹. Allow ECF from their gradient. Using k = m × gradient (no square): [0] for this mark.
Part (b)(ii)
% uncertainty in gradient = 0.10/2.4 × 100 ≈ 4 %; % uncertainty in k = 2 × 4 % + (0.005/0.240) × 100 = 8 % + 2 % ≈ 10 %	✓ 1	Doubling for the square is required; 4 % + 2 % = 6 % scores [0] for this mark.
Δk = 0.10 × 1.56 ≈ ±0.2 N m⁻¹	✓ 1	Allow ECF from (b)(i). Absolute uncertainty to 1 s.f.

Answers: (a)(i) 8.94 m s⁻¹ · (a)(ii) supported: v/L ≈ 2.52 s⁻¹ (constant) · (b)(i) 2.55 s⁻¹; 1.56 N m⁻¹ · (b)(ii) ±0.2 N m⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — transverse and longitudinal travelling waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; A.2 — elastic restoring force F_H = −kx; Tools — testing a hypothesis from a model, gradient with units, propagating uncertainty through a power Command term: Deduce

30C-1B-31

Measuring a travelling wave·C.2 Wave model

Paper 1BEasy7 marks

Data-based question6 steps to full marksDetermine

A vibrator drives a transverse wave along a long horizontal rubber cord. A video camera records the cord against a metre rule fixed behind it. Working frame by frame, a student follows one crest and records its position x along the rule at times t. Each position reading has an uncertainty of ±0.02 m; the uncertainty in the times is negligible.

A small piece of tape is fixed to the cord. The graph shows the displacement y of the tape against time t.

t / s	0.00	0.10	0.20	0.30	0.40	0.50
x / m	0.29	0.54	0.78	1.02	1.26	1.49

Displacement y of the tape against time t (drawn to scale).

(a)

(i)

Using the full range of the data, determine the speed of the wave.

(2)

(b)

(i)

Use the graph to determine the period of the wave.

(1)

(ii)

Hence determine the wavelength of the wave.

(1)

(c)

(i)

Determine the percentage uncertainty in your answer to (a)(i).

(2)

(ii)

Suggest how the student could reduce this percentage uncertainty using the same camera and metre rule.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Distance = 1.49 − 0.29 = 1.20 m in 0.50 s	✓ 1	Adjacent rows only: [1 max].
v = 1.20/0.50 = 2.40 m s⁻¹	✓ 1	Accept a gradient of an x–t graph in the range 2.35–2.45 m s⁻¹.
Part (b)(i)
Four cycles between crests at 0.11 s and 1.11 s take 1.00 s, so T = 0.250 s	✓ 1	Accept 0.24–0.26 s. One cycle read directly: accept if within range.
Part (b)(ii)
λ = vT = 2.40 × 0.250 = 0.60 m	✓ 1	Allow ECF from (a)(i) and (b)(i).
Part (c)(i)
The distance is the difference of two readings, so its uncertainty is ±0.04 m	✓ 1
0.04/1.20 × 100 = 3.3 %	✓ 1	Allow ECF from (a)(i). Award [1 max] for 1.7 % (one reading).
Part (c)(ii)
Follow the crest over a longer distance (more frames, a longer cord within the field of view), so that the same ±0.04 m is a smaller fraction of the distance	✓ 1	"Repeat the readings" or "use a better camera": [0].

Answers: (a)(i) 2.40 m s⁻¹ · (b)(i) 0.250 s · (b)(ii) 0.60 m · (c)(i) 3.3 % (the remaining parts are explanations — see the table above)

31C-2-22

Transverse and longitudinal waves·C.2 Wave model

Paper 2Medium11 marks

Short answer & extended response7 steps to full marksDetermine

An earthquake produces longitudinal P waves and transverse S waves that travel through the Earth's crust. The graph shows the travel times of the two waves to six seismic stations at different distances from the source, with lines of best fit.

Travel time against distance for P waves and S waves (drawn to scale).

(a)

(i)

Use the graph to determine the speeds of the P waves and of the S waves.

(2)

(ii)

At another station the S wave arrives 30.0 s after the P wave. Determine the distance of this station from the source.

(3)

(iii)

The P wave reaches this station at 14:02:50. Determine the time at which the earthquake occurred.

