Summary to follow. 5 syllabus statements · 21 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Travelling (progressive) wave
Travelling (progressive) wave
A disturbance that moves through a medium or through space and transfers energy from one place to another. In a mechanical travelling wave each particle of the medium oscillates about its own fixed equilibrium position, passing the oscillation on to its neighbours, so energy moves along the wave while the medium as a whole undergoes no resultant (net) displacement.
Transverse wave
A travelling wave in which the oscillations are perpendicular to the direction in which the wave transfers energy. Examples: a wave on a stretched string or rope, a seismic S-wave, and every electromagnetic wave (in which the oscillating quantities are electric and magnetic fields rather than particles).
Longitudinal wave
A travelling wave in which the particles of the medium oscillate parallel to the direction in which the wave transfers energy, producing alternate compressions and rarefactions. Examples: sound, a seismic P-wave, and a wave sent along a slinky spring by pushing and pulling its end.
Displacement (of a particle in a wave)
The distance and direction of a particle of the medium from its equilibrium (undisturbed) position at a given instant; a vector, positive or negative along a chosen axis. For a transverse wave it is measured perpendicular to the direction of travel, for a longitudinal wave along it. SI unit: metre (m).
Displacement–position graph
A graph of the displacement of every particle of the medium against its equilibrium position along the wave, at ONE instant (a 'snapshot'). The distance between adjacent points in the same phase (e.g. adjacent crests) is the wavelength. It is not a picture of the path of any particle; for a longitudinal wave it is not even a picture of the wave's shape, because the displacements plotted upwards are really along the direction of travel.
Displacement–time graph
A graph of the displacement of ONE particle of the medium against time as the wave passes it. The time between adjacent points in the same phase (e.g. successive maxima) is the time period T. Where the curve crosses zero the particle is passing through its equilibrium position at its maximum speed; at a maximum or minimum it is momentarily at rest.
Compression and rarefaction
In a longitudinal wave, a compression is a region where the particles are closer together than at equilibrium (higher density and pressure) and a rarefaction is a region where they are further apart (lower density and pressure). On a displacement–position graph both occur at points of zero displacement: a compression where the particles on either side are displaced towards the point, a rarefaction where they are displaced away from it. Adjacent compressions are one wavelength apart.
Students often think The particles of the medium are carried along by the wave, so a wave transports material from the source towards the receiver. In fact No. Each particle oscillates about its own fixed equilibrium position. The wave shape and the energy travel; the medium has no resultant displacement.
Students often think In a longitudinal wave the particles oscillate perpendicular to the direction of energy transfer; the definitions of transverse and longitudinal are swapped. In fact Parallel to the direction in which the wave transfers energy (back and forth along the line of travel).
Amplitude
Amplitude
The maximum displacement of a particle of the medium from its equilibrium position. SI unit: metre (m). The amplitude of a mechanical wave is set by the source; it affects the energy the wave transfers but not the wave speed, which is set by the medium.
Wavelength λ
The distance between two adjacent points on a wave that are oscillating in phase, e.g. adjacent crests or adjacent compressions; the distance the wave travels in one time period. SI unit: metre (m).
Frequency f
The number of complete oscillations made by a particle of the medium (or by the source) per unit time; f = 1/T. SI unit: hertz (Hz), where 1 Hz = 1 s⁻¹. The frequency of a wave is set by its source.
Time period T
The time taken for one complete oscillation of a particle of the medium; the time for the wave to advance by one wavelength; T = 1/f. SI unit: second (s).
Wave speed v and the wave equation
The speed at which the wave (and the energy it carries) travels. Because the wave advances one wavelength λ in one time period T, v = λ/T = fλ. SI unit: m s⁻¹. For a mechanical wave the speed is set by the medium (e.g. the tension and mass per unit length of a string); changing the source's frequency changes the wavelength, not the speed.
Students often think The time period is the time between one zero-displacement point on a displacement–time graph and the next. In fact No. Successive zero crossings are half a period apart: the particle passes through equilibrium twice in each oscillation, once in each direction. The period is the time between points in the same phase, e.g. successive maxima.
Students often think A time given in milliseconds can be substituted into the wave equation without converting it to seconds. In fact No. SI units must be used: 20 ms = 0.020 s. Otherwise the speed comes out 1000 times too small (or the frequency 1000 times too small).
Sound wave
Sound wave
A longitudinal mechanical wave: the particles of the medium (a gas, liquid or solid) oscillate parallel to the direction of energy transfer, forming compressions and rarefactions, i.e. variations of pressure about the undisturbed value. Sound needs a medium and cannot travel through a vacuum. Its speed depends on the medium: about 340 m s⁻¹ in air at room temperature, about 1500 m s⁻¹ in water and about 5000 m s⁻¹ in steel.
