IB Physics C.2 Wave Model — Notes (SL & HL)
Written by Evan Kotronis for the IB Diploma Physics course (first assessment 2025). C.2 is identical for SL and HL: the guide lists no additional higher level content, so everything on this page is examinable at both levels.
What this page covers
- What a travelling wave is, and why it transfers energy without carrying the medium along with it
- Transverse and longitudinal waves, and how the particles of the medium move in each
- Wavelength, frequency, period, amplitude and wave speed, linked by \(v=f\lambda=\dfrac{\lambda}{T}\)
- Reading λ, T and A from displacement–distance and displacement–time graphs
- Finding the direction in which a particle of the medium is moving, and locating compressions and rarefactions
- What happens to the wavelength when the frequency or the medium changes
- The nature of sound waves
- The nature of electromagnetic waves and the orders of magnitude of their wavelengths
- How mechanical waves and electromagnetic waves differ
In one minute
- A travelling wave is a disturbance that moves through a medium (or through space) and transfers energy. Each particle of the medium only oscillates about its own equilibrium position.
- Transverse: oscillations perpendicular to the direction of energy transfer (rope, string, all electromagnetic waves). Longitudinal: oscillations parallel to it, forming compressions and rarefactions (sound).
- \(v=f\lambda=\lambda/T\). The source sets the frequency; the medium sets the speed; the wavelength follows from \(\lambda=v/f\).
- A displacement–distance graph gives λ and A; a displacement–time graph gives T and A. Always read the axis label first.
- Sound is a longitudinal mechanical wave that needs a medium. Electromagnetic waves are transverse oscillations of electric and magnetic fields that travel through a vacuum at \(c=3.00\times10^8\ \text{m s}^{-1}\).
What is a travelling wave?
In C.1 you met a single oscillator. A wave appears when many oscillators are coupled: each part of the medium pulls or pushes on its neighbours, so a local disturbance at the source is passed on, one particle to the next, slightly later each time. That passing-on is the wave.
- Every particle of the medium oscillates about its own equilibrium position and ends up where it started. What travels is the pattern of the disturbance and the energy.
- A single flick of a rope sends a pulse; a source that keeps oscillating sends a continuous wave. Every particle then oscillates with the frequency of the source.
- Each particle repeats the motion of the particle next to it on the source side, with a short delay. Particles further from the source are further behind in their cycle.
- Mechanical waves (waves on ropes and springs, water waves, sound) need a material medium. Electromagnetic waves do not: what oscillates is an electric and a magnetic field.
What is the difference between transverse and longitudinal waves?
| Transverse | Longitudinal | |
|---|---|---|
| Direction of oscillation | perpendicular to the direction of travel | parallel to the direction of travel |
| Features | crests (peaks) and troughs | compressions (particles closer together, higher pressure and density) and rarefactions (particles further apart, lower pressure and density) |
| Wavelength measured | crest to crest, or between any two adjacent points oscillating in phase | compression to compression, or between any two adjacent points oscillating in phase |
| Examples | waves on a rope or a string, a slinky shaken sideways, all electromagnetic waves (light, radio, X-rays…) | sound and ultrasound, a slinky pushed and pulled along its length |
- The same slinky can carry either type: shake the end sideways for a transverse wave, push and pull it along its axis for a longitudinal wave. The type depends on the direction of the oscillation, not on the medium.
- “Perpendicular to the direction of travel” is the clean wording. Avoid “perpendicular to the wave”, which does not say which direction you mean.
What are wavelength, frequency, period and wave speed?
| Quantity | Symbol and unit | Meaning |
|---|---|---|
| Displacement | \(y\) or \(x\) (m) | distance and direction of a particle from its equilibrium position |
| Amplitude | \(A\) (m) | maximum displacement of a particle from its equilibrium position |
| Wavelength | \(\lambda\) (m) | the distance between two adjacent points on the wave that oscillate in phase, e.g. crest to crest or compression to compression |
| Period | \(T\) (s) | the time for one complete oscillation of a particle of the medium |
| Frequency | \(f\) (Hz = s−1) | the number of complete oscillations per second; \(f=1/T\) |
| Wave speed | \(v\) (m s−1) | the speed at which the energy (the wave pattern) travels through the medium |
- \(v\) = wave speed (m s−1), \(f\) = frequency (Hz), \(\lambda\) = wavelength (m), \(T\) = period (s). In the data booklet under C.2; \(T=1/f\) is in the data booklet under C.1.
- The equation applies to every wave: ropes, sound, water waves and electromagnetic waves. For electromagnetic waves in a vacuum it becomes \(c=f\lambda\).
- Do not confuse the wave speed (constant in a given medium) with the speed of a particle of the medium, which changes continuously during each oscillation (C.1): zero at maximum displacement, maximum at the equilibrium position.
- Watch prefixes: kHz (103), MHz (106), GHz (109); mm, μm (10−6), nm (10−9).
A student holds one end of a long rope and moves her hand up and down three times every second. A crest is observed to move 4.5 m along the rope in 1.5 s.