(1)

(b)

(i)

Describe the motion of a small region of rock at the station as (1) the P wave and (2) the S wave passes.

(2)

(ii)

The S wave has a frequency of 0.80 Hz. Calculate its wavelength.

(1)

(iii)

The S wave makes the ground at the station move in simple harmonic motion with amplitude 2.0 cm. Determine the maximum horizontal force needed to make a building of mass 2.0 × 10⁶ kg move with the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
P waves: speed = 1/gradient = 400 km/64.5 s = 6.2 km s⁻¹	✓ 1	Accept 6.0–6.4 km s⁻¹. Gradient itself (0.16 s km⁻¹) quoted as speed: [0].
S waves: 400 km/111 s = 3.6 km s⁻¹	✓ 1	Accept 3.5–3.7 km s⁻¹.
Part (a)(ii)
d/v_S − d/v_P = 30.0 s	✓ 1
d = 30.0/(1/3.6 − 1/6.2)	✓ 1	Allow ECF from (a)(i).
d = 258 km	✓ 1	Accept 240–275 km. Reading the graph where the lines are 30 s apart is an acceptable alternative.
Part (a)(iii)
Travel time of the P wave = 258/6.2 = 42 s, so the earthquake occurred at about 14:02:08	✓ 1	Allow ECF from (a)(ii).
Part (b)(i)
(1) It oscillates about its equilibrium position parallel to the direction in which the wave travels	✓ 1
(2) It oscillates about its equilibrium position perpendicular to the direction in which the wave travels	✓ 1	Reference to oscillation about a fixed position is needed once.
Part (b)(ii)
λ = v/f = 3.6 × 10³/0.80 = 4.5 × 10³ m	✓ 1	Allow ECF from (a)(i).
Part (b)(iii)
a_max = ω²x₀ = (2π × 0.80)² × 0.020 = 0.51 m s⁻²	✓ 1
F = ma_max = 2.0 × 10⁶ × 0.51 = 1.0 × 10⁶ N	✓ 1

Answers: (a)(i) 6.2 km s⁻¹; 3.6 km s⁻¹ · (a)(ii) 258 km · (a)(iii) 14:02:08 · (b)(ii) 4.5 × 10³ m · (b)(iii) 1.0 × 10⁶ N (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — transverse and longitudinal travelling waves; v = fλ = λ/T; describing the motion of particles of a medium when a wave passes through it; C.1 — a = −ω²x; A.2 — Newton's second law Command term: Determine

32C-2-23

Sound waves·C.2 Wave model

Paper 2Easy11 marks

Short answer & extended response8 steps to full marksDetermine

During a storm an observer sees a flash of lightning and hears the thunder 6.0 s later. The air temperature is 30 °C. The speed of sound in air at kelvin temperature T is v = √(γRT/M), where γ = 1.40 and M = 0.0290 kg mol⁻¹ is the molar mass of air.

(a)

(i)

Show that the speed of sound in the air is about 350 m s⁻¹.

(2)

(ii)

Calculate the distance to the lightning.

(1)

(iii)

Explain why the travel time of the light can be ignored.

(1)

(b)

(i)

State two differences between the light and the sound, as waves.

(2)

(ii)

The thunder contains frequencies from 50 Hz to 250 Hz. Determine the range of wavelengths and explain why the thunder can be heard behind a large building that blocks the view of the lightning.

(2)

(iii)

The lightning also emits electromagnetic waves of frequency 30 kHz. State the region of the electromagnetic spectrum to which they belong.

(1)

(c)

(i)