Students often think Sound is a transverse wave in which the air particles oscillate perpendicular to the direction in which the sound travels. In fact No. Sound in air is longitudinal: the air particles oscillate parallel to the direction of energy transfer, producing compressions and rarefactions.
Students often think Sound is a flow of air (or of a 'sound substance') from the source to the ear, carrying the energy with it. In fact No. Sound is a wave: each air particle oscillates about a fixed position, and the pattern of compressions and rarefactions, with the energy, travels. No air flows from source to listener.
Electromagnetic wave
Electromagnetic wave
A transverse wave consisting of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel. It needs no medium and travels through a vacuum at c = 3.00 × 10⁸ m s⁻¹ whatever its frequency; in a material medium it travels more slowly. The wave equation c = fλ applies.
Electromagnetic spectrum
The range of electromagnetic waves, which differ only in frequency and wavelength. In order of increasing wavelength (decreasing frequency): gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves. Approximate orders of magnitude (the regions overlap and have no sharp boundaries): gamma rays about 10⁻¹² m and shorter; X-rays about 10⁻¹⁰ m; ultraviolet about 10⁻⁸ m; visible light 4 × 10⁻⁷ m to 7 × 10⁻⁷ m (400–700 nm); infrared about 10⁻⁵ m; microwaves about 10⁻² m; radio waves about 1 m and longer. The Physics data booklet gives the approximate orders of magnitude for each region.
Students often think Electromagnetic waves of higher frequency (or higher energy) travel faster, so gamma rays are the fastest and radio waves the slowest. In fact No. All electromagnetic waves travel through a vacuum at the same speed, c = 3.00 × 10⁸ m s⁻¹, whatever their frequency.
Students often think Electromagnetic radiation is a stream of tiny charged particles given out by the source, the same kind of thing as alpha and beta radiation. In fact No. Electromagnetic waves are oscillations of electric and magnetic fields. They carry no charge and are not streams of charged particles, unlike alpha and beta radiation.
Mechanical wave
Mechanical wave
A wave that is an oscillation of the particles of a material medium (solid, liquid or gas), such as sound, a wave on a string, water waves or seismic waves. It needs a medium, cannot cross a vacuum, and its speed depends on the medium. Mechanical waves can be transverse (string, S-wave) or longitudinal (sound, P-wave).
Differences between mechanical and electromagnetic waves
Mechanical waves are oscillations of particles of a medium and need a medium; electromagnetic waves are oscillations of electric and magnetic fields and can also travel through a vacuum. Mechanical waves can be transverse or longitudinal; electromagnetic waves are transverse. In a vacuum all electromagnetic waves travel at c; mechanical waves travel far more slowly, at speeds set by the medium. Both transfer energy without net displacement of any medium, and both obey v = fλ = λ/T.
Students often think Transverse waves are electromagnetic and longitudinal waves are mechanical, so whether a wave needs a medium depends on whether it is longitudinal. In fact No. Waves on strings and seismic S-waves are transverse but mechanical, and need a medium. What makes a wave need a medium is that it is an oscillation of particles, not whether it is longitudinal.
Students often think Electromagnetic waves travel at the same speed, c, in every medium, so only mechanical waves change speed from one medium to another. In fact No. c = 3.00 × 10⁸ m s⁻¹ is the speed in a vacuum (and almost the same in air). In glass or water light travels more slowly, which is why it refracts.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement defines a longitudinal wave?
Answer and reasoning
A wave in which the particles of the medium oscillate parallel to the direction of energy transfer — This is the definition. In sound, for example, the air particles move back and forth along the line from source to listener, forming compressions and rarefactions.
A wave in which the particles of the medium oscillate perpendicular to the direction of energy transfer — A student who has swapped the two definitions picks this. Oscillation perpendicular to the direction of energy transfer defines a transverse wave, such as a wave on a string.
A wave in which the particles of the medium travel along with it, carrying the energy with them — A student who thinks waves carry the medium with them picks this. In every travelling wave the particles only oscillate about fixed positions; the energy travels, the medium does not.
Any wave that travels along the length of a long medium, such as a stretched spring or a rope — A student who reads 'longitudinal' as 'along something long' picks this. Every wave travels along its medium; what defines a longitudinal wave is the direction of oscillation. A rope shaken sideways carries a transverse wave.
2 A transverse wave travels along a stretched string. Adjacent crests are 1.2 m apart. A graph of displacement y against time t for one point on the string is a sine curve: y = 0 at t = 0, y is a maximum at t = 5 ms, y = 0 at t = 10 ms, y is a minimum at t = 15 ms and y = 0 at t = 20 ms, after which the pattern repeats. What is the wave speed?
Answer and reasoning
120 m s⁻¹ — A student who takes the time between successive zero points (10 ms) as the period gets 1.2 ÷ 0.010 = 120 m s⁻¹. The point passes through zero twice in each oscillation; a full cycle includes both the maximum and the minimum, so T = 20 ms.
0.06 m s⁻¹ — A student who substitutes the period in milliseconds gets 1.2 ÷ 20 = 0.06. The period must be converted to seconds, 20 ms = 0.020 s, which gives 60 m s⁻¹.
60 m s⁻¹ — One full oscillation (maximum, zero, minimum, back to zero) takes T = 20 ms = 0.020 s, and λ = 1.2 m. v = λ/T = 1.2 m ÷ 0.020 s = 60 m s⁻¹.
0.024 m s⁻¹ — A student who uses the period in place of the frequency in v = fλ gets 0.020 × 1.2 = 0.024. The frequency is f = 1/T = 1/0.020 s = 50 Hz, so v = fλ = 50 × 1.2 = 60 m s⁻¹. Check units: s × m is not a speed.
Working From the displacement–time graph, one complete oscillation (0 → maximum → 0 → minimum → 0) takes T = 20 ms = 0.020 s. Adjacent crests give λ = 1.2 m. v = λ/T = 1.2 m / 0.020 s = 60 m s⁻¹ (equivalently f = 1/T = 50 Hz and v = fλ = 50 Hz × 1.2 m = 60 m s⁻¹).
3 Which statement describes the nature of a sound wave travelling through air?
Answer and reasoning
A transverse wave in which air particles oscillate perpendicular to the direction of energy transfer — A student who takes the oscilloscope trace or sine-curve drawing of sound literally picks this. The trace is a graph against time; in the air itself the particles oscillate along the direction of travel.
A flow of air particles from the source to the ear, carrying the energy along with them — A student who pictures sound as moving air picks this. No air flows from source to listener; each particle oscillates about a fixed position and only the disturbance, with its energy, travels.
A longitudinal wave that, like light, can travel through a vacuum as well as through air — A student who thinks all waves can cross empty space picks this. Sound is an oscillation of particles, so it needs a medium; in a vacuum there is nothing to oscillate.
A longitudinal wave in which air particles oscillate parallel to the direction of energy transfer — Sound is a longitudinal mechanical wave. Air particles oscillate back and forth along the direction of travel about fixed positions, forming compressions and rarefactions (pressure variations) that travel outward.
4 Which statement describes the nature of electromagnetic waves?
Answer and reasoning
Oscillations of the particles of the medium they travel through, so they cannot cross a vacuum — A student who thinks every wave needs a medium picks this. Light reaches Earth from the Sun through the near-vacuum of space; what oscillates in an electromagnetic wave is the electric and magnetic field, not particles.
Oscillating electric and magnetic fields that travel faster in a vacuum the higher their frequency — A student who thinks more energetic radiation travels faster picks this. The fields are right, but in a vacuum every electromagnetic wave travels at the same speed c; higher frequency means shorter wavelength, with fλ = c.
Streams of tiny charged particles given out by the source, which can travel through a vacuum — A student who groups all 'radiation' together picks this. Alpha and beta radiation are charged particles; electromagnetic waves carry no charge and are oscillations of electric and magnetic fields.
Oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel — Electromagnetic waves are transverse oscillations of electric and magnetic fields, perpendicular to each other and to the direction of travel. They need no medium and travel through a vacuum at c = 3.00 × 10⁸ m s⁻¹.
5 Which statement describes a difference between mechanical waves and electromagnetic waves?
Answer and reasoning
Mechanical waves are longitudinal waves, whereas electromagnetic waves are transverse. — A student who merges the two classifications picks this. Electromagnetic waves are transverse, but mechanical waves can be transverse too: waves on strings and seismic S-waves are transverse mechanical waves.
Mechanical waves need a material medium; electromagnetic waves can also travel through a vacuum. — A mechanical wave is an oscillation of particles of a medium, so it needs one. An electromagnetic wave is an oscillation of electric and magnetic fields and travels through a vacuum as well as through many materials.
Mechanical waves change speed between media; electromagnetic waves keep speed c in all. — A student who thinks c is the speed of light everywhere picks this. c is the speed in a vacuum; light travels more slowly in water and glass, which is why it refracts.
Mechanical waves transfer matter from place to place, whereas electromagnetic waves transfer energy. — A student who thinks waves carry their medium along picks this. Mechanical waves, like electromagnetic waves, transfer energy; they cause no net displacement of their medium.
6 A transverse wave travels in the +x direction along a stretched rope. At one instant, the graph of the displacement y of the rope (upwards positive) against position x is a sine curve: y = 0 at x = 0, y = +2.0 cm at x = 0.25 m, y = 0 at x = 0.50 m, y = −2.0 cm at x = 0.75 m and y = 0 at x = 1.00 m. How is the point of the rope at x = 0.50 m moving at this instant?