The lightning flash transfers a charge of 5.0 C in 30 μs through a potential difference of 1.0 × 10⁸ V. Determine the mean current and the energy transferred.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = 30 + 273 = 303 K	✓ 1	Use of 30 K: [0] for the question part.
v = √(1.40 × 8.31 × 303/0.0290) = 349 m s⁻¹	✓ 1	Answer to at least 3 s.f. required.
Part (a)(ii)
d = 349 × 6.0 = 2.1 × 10³ m	✓ 1	Allow ECF from (a)(i).
Part (a)(iii)
The light takes 2.1 × 10³/(3.00 × 10⁸) ≈ 7 μs, which is negligible compared with 6.0 s	✓ 1	A numerical comparison or "about a million times faster" required. Allow ECF from (a)(ii).
Part (b)(i)
Light is an electromagnetic wave (oscillating electric and magnetic fields) and can travel through a vacuum; sound is a mechanical wave and needs a medium	✓ 1	Any two distinct differences, one mark each.
Light is transverse; sound is longitudinal (or: light travels about 10⁶ times faster)	✓ 1
Part (b)(ii)
λ = v/f: from 1.4 m (250 Hz) to 7.0 m (50 Hz)	✓ 1	Allow ECF from (a)(i).
These wavelengths are comparable with the size of buildings and gaps, so the sound diffracts around the building	✓ 1	Reference to diffraction and comparable size needed.
Part (b)(iii)
Radio waves (λ = 1.0 × 10⁴ m)	✓ 1
Part (c)(i)
I = Δq/Δt = 5.0/30 × 10⁻⁶ = 1.7 × 10⁵ A	✓ 1
Energy = qV = 5.0 × 1.0 × 10⁸ = 5.0 × 10⁸ J	✓ 1

Answers: (a)(i) 349 m s⁻¹ · (a)(ii) 2.1 × 10³ m · (b)(ii) 1.4 m to 7.0 m · (b)(iii) radio · (c)(i) 1.7 × 10⁵ A; 5.0 × 10⁸ J (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of sound waves; the nature of electromagnetic waves; the differences between mechanical waves and electromagnetic waves; orders of magnitude of the wavelengths of the electromagnetic spectrum; C.3 — wave diffraction around a body; B.3 — the kelvin temperature and the molar gas constant; B.5 — electric current; D.2 — work done moving charge through a potential difference Command term: Determine

33C-2-36

Wave graphs·C.2 Wave model

Paper 2Easy12 marks

Short answer & extended response8 steps to full marksDetermine

A transverse wave travels to the right along a long horizontal rope. The graph shows the displacement y of the rope against the position x along it at one instant, t = 0. P and Q are two points on the rope.

Displacement y of the rope against position x at t = 0 (drawn to scale). The wave travels in the positive x direction.

(a)

(i)

State the amplitude and the wavelength of the wave.

(1)

(ii)

A point on the rope completes 5 oscillations in 2.0 s. Calculate the speed of the wave.

(1)

(b)

(i)

State and explain the direction of motion of Q at t = 0.

(2)

(ii)

Determine the phase difference between the oscillations of P and Q. Give your answer in radians.

(2)

(c)

(i)

Determine the displacement of Q at t = 0.10 s.

(3)

(ii)

Calculate the maximum speed of a point on the rope, and explain why it is different from the speed of the wave.

(2)

(d)

(i)

Outline how the wave transfers energy along the rope even though no part of the rope moves along it.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Amplitude 4.0 cm and wavelength 1.60 m	✓ 1	Both needed.
Part (a)(ii)
f = 5/2.0 = 2.5 Hz, so v = fλ = 2.5 × 1.60 = 4.0 m s⁻¹	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
Q is moving upwards (in the positive y direction)	✓ 1
The wave profile moves to the right, so Q next takes the displacement of the rope just to its left, which is higher (less negative) than Q	✓ 1	Accept a sketch of the profile shifted slightly to the right. Award [0] for "upwards" with no valid reason.
Part (b)(ii)
Separation = 1.00 − 0.40 = 0.60 m = (0.60/1.60)λ	✓ 1
Phase difference = 2π × 0.60/1.60 = 3π/4 = 2.36 rad	✓ 1	Allow ECF from (a)(i).
Part (c)(i)
In 0.10 s the wave moves vt = 4.0 × 0.10 = 0.40 m (a quarter of a wavelength)	✓ 1	Allow ECF from (a)(ii).
Q then has the displacement that the point at x = 1.00 − 0.40 = 0.60 m had at t = 0	✓ 1	ALT: 0.10 s = T/4, and Q moves up from −2.8 cm through equilibrium.
y = +2.8 cm	✓ 1	Accept +2.7 to +2.9 cm from the graph. Sign required.
Part (c)(ii)
v_max = 2πfA = 2π × 2.5 × 0.040 = 0.63 m s⁻¹	✓ 1	Allow ECF from (a).
The wave speed is the speed at which the profile (and energy) moves along the rope; the point only oscillates transversely about a fixed position, with a speed that depends on the amplitude and frequency	✓ 1	Some reference to the different meanings is needed.
Part (d)(i)
Each part of the rope pulls on (does work on) the next, so oscillation energy is passed along while each part only oscillates about its equilibrium position	✓ 1	Accept "the disturbance, not the material, travels".