Answer and reasoning
Downwards (−y), at its maximum speed — A student who treats the graph as a track that the particle follows moves along the curve in the direction of travel, from x = 0.50 m down towards the trough at 0.75 m. The graph is a snapshot of many particles, not a path. Shifting the curve forwards brings the crest from x = 0.25 m to this point, so it is moving upwards.
Neither way: it is momentarily at rest — A student who thinks zero displacement means zero velocity picks this. At zero displacement the point is passing through equilibrium at its maximum speed; it is momentarily at rest only at maximum displacement, at x = 0.25 m and x = 0.75 m.
In the +x direction, along with the wave — A student who thinks the medium travels with the wave picks this. In a transverse wave each part of the rope moves only perpendicular to the direction of travel (in ±y) about its equilibrium position; it has no motion along x.
Upwards (+y), at its maximum speed — Imagine the whole curve shifted slightly in the +x direction: the crest at x = 0.25 m moves towards x = 0.50 m, so the displacement there is about to become positive. The point is passing through its equilibrium position, where an oscillating particle moves fastest, so it is moving in the +y direction at its maximum speed.
7 A vibrator sends a continuous transverse wave along a stretched string. The student increases the amplitude of the vibrator's oscillation but keeps its frequency and the tension in the string the same. What happens to the wave speed and to the wavelength?
Answer and reasoning
The speed and the wavelength both stay the same, as f and the string are unchanged — The wave speed is set by the string (its tension and mass per unit length), not by the amplitude. The frequency is unchanged, so λ = v/f is unchanged. Only the energy transferred increases.
Both increase, as a larger amplitude makes the wave move faster along the string — A student who thinks a larger amplitude makes a wave travel faster picks this, and then, with f fixed, λ = v/f increases too. The speed of a wave on a string depends on the string and its tension, not on the amplitude.
The speed is unchanged; the wavelength is longer, as the wave is now bigger — A student who thinks a taller wave is also a longer wave picks this. Amplitude and wavelength are independent: with v and f unchanged, λ = v/f is unchanged.
The speed is unchanged; the wavelength is shorter, as the vibrator now moves faster — A student who confuses how fast the vibrator moves with how often it oscillates picks this. With a larger amplitude at the same frequency the vibrator does move faster, but it completes the same number of oscillations per second, so f, and hence λ = v/f, is unchanged.
8 A loudspeaker emits a steady sound of frequency 500 Hz towards a listener 10 m away. Ignoring random thermal motion, which describes the motion of a particle of air midway between them while the sound passes?
Answer and reasoning
It oscillates about a fixed position, back and forth along the line to the listener. — Sound is longitudinal, so each air particle oscillates back and forth along the direction of travel, towards and away from the listener, about its own equilibrium position. It has no net displacement: only the disturbance and the energy travel. It completes 500 oscillations per second (T = 1/f = 2.0 × 10⁻³ s).
It oscillates about a fixed position, at right angles to the line to the listener. — A student who thinks sound is a transverse wave picks this. Sound in air is longitudinal: the air particles oscillate along the line from loudspeaker to listener, producing compressions and rarefactions, not across it.
It travels from the loudspeaker to the listener at the speed of sound in air. — A student who pictures sound as a flow of air picks this. The disturbance and the energy travel at the speed of sound; each air particle only oscillates about its equilibrium position and has no net displacement.
It oscillates about a fixed position, and completes one oscillation every 500 s. — A student who treats frequency and time period as interchangeable reads 500 Hz as a period of 500 s. The period is T = 1/f = 1/(500 Hz) = 2.0 × 10⁻³ s: the particle completes 500 oscillations every second, not one every 500 s.
Working Sound in air is longitudinal, so each air particle oscillates about its own equilibrium position along the direction of travel (along the line to the listener), with no net displacement. It oscillates at the source frequency, 500 Hz, so T = 1/f = 1/(500 Hz) = 2.0 × 10⁻³ s (not 500 s).
9 Which list places the regions of the electromagnetic spectrum in order of increasing wavelength?
Answer and reasoning
radio, microwave, infrared, visible, ultraviolet, X-ray, gamma — A student who assumes wavelength rises along the familiar radio-to-gamma list picks this. That list is in order of increasing frequency; since fλ = c, wavelength decreases along it, so this is the order of decreasing wavelength.
gamma, X-ray, ultraviolet, visible, infrared, microwave, radio — Gamma rays have the shortest wavelengths (about 10⁻¹² m and less) and radio waves the longest (about 1 m and more). X-rays (about 10⁻¹⁰ m), ultraviolet (about 10⁻⁸ m), visible (400–700 nm), infrared (about 10⁻⁵ m) and microwaves (about 10⁻² m) lie between, in that order.
gamma, X-ray, infrared, visible, ultraviolet, microwave, radio — A student who has swapped infrared and ultraviolet picks this. Ultraviolet lies beyond violet, with shorter wavelengths than visible light; infrared lies beyond red, with longer wavelengths.
gamma, X-ray, ultraviolet, visible, microwave, infrared, radio — A student who takes "micro" to mean micrometre-scale wavelengths places microwaves before infrared. Microwave wavelengths are of order 10⁻² m, longer than infrared (about 10⁻⁵ m); only radio waves are longer.
10 An electric bell rings inside a sealed glass jar and can be both seen and heard. A pump gradually removes the air from the jar while the bell keeps working. Which prediction, with its explanation, is correct?
Answer and reasoning
The sound fades and the bell vanishes from view, since both waves need air to travel. — A student who thinks every wave needs a medium picks this. Light is an electromagnetic wave and needs no medium; sunlight crosses the vacuum of space, so the bell stays visible.
Neither changes, because sound, like light, can travel through an evacuated jar. — A student who thinks sound can cross a vacuum picks this. Sound is an oscillation of particles of a medium; as the air is removed there is less and less to carry it, so the sound fades.
The sound fades but the bell stays visible: light, unlike sound, needs no medium. — Sound is a mechanical wave and needs particles to oscillate, so it fades as the air is removed. Light is an electromagnetic wave, an oscillation of electric and magnetic fields, and travels through a vacuum.
The sound fades but the bell stays visible, since only longitudinal waves need a medium. — A student who merges 'transverse' with 'electromagnetic' picks this. The prediction is right but the reason is wrong: transverse waves on a string or seismic S-waves also need a medium. Light needs none because it is electromagnetic, not because it is transverse.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
11 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A longitudinal wave travels in the +x direction along a slinky spring. Displacements of the coils in the +x direction are positive. At one instant, the graph of displacement s against equilibrium position x is a sine curve: s = 0 at x = 0, s = +3.0 mm at x = 10 cm, s = 0 at x = 20 cm, s = −3.0 mm at x = 30 cm and s = 0 at x = 40 cm. Where, between x = 0 and x = 40 cm, is the centre of a compression?
Answer and reasoning
At x = 10 cm, where the displacement is greatest and positive — A student who matches compressions to the crests of the graph picks this, a habit taken from diagrams that align compressions with the crests of a pressure curve. At x = 10 cm the coils on both sides are displaced forwards by nearly the same amount, so their spacing is almost normal.
At x = 10 cm and x = 30 cm, where s has its greatest magnitude — A student who thinks the spring is most squashed where coils have moved furthest picks this. Compression depends on neighbouring coils being displaced by different amounts, towards each other. At x = 10 cm and 30 cm neighbouring coils have almost the same displacement, so the spacing is nearly normal.
At x = 20 cm, where coils on both sides are displaced towards it — Coils just before x = 20 cm have positive displacement (moved forwards, towards 20 cm) and coils just after have negative displacement (moved backwards, also towards 20 cm). The coils crowd together there: a compression. At x = 0 and x = 40 cm the neighbours are displaced away from the point, so those are rarefactions.
At x = 0, 20 cm and 40 cm, where the displacement s is zero — A student who applies 'compressions are at zero displacement' to every zero point picks this. Zero points alternate between compressions and rarefactions. At x = 0 and 40 cm the coils on either side are displaced away from the point, making rarefactions; adjacent compressions are one wavelength (40 cm) apart.
Working Wavelength = 40 cm (s = 0 → +max → 0 → −max → 0 between x = 0 and 40 cm). Just below x = 20 cm, s > 0: coils displaced in +x, towards 20 cm. Just above x = 20 cm, s < 0: coils displaced in −x, also towards 20 cm. So the coils crowd together at x = 20 cm (compression). At x = 0 and x = 40 cm the neighbours are displaced away from the point (rarefactions). At x = 10 cm and 30 cm neighbouring coils have almost equal displacements, so the spacing there is almost normal.
2 A student flicks one end of a long, horizontal, stretched rope once, and a single transverse pulse travels along it. A ribbon tied to the rope halfway along moves up and then back down to its original position as the pulse passes. Which statement about the pulse is correct?
Answer and reasoning
It transferred no energy past the ribbon, since the rope there ended where it started. — A student who thinks energy transfer needs something to be moved picks this. The ribbon gained kinetic and potential energy as it rose and passed it on as the pulse moved on; energy reached the far end although no part of the rope was displaced overall.
It transferred energy along the rope, although the rope had no net displacement. — Travelling waves transfer energy even when there is no resultant displacement of the medium. Each section of rope is pulled up by the section behind it and passes the disturbance, with its energy, to the section ahead, then returns to rest where it started.
It transferred energy by carrying sections of the rope along with it to the far end. — A student who thinks waves carry the medium with them picks this. The ribbon shows the rope moving only up and down about its original position; no rope travels along with the pulse.