Answers: (a)(i) 4.0 cm; 1.60 m · (a)(ii) 4.0 m s⁻¹ · (b)(ii) 3π/4 rad (2.36 rad) · (c)(i) +2.8 cm · (c)(ii) 0.63 m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — transverse travelling waves; wavelength, frequency, time period and wave speed applied to wave motion as given by v = fλ = λ/T; the motion of the particles of a medium with position along the wave and with time; travelling waves transfer energy; C.1 (HL) — v_max = ωx₀ Command term: Determine

34C-2-45

Sound and radio broadcasting·C.2 Wave model

Paper 2Easy13 marks

Short answer & extended response9 steps to full marksDetermine

A singer performs in a concert hall. Listener H sits in the hall 34 m from the singer. The concert is also broadcast live: a microphone 0.30 m from the singer feeds a radio transmitter that broadcasts at a frequency of 98.0 MHz. Listener R is at home, 150 km from the transmitter, sitting 2.0 m from the loudspeaker of a radio.

The speed of sound in air is 343 m s⁻¹. Ignore any delays in the electronic circuits.

(a)

(i)

Calculate the wavelength of the radio waves.

(1)

(ii)

State the region of the electromagnetic spectrum to which these waves belong.

(1)

(b)

(i)

Determine the time taken for a note sung by the singer to reach R.

(3)

(ii)

Determine how much earlier R hears the note than H.

(1)

(c)

One note has a frequency of 440 Hz.

(i)

Describe the motion of an air molecule near H as this note passes.

(2)

(ii)

Calculate the wavelength of this note in air.

(1)

(iii)

The note and the radio wave have wavelengths of the same order of magnitude. Explain, with a calculation, why their frequencies are so different.

(2)

(d)

(i)

The transmitter radiates 20 kW, spread uniformly over a hemisphere above the ground. Determine the intensity of the radio waves at R.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ = c/f = 3.00 × 10⁸/98.0 × 10⁶ = 3.06 m	✓ 1
Part (a)(ii)
Radio	✓ 1	Allow ECF from (a)(i): the region must match their wavelength.
Part (b)(i)
Sound from singer to microphone: 0.30/343 = 0.875 ms; sound from loudspeaker to R: 2.0/343 = 5.83 ms	✓ 1	Both sound legs needed.
Radio wave: 150 × 10³/3.00 × 10⁸ = 0.500 ms	✓ 1	Use of 343 m s⁻¹ for the radio wave: [0] for this mark.
Total = 7.2 ms	✓ 1	Accept 7.2 ms.
Part (b)(ii)
t_H = 34/343 = 99.1 ms, so R hears it 92 ms earlier	✓ 1	Allow ECF from (b)(i). Accept 91–93 ms.
Part (c)(i)
The molecule oscillates about a fixed (equilibrium) position, parallel to the direction in which the sound travels	✓ 1	Longitudinal motion must be clear; "vibrates" alone is not enough.
It makes 440 oscillations per second and has no resultant movement towards H	✓ 1	Accept "the energy, not the air, travels to H".
Part (c)(ii)
λ = 343/440 = 0.780 m	✓ 1	Accept 0.78 m.
Part (c)(iii)
The radio wave is electromagnetic and travels at 3.00 × 10⁸ m s⁻¹; the sound is a mechanical wave travelling at 343 m s⁻¹, about 10⁶ times slower	✓ 1	Some reference to the different natures of the waves is needed.
f = v/λ, so the frequency ratio = 98.0 × 10⁶/440 = 2.2 × 10⁵, i.e. the speed ratio divided by the wavelength ratio (3.06/0.78)	✓ 1	Allow ECF from (a)(i) and (c)(ii).
Part (d)(i)
Area = 2πr² = 2π × (1.50 × 10⁵)²	✓ 1	Using 4πr²: [1 max] (7.1 × 10⁻⁸ W m⁻²).
I = 2.0 × 10⁴/(2π × (1.50 × 10⁵)²) = 1.4 × 10⁻⁷ W m⁻²	✓ 1