It transferred energy, and a bigger flick would have made it travel along faster. — A student who thinks amplitude affects wave speed picks this. A bigger flick gives a larger amplitude and transfers more energy, but the speed of the pulse is set by the rope and its tension, so it would travel at the same speed.
3 A student measures the wavelength λ of waves on one stretched string at three frequencies f of the vibrator, keeping the tension constant: f = 20 Hz, λ = 1.50 m; f = 40 Hz, λ = 0.75 m; f = 60 Hz, λ = 0.50 m. Which conclusion do the data support?
Answer and reasoning
The speed rises with frequency, because v = fλ and f was increased each time. — A student who reads v = fλ as v ∝ f picks this without calculating fλ. The wavelength fell in exact proportion as f rose, so the product fλ, the speed, stayed at 30 m s⁻¹.
The speed is the same at each frequency, so it does not depend on the frequency. — fλ = 20 × 1.50 = 40 × 0.75 = 60 × 0.50 = 30 m s⁻¹ in every case. The string sets the speed; raising the frequency shortens the wavelength in inverse proportion.
The speed falls as the frequency rises, because the period rises as frequency does. — A student who treats period and frequency as increasing together picks this and uses v = λ/T with a growing T. In fact T = 1/f falls as f rises: T = 0.050 s, 0.025 s, 0.017 s, and λ/T = 30 m s⁻¹ each time.
No conclusion is possible, as the amplitude, which sets the speed, was not measured. — A student who thinks amplitude affects wave speed picks this. The speed of a wave on a string depends on the string and its tension, not on the amplitude, and the data already show fλ constant at 30 m s⁻¹.
Working v = fλ: 20 Hz × 1.50 m = 30 m s⁻¹; 40 Hz × 0.75 m = 30 m s⁻¹; 60 Hz × 0.50 m = 30 m s⁻¹. The speed is constant; λ is inversely proportional to f.
4 A student at one end of a long steel railway track puts an ear to the rail. At the other end, 1.0 km away, the rail is struck once with a hammer. The student hears two separate sounds, 0.20 s and 2.9 s after the blow. Which conclusion do these observations support?
Answer and reasoning
Sound is slowed by closely packed particles: the first sound came through the air. — A student who thinks solids obstruct sound picks this. If the first sound had come through air, the speed in air would be 5000 m s⁻¹, far above its known value of about 340 m s⁻¹. Strong coupling between particles in a solid passes sound on faster.
Sound has one fixed speed, so the second sound was an echo of the first sound. — A student who thinks 'the speed of sound' is the same in every medium picks this. The speed depends on the medium; two media (steel and air) connect hammer and ear, and the two arrival times match the speeds in steel and in air.
Sound travels faster in steel than in air: the first sound came through the rail. — Both paths are 1.0 km long. 1000 m ÷ 0.20 s = 5.0 × 10³ m s⁻¹, a typical speed of sound in steel; 1000 m ÷ 2.9 s ≈ 3.4 × 10² m s⁻¹, the speed of sound in air. The sound through the solid rail arrives first.
The blow pushed steel along the rail to the ear, so the sound through the rail came first. — A student who thinks a wave carries its medium along with it picks this. No steel travels along the rail; its particles oscillate about fixed positions, and only the disturbance, with its energy, moves along the rail at about 5000 m s⁻¹.
Working Distance = 1.0 × 10³ m by both paths. Speed via first sound = 1.0 × 10³ m / 0.20 s = 5.0 × 10³ m s⁻¹ (steel). Speed via second sound = 1.0 × 10³ m / 2.9 s ≈ 3.4 × 10² m s⁻¹ (air).
5 In a vacuum, the radio waves from an FM radio station have a wavelength of 3.0 m, and the X-rays used in medical imaging have a wavelength of about 1.0 × 10⁻¹⁰ m. How do their frequencies compare?
Answer and reasoning
The X-ray frequency is 3 × 10¹⁰ times the radio frequency, since f = c/λ. — Both travel at c in a vacuum, so f = c/λ and the ratio of frequencies is the inverse ratio of wavelengths: 3.0 m ÷ 1.0 × 10⁻¹⁰ m = 3 × 10¹⁰. (f_radio = 1.0 × 10⁸ Hz, f_X-ray = 3.0 × 10¹⁸ Hz.)
The radio frequency is 3 × 10¹⁰ times the X-ray frequency, since f rises with λ. — A student who thinks frequency and wavelength increase together picks this. With the same speed c, fλ = c, so the wave with the shorter wavelength, the X-ray, has the higher frequency.
The X-ray frequency is higher, because X-rays travel faster than radio waves. — A student who thinks higher-energy electromagnetic waves are faster picks this. X-rays do have the higher frequency, but not because they are faster: in a vacuum both travel at c, and the frequency is higher because the wavelength is shorter.