Answers: (a)(i) 3.06 m · (a)(ii) radio · (b)(i) 7.2 ms · (b)(ii) 92 ms · (c)(ii) 0.780 m · (c)(iii) 2.2 × 10⁵ · (d)(i) 1.4 × 10⁻⁷ W m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of sound waves; the nature of electromagnetic waves; the differences between mechanical waves and electromagnetic waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; Guidance: the approximate orders of magnitude of the wavelengths of radio, microwave, infrared, visible, ultraviolet, X-rays and gamma rays (data booklet); C.2 (linking question) — the intensity of an electromagnetic wave decreases with distance according to the inverse square law Command term: Determine

35C-2-46

Ultrasound in air·C.2 Wave model

Paper 2Medium14 marks

Short answer & extended response8 steps to full marksDetermine

An ultrasonic parking sensor on the rear bumper of a car emits short pulses of ultrasound of frequency 40.0 kHz and detects the echoes from an obstacle behind the car. The speed of sound in air at a temperature θ (in °C) is v = (331 + 0.60θ) m s⁻¹. The sensor always calculates the distance to the obstacle assuming v = 343 m s⁻¹.

(a)

(i)

Show that the wavelength of the ultrasound at 20 °C is about 9 mm.

(1)

(ii)

Suggest two reasons why ultrasound, rather than audible sound, is used in the sensor.

(2)

(b)

(i)

At 20 °C an echo from an obstacle is detected 8.75 ms after a pulse is emitted. Calculate the distance to the obstacle.

(2)

(ii)

On a winter day the air temperature is −15 °C and the obstacle is at the distance found in (b)(i). Determine the distance displayed by the sensor, and state whether the obstacle is closer or further than displayed.

(3)

(c)

(i)

The sensor must receive the echo from an obstacle 3.0 m away before it emits the next pulse. Determine the greatest number of pulses that it can emit per second at 20 °C.

(2)

(ii)

Each pulse lasts 0.25 ms. Determine the number of oscillations in each pulse and the length of each pulse in the air at 20 °C.

(2)

(d)

(i)

Some cars use radar waves of frequency 77 GHz instead. Calculate the echo time from an obstacle 1.50 m away, and suggest why the distance measured by the radar does not depend on the air temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 331 + 0.60 × 20 = 343 m s⁻¹; λ = 343/40.0 × 10³ = 8.58 mm	✓ 1	Answer to at least 2 s.f. or full substitution required.
Part (a)(ii)
The wavelength (≈ 9 mm) is much smaller than the size of the sensor/obstacles, so there is little diffraction: the pulse travels as a narrow beam and its direction is well defined	✓ 1	Reference to diffraction (or spreading) and wavelength needed.
Small obstacles reflect a detectable echo / the pulses are inaudible, so they do not disturb people	✓ 1	Any second valid reason; [2 max].
Part (b)(i)
The pulse travels to the obstacle and back: 2d = 343 × 8.75 × 10⁻³	✓ 1	Award [1 max] for 3.00 m (factor 2 omitted).
d = 1.50 m	✓ 1
Part (b)(ii)
v = 331 + 0.60 × (−15) = 322 m s⁻¹; echo time = 2 × 1.50/322 = 9.32 ms	✓ 1	Allow ECF from (b)(i).
Displayed distance = 343 × 9.32 × 10⁻³/2 = 1.60 m	✓ 1	Accept 1.59–1.60 m.
The obstacle is closer than displayed (by about 0.1 m)	✓ 1	Must be consistent with their displayed value.
Part (c)(i)
Round-trip time = 2 × 3.0/343 = 17.5 ms	✓ 1
Pulse rate = 1/17.5 ms = 57 s⁻¹	✓ 1	Accept 57 s⁻¹.
Part (c)(ii)
Number of oscillations = 0.25 × 10⁻³ × 40.0 × 10³ = 10	✓ 1
Length = 10λ = 10 × 8.58 mm = 8.6 cm	✓ 1	Allow ECF from (a)(i). ALT: 343 × 0.25 × 10⁻³.
Part (d)(i)
Echo time = 2 × 1.50/3.00 × 10⁸ = 10 ns	✓ 1
Radar waves are electromagnetic: they are not passed on by the air molecules, and their speed in air is c (to a very good approximation) whatever the temperature	✓ 1	Contrast with sound as a mechanical wave in the air is expected.

Answers: (a)(i) 8.58 mm · (b)(i) 1.50 m · (b)(ii) 1.60 m; closer than displayed · (c)(i) 57 s⁻¹ · (c)(ii) 10; 8.6 cm · (d)(i) 10 ns (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of sound waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; the nature of electromagnetic waves; the differences between mechanical waves and electromagnetic waves; C.3 — wave diffraction around a body and through an aperture Command term: Determine

36C-2-47

Wave graphs·C.2 Wave model

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

A transverse wave travels along a long horizontal rope from point A towards point B. A and B are 1.50 m apart. Motion sensors record the vertical displacement y of the rope at A and at B. The graph shows y against time t for A (solid line) and for B (dashed line).

Displacement y against time t at A (solid) and at B (dashed), drawn to scale.

(a)

(i)

State the period and the amplitude of the oscillations.

(1)

(ii)

Deduce the phase angle, between 0 and 2π rad, by which the oscillation of B lags behind that of A.

(1)

(b)

The distance AB is greater than one wavelength but less than two wavelengths.

(i)

Determine the speed of the wave.

(3)

(ii)

Calculate the wavelength.

(1)

(c)

Each small part of the rope moves with simple harmonic motion.

(i)

Calculate the maximum acceleration of a point on the rope.

(2)

(ii)

The tension in the rope is 1.35 N. The speed of the wave is given by v = √(F/μ), where μ is the mass per unit length of the rope. Show that the mass of a 2.0 cm length of the rope is about 3 g.

(2)

(iii)

Determine the maximum kinetic energy of this 2.0 cm length of rope.

(2)

(d)

(i)

Point C on the rope is midway between A and B. Determine the displacement of C at t = 0.60 s.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T = 0.40 s and amplitude = 2.5 cm	✓ 1	Both needed.
Part (a)(ii)
B reaches each crest 0.10 s after A; phase = 2π × 0.10/0.40 = π/2 rad (1.57 rad)	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
B lags A by a whole period plus 0.10 s, because AB is between one and two wavelengths	✓ 1	The hidden step. Allow ECF from (a)(i).
Time for the wave to travel from A to B = 0.40 + 0.10 = 0.50 s	✓ 1
v = 1.50/0.50 = 3.0 m s⁻¹	✓ 1	Award [1 max] for 15 m s⁻¹ (0.10 s used as the travel time).
Part (b)(ii)
λ = vT = 3.0 × 0.40 = 1.20 m	✓ 1	Allow ECF from (b)(i); check that 1 < 1.50/λ < 2.
Part (c)(i)
ω = 2π/T = 2π/0.40 = 15.7 rad s⁻¹	✓ 1	Allow ECF from (a)(i).
a_max = ω²x₀ = 15.7² × 0.025 = 6.2 m s⁻²	✓ 1	Amplitude in cm: [1 max].
Part (c)(ii)
μ = F/v² = 1.35/3.0² = 0.150 kg m⁻¹	✓ 1	Allow ECF from (b)(i).
Mass = 0.150 × 0.020 = 3.0 × 10⁻³ kg	✓ 1	Full substitution or 3.0 g required.
Part (c)(iii)
v_max = ωx₀ = 15.7 × 0.025 = 0.393 m s⁻¹	✓ 1	Allow ECF from (c)(i).
E_k = ½ × 3.0 × 10⁻³ × 0.393² = 2.3 × 10⁻⁴ J	✓ 1	Allow ECF from (c)(ii).
Part (d)(i)
The wave takes 0.75/3.0 = 0.25 s to travel from A to C, so C at 0.60 s has the displacement that A had at 0.35 s	✓ 1	Allow ECF from (b)(i).
y = 2.5 sin(2π × 0.35/0.40) = −1.8 cm	✓ 1	Accept −1.6 to −1.9 cm read from the graph. Sign required.