They are equal, because both waves travel at the same speed, c, in a vacuum. — A student who thinks wave speed and frequency go together picks this: same speed, so same frequency. v = fλ has two factors; at the same speed, a wavelength 3 × 10¹⁰ times shorter means a frequency 3 × 10¹⁰ times higher.
Working In a vacuum both travel at c, so f = c/λ. f(radio) = 3.00 × 10⁸ m s⁻¹ / 3.0 m = 1.0 × 10⁸ Hz. f(X-ray) = 3.00 × 10⁸ m s⁻¹ / 1.0 × 10⁻¹⁰ m = 3.0 × 10¹⁸ Hz. Ratio = 3.0 × 10¹⁸ / 1.0 × 10⁸ = 3.0 × 10¹⁰ (= 3.0 m / 1.0 × 10⁻¹⁰ m).
6 The graph shows the displacement y of a stretched rope against position x at one instant, as a transverse wave travels along the rope in the direction of the arrow (towards −x). P is a point on the rope. How is P moving at this instant?
Answer and reasoning
Upwards, in the +y direction — This follows the curve from P towards the crest on its left, as if the point rode along the profile in the direction the wave travels. Points of the rope do not move along the curve: each oscillates in y about a fixed x. Shift the whole profile towards −x and read the new displacement at P's own position: it is negative, so P is moving down.
Downwards, in the −y direction — Correct. The wave travels towards −x, so a moment later the whole profile has shifted to the left and the displacement now found just to the right of P (negative, on the way down to the trough) arrives at P's position. P's displacement is decreasing, so P moves in the −y direction, and because it is passing through zero displacement it does so at its maximum speed.
At rest: its displacement is zero — Zero displacement is where a point of the rope moves fastest, not where it stops. P is passing through its equilibrium position on the way from the crest to the trough; a point is momentarily at rest only when it is at a crest or a trough.
To the left, carried by the wave — The wave transfers energy towards −x, but no part of the rope travels along with it. Each point of the rope oscillates about a fixed position along x; the only motion of P is in the y direction.
7 The graph shows the displacement y of one point on a stretched rope against time t as a transverse wave passes along the rope. Which describes the motion of this point at the instant t₁ marked on the graph?
Answer and reasoning
Moving upwards at its maximum speed — Correct. This is a displacement–time graph for one point, so the point's velocity is the gradient of the curve. At t₁ the curve crosses zero going from the trough towards the crest: the gradient is positive and at its steepest, so y is increasing at the fastest rate. The point is moving in +y at its maximum speed.
Moving downwards at its maximum speed — This treats the graph as a snapshot of the rope with the wave moving to the right and applies the 'shift the profile' rule. But the horizontal axis is time, not position: the curve is the history of one point, and there is no profile to shift. The point's velocity is the gradient of this curve, which is positive at t₁, so y is increasing.
At rest, because its displacement is zero — On a displacement–time graph the velocity is the gradient, and at t₁ the gradient is at its steepest. Zero displacement is where the point moves fastest; it is at rest only at the crests and troughs of this graph, where the gradient is zero.
Moving along the rope at the wave speed — The wave moves along the rope, but the point does not: it oscillates in y about a fixed position. The graph records that oscillation; nothing on it represents a velocity along the rope, and no part of the rope travels with the wave.
8 A transverse wave travels along a string. Graph 1 shows the displacement y against position x along the string at one instant. Graph 2 shows the displacement y of one point on the string against time t. What is the speed of the wave?
Answer and reasoning
12 m s⁻¹ — This uses T = 0.05 s, the time between one zero crossing and the next on Graph 2. Successive zero crossings are half a period apart: the point is at zero once on the way up and once on the way down in each cycle. The full repeat time is 0.10 s, giving v = 0.60/0.10 = 6.0 m s⁻¹.
0.17 m s⁻¹ — This reads the wavelength from Graph 2 (0.10) and the period from Graph 1 (0.60). Graph 2 has time on its horizontal axis, so its repeat distance is the period T = 0.10 s; Graph 1 has position on its axis, so its repeat distance is the wavelength λ = 0.60 m. v = 0.60/0.10 = 6.0 m s⁻¹.
6.0 m s⁻¹ — Correct. From Graph 1 (displacement against position) the wavelength is the repeat distance, λ = 0.60 m. From Graph 2 (displacement against time) the period is the repeat time, T = 0.10 s. v = λ/T = 0.60/0.10 = 6.0 m s⁻¹.
0.060 m s⁻¹ — This multiplies λ by T, as though the period could be used in place of the frequency in v = fλ. Frequency is the reciprocal of the period: f = 1/T = 1/0.10 = 10 Hz, so v = fλ = 10 × 0.60 = 6.0 m s⁻¹, or directly v = λ/T.