Answers: (a)(i) 0.40 s; 2.5 cm · (a)(ii) π/2 rad · (b)(i) 3.0 m s⁻¹ · (b)(ii) 1.20 m · (c)(i) 6.2 m s⁻² · (c)(iii) 2.3 × 10⁻⁴ J · (d)(i) −1.8 cm (the remaining parts are explanations — see the table above)

37C-2-48

Ocean swell·C.2 Wave model

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

A storm far out in the ocean creates water waves with a wide range of periods, all at about the same time t₀ and in the same small region. In deep water, the energy of waves of period T travels at a speed v = gT/(4π); the crests themselves travel at twice this speed.

A buoy at distance D from the storm records the frequency f of the swell arriving at time t. The graph shows f against t, where t is measured in hours from midnight on the day the swell first arrives.

Frequency f of the swell arriving at the buoy against time t, with the line of best fit (drawn to scale).

(a)

(i)

Show that the energy of waves of period 16.0 s travels at a speed of about 12 m s⁻¹.

(1)

(ii)

Calculate the wavelength of these waves.

(1)

(b)

(i)

Explain why the swell that arrives first at the buoy has the longest period, and why the period of the arriving swell then decreases.

(2)

(c)

(i)

Show that f = g(t − t₀)/(4πD).

(2)

(ii)

Use the graph to determine D.

(3)

(iii)

Determine how many hours before t = 0 the storm produced the swell.

(2)

(d)

(i)

As the 16.0 s swell passes, the buoy moves up and down with simple harmonic motion of amplitude 0.80 m. Determine the maximum acceleration of the buoy.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = 9.81 × 16.0/(4π) = 12.49 m s⁻¹	✓ 1	Answer to at least 3 s.f. or full substitution required.
Part (a)(ii)
Crest speed = 2 × 12.49 = 25.0 m s⁻¹, so λ = 25.0 × 16.0 = 400 m	✓ 1	Allow ECF from (a)(i). Using v itself (200 m): [0].
Part (b)(i)
The speed is proportional to the period, so long-period waves travel fastest	✓ 1
All the waves left the storm at the same time and cover the same distance, so the longest-period waves arrive first and waves of shorter and shorter period arrive later	✓ 1	Reference to the common starting time and distance is needed.
Part (c)(i)
The waves arriving at t have travelled D in time t − t₀, so v = D/(t − t₀)	✓ 1
v = g/(4πf) and equating gives f = g(t − t₀)/(4πD)	✓ 1	f = 1/T must be used.
Part (c)(ii)
Gradient of the line = 0.703 mHz h⁻¹	✓ 1	Accept 0.66–0.74 mHz h⁻¹.
In SI units: 0.703 × 10⁻³/3600 = 1.95 × 10⁻⁷ s⁻²	✓ 1	Unit conversion of both mHz and hours needed.
D = g/(4π × gradient) = 9.81/(4π × 1.95 × 10⁻⁷) = 4.00 × 10⁶ m	✓ 1	Allow ECF from (c)(i). Accept (3.8–4.3) × 10⁶ m.
Part (c)(iii)
t = t₀ where f = 0: extrapolate the line or use t₀ = t − f/gradient, e.g. 6 − 44.9/0.703	✓ 1	Allow ECF from (c)(ii).
t₀ = −58 h, i.e. about 58 hours before t = 0	✓ 1	Accept 52–64 hours.
Part (d)(i)
ω = 2π/16.0 = 0.393 rad s⁻¹	✓ 1
a_max = ω²x₀ = 0.393² × 0.80 = 0.123 m s⁻²	✓ 1	Accept 0.12 m s⁻².

Answers: (a)(i) 12.49 m s⁻¹ · (a)(ii) 400 m · (c)(ii) 4.00 × 10⁶ m · (c)(iii) 58 h · (d)(i) 0.123 m s⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; transverse and longitudinal travelling waves; Guidance: travelling waves transfer energy, even if there is no resultant displacement of the medium; A.1 — velocity as the rate of change of position; C.1 — the defining equation of SHM a = −ω²x Command term: Determine

38C-2-49

Laser ranging to the Moon·C.2 Wave model

Paper 2Hard14 marks

Short answer & extended response9 steps to full marksDetermine

A telescope on the Earth sends pulses of laser light of wavelength 532 nm towards an array of reflectors on the Moon, which send light straight back. Each pulse lasts 200 ps and carries an energy of 0.10 J. The time between the emission of a pulse and the return of its reflection is 2.51 s.