Working Graph 1 is a snapshot: its repeat distance is the wavelength. The pattern repeats every 0.60 m (zero crossings at 0, 0.30 m and 0.60 m are half a wavelength apart), so λ = 0.60 m. Graph 2 is the history of one point: its repeat time is the period. The displacement repeats every 0.10 s (zero crossings at 0.05 s and 0.10 s are half a period apart), so T = 0.10 s. v = λ/T = 0.60 m / 0.10 s = 6.0 m s⁻¹ (equivalently f = 1/T = 10 Hz and v = fλ = 10 Hz × 0.60 m = 6.0 m s⁻¹).
9 The graph shows the displacement s of the air particles against position x at one instant as a sound wave travels through air along the x direction. Displacements in the +x direction are positive. Four positions P, Q, R and S are marked on the graph. Which position is at the centre of a rarefaction?
Answer and reasoning
Point P — Q and S are not both compressions, so the rarefaction is not forced onto a crest. P is a crest: the particles there are displaced furthest in +x, but their neighbours on either side are displaced by almost the same amount, so the spacing at P is normal. Where s falls through zero (Q) neighbours have moved together; where it rises through zero (S) they have moved apart. The rarefaction is at S.
Point Q — Q is the other kind of zero crossing. Just to the left of Q the displacement is positive, so those particles have moved in +x, towards Q; just to the right it is negative, so those particles have moved in -x, also towards Q. Neighbours have moved together, so Q is the centre of a compression, the opposite of what is asked. Where s rises through zero (S) the neighbours have moved apart.
Point S — Correct. Just to the left of S the displacement is negative (those particles have moved in −x, away from S) and just to the right it is positive (those particles have moved in +x, also away from S). Neighbouring particles have moved apart, so S is the centre of a rarefaction. Compressions and rarefactions sit at the zero crossings of a displacement–position graph, not at its crests and troughs.
Point R — R is the trough of the graph, but a trough on the displacement graph of a longitudinal wave is not a rarefaction. The particles at R are simply displaced furthest in −x, and their neighbours on either side are displaced by nearly the same amount, so the spacing there is normal. The rarefaction is at the zero crossing where s rises through zero, S.
10 The diagram shows the regions of the electromagnetic spectrum on a logarithmic scale of wavelength λ. A wave has a wavelength of 2 × 10⁻⁵ m. In which region of the spectrum shown does it lie?
Answer and reasoning
the ultraviolet region — Ultraviolet lies on the short-wavelength side of the visible band, below about 4 × 10⁻⁷ m: on the diagram it is to the left of the visible sliver. 2 × 10⁻⁵ m is about thirty times longer than red light, on the other side of the visible band, in the infrared.
the microwave region — The name suggests micrometre wavelengths, but on the scale the microwave band runs from about 10⁻³ m to 10⁻¹ m, i.e. millimetres to decimetres. 2 × 10⁻⁵ m is nearly two decades shorter than the bottom of that band, in the infrared.
the infrared region — Correct. On the scale, 2 × 10⁻⁵ m lies between the 10⁻⁶ m and 10⁻³ m marks, a little over one decade above 10⁻⁶ m. That is inside the infrared band, which runs from just above the red end of the visible (about 7 × 10⁻⁷ m) up to about 10⁻³ m.
the visible region — The visible band is a sliver on this scale, roughly 4 × 10⁻⁷ m to 7 × 10⁻⁷ m, less than one decade wide. 2 × 10⁻⁵ m is about thirty times the wavelength of red light and lies well inside the infrared band.
11 A small loudspeaker emitting a steady note and a lamp are sealed inside a glass jar. Outside the jar a microphone and a light sensor face them. At t = 0 a pump starts removing the air from the jar. The graph shows the output of each sensor against time. Which conclusion do the two traces support?
Answer and reasoning
The sound wave needs a material medium to reach the microphone, but the light wave does not. — Correct. As the air is removed the microphone output falls to zero while the light sensor output stays constant. Sound is a mechanical wave, carried by oscillations of the air particles, so with no air there is nothing to carry it. Light is an electromagnetic wave and needs no medium, so removing the air leaves it unchanged.
The pump drew the sound out of the jar along with the air that was carrying it. — Sound is not a substance mixed into the air that a pump can extract: it is an oscillation of the air particles about fixed positions, and the air itself does not flow from the loudspeaker to the microphone. The trace falls because there are fewer and fewer particles left to pass the oscillation on.
Both waves need air, but the light needed only the trace of air the pump left behind. — The light trace does not change at all while the microphone trace falls to zero; if light also needed air, its output would fall as the air pressure dropped. Light travels through a vacuum: it is an oscillating electric and magnetic field and needs no particles at all.
Only longitudinal waves need a medium; the light passed because it is a transverse wave. — Light is transverse, but that is not why it passes: transverse mechanical waves, such as waves on a rope, need a medium too. The distinction the traces show is between a mechanical wave (sound) and an electromagnetic wave (light), not between longitudinal and transverse.
That was your twenty minutes. Real practice on C.2 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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