Assume the laser light travels at c = 3.00 × 10⁸ m s⁻¹ in the atmosphere as in space.

(a)

(i)

Calculate the distance from the telescope to the reflectors.

(1)

(ii)

Show that the length of each pulse in space is about 6 cm.

(1)

(iii)

Calculate the number of wavelengths in one pulse.

(1)

(b)

(i)

Explain why a pulse of light can be used to measure this distance but a pulse of sound could not.

(2)

(ii)

Determine the number of photons in each pulse.

(2)

(c)

When a pulse reaches the Moon its light is spread uniformly over a circle of diameter 6.0 km. The reflector array has an area of 0.50 m².

(i)

Determine the number of photons in each pulse that strike the reflector array.

(3)

(ii)

The reflected light is spread uniformly over a circle of diameter 15 km when it returns to the Earth. The telescope collects light over an area of 3.0 m². Determine the number of photons from each pulse that the telescope collects.

(2)

(d)

(i)

The Moon moves in a circular orbit of radius 3.84 × 10⁸ m with a period of 27.3 days. Determine how far the Moon moves along its orbit while a pulse travels from the Earth to the Moon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
d = ct/2 = 3.00 × 10⁸ × 2.51/2 = 3.77 × 10⁸ m	✓ 1	Accept 3.76–3.77 × 10⁸ m. Award [0] for 7.53 × 10⁸ m (round trip not halved).
Part (a)(ii)
Length = c × duration = 3.00 × 10⁸ × 200 × 10⁻¹² = 0.0600 m	✓ 1	Full substitution required.
Part (a)(iii)
N = 0.0600/532 × 10⁻⁹ = 1.1 × 10⁵	✓ 1	Allow ECF from (a)(ii).
Part (b)(i)
Light is an electromagnetic wave: oscillating electric and magnetic fields that need no medium, so it travels through the vacuum of space	✓ 1
Sound is a mechanical wave, passed on by the oscillation of particles of a medium; there is no medium between the Earth and the Moon	✓ 1	Answers that only state "sound needs air": [1 max].
Part (b)(ii)
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/532 × 10⁻⁹ = 3.74 × 10⁻¹⁹ J	✓ 1
Number = 0.10/3.74 × 10⁻¹⁹ = 2.7 × 10¹⁷	✓ 1
Part (c)(i)
Area of the circle = π × (3.0 × 10³)² = 2.8 × 10⁷ m²	✓ 1	Using the diameter as the radius: [2 max] (1.2 × 10⁹ photons).
Fraction striking the array = 0.50/2.8 × 10⁷ = 1.8 × 10⁻⁸	✓ 1
Number = 2.7 × 10¹⁷ × 1.8 × 10⁻⁸ = 4.7 × 10⁹	✓ 1	Allow ECF from (b)(ii).
Part (c)(ii)
Fraction collected = 3.0/(π × 7500²) = 1.7 × 10⁻⁸	✓ 1
Number = 4.7 × 10⁹ × 1.7 × 10⁻⁸ ≈ 80	✓ 1	Allow ECF from (c)(i). Accept 70–90.
Part (d)(i)
v = 2πr/T = 2π × 3.84 × 10⁸/(27.3 × 86 400) = 1023 m s⁻¹	✓ 1
Distance = 1023 × 2.51/2 = 1.28 km	✓ 1	Using the round-trip time 2.51 s: [1 max] (2.57 km).

Answers: (a)(i) 3.77 × 10⁸ m · (a)(ii) 0.0600 m · (a)(iii) 1.1 × 10⁵ · (b)(ii) 2.7 × 10¹⁷ · (c)(i) 4.7 × 10⁹ · (c)(ii) 80 · (d)(i) 1.28 km (the remaining parts are explanations — see the table above)

Syllabus understandingC.2 — the nature of electromagnetic waves; the differences between mechanical waves and electromagnetic waves; wavelength λ, frequency f, time period T and wave speed v applied to wave motion as given by v = fλ = λ/T; E.1 — photons and E = hf; A.2 — circular motion, v = 2πr/T Command term: Determine

C.2 Wave model: IB Physics HL exam-style questions

13 practice questions on C.2 Wave model

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