Calculate (a) the frequency and the period of the wave, (b) the wave speed, (c) the wavelength.
(a) Three oscillations per second: f = 3.0 Hz; \(T=\dfrac1f=\dfrac{1}{3.0}=\) 0.33 s
(b) The crest moves with the wave: \(v=\dfrac{\text{distance}}{\text{time}}=\dfrac{4.5}{1.5}=\) 3.0 m s−1
(c) \(\lambda=\dfrac{v}{f}=\dfrac{3.0}{3.0}=\) 1.0 m
How do you read λ, T and A from wave graphs?
Exam questions describe the motion of the medium in two ways: how the displacement varies with position along the wave at one instant, and how the displacement of one particle varies with time. The two graphs look alike, so the axis label decides what you can read from them.
| Displacement–distance graph | Displacement–time graph | |
|---|---|---|
| What it shows | a “snapshot” of the whole wave at one instant | the history of one particle at a fixed position |
| Horizontal axis | distance \(x\) along the wave (m) | time \(t\) (s) |
| Repeat length gives | wavelength \(\lambda\) | period \(T\) (then \(f=1/T\)) |
| Height of peak gives | amplitude \(A\) | amplitude \(A\) |
| Cannot give on its own | \(T\) or \(f\) (you need \(v\) as well) | \(\lambda\) (you need \(v\) as well) |
- Measure a repeat length across several cycles and divide, to reduce the reading uncertainty.
- The amplitude is measured from the equilibrium line (\(y=0\)) to a peak — not from a trough to a crest, which is \(2A\).
- For a longitudinal wave the vertical axis still shows displacement, but the displacement is along the direction of travel (positive = one way, negative = the other). The graph is not a picture of the wave.
The two graphs above describe the same transverse wave travelling along a long spring. The first shows the displacement \(y\) against distance \(x\) at \(t=0\); the second shows the displacement of particle P (at \(x=0.80\) m) against time.
(a) State the amplitude, the wavelength and the period. (b) Determine the frequency and the speed of the wave. (c) Determine the distance moved by the wave, and the total distance travelled by P, in 3.0 s. (d) Deduce, using both graphs, the direction in which the wave is travelling.
(a) A = 2.5 cm; from the first graph λ = 1.6 m (crests at 0.40 m and 2.0 m); from the second graph T = 0.40 s (crests at 0.10 s and 0.50 s).
(b) \(f=\dfrac1T=\dfrac{1}{0.40}=\) 2.5 Hz; \(v=f\lambda=2.5\times1.6=\) 4.0 m s−1
(c) Wave: \(4.0\times3.0=\) 12 m.
P: in 3.0 s it makes \(3.0/0.40=7.5\) oscillations, and in each oscillation it travels \(4A=10\ \text{cm}\). Total distance \(=7.5\times10=\) 75 cm (0.75 m) — up and down, while staying at \(x=0.80\) m.
(d) The second graph shows P moving upwards at \(t=0\) (its displacement becomes positive). On the first graph, the particles just to the left of P are displaced upwards. P copies the particles on the source side, so the source side is to the left: the wave travels in the +x direction (to the right).
How do the particles of the medium move when a wave passes?
Transverse waves: which way is a particle moving?
- Particles at a crest or a trough are momentarily at rest (maximum displacement).
- Particles at the equilibrium position (\(y=0\)) move with their maximum speed.
- For a wave travelling to the right, the particles just ahead of a crest (on its right) are moving up, the particles just behind it (on its left) are moving down. Reverse the wave direction and every particle's velocity reverses.
Longitudinal waves: locating compressions and rarefactions
For a longitudinal wave, the displacement–distance graph has the same shape, but a positive displacement means “displaced in the direction of travel” (or whichever direction the question defines as positive). To find the compressions, ask where neighbouring particles move towards each other.
- With “positive = to the right”: at a compression the graph crosses zero going down (particles on the left are displaced right, particles on the right are displaced left — both towards the point).
- At a rarefaction the graph crosses zero going up (particles on both sides are displaced away).
- The particle at the centre of a compression or rarefaction has zero displacement, but it is moving at its maximum speed.
- Adjacent compressions are one wavelength apart; a compression and the next rarefaction are half a wavelength apart.
A sound wave of frequency 425 Hz travels to the left through air. The graph above shows the displacement \(y\) of the air molecules against position \(x\) at \(t=0\); a positive displacement is to the right. Take the speed of sound in air as 340 m s−1.
(a) Show that the graph is consistent with the speed and frequency given. (b) Identify the positions of the centres of the compressions at \(t=0\). (c) Deduce the direction of motion of the molecule whose equilibrium position is at \(x=0.40\) m, at \(t=0\). (d) Determine the shortest time after \(t=0\) at which there is a compression at \(x=0\). (e) A student says the molecule at \(x=0.40\) m will reach \(x=0\) at that time. Comment.
(a) From the graph \(\lambda=0.80\ \text{m}\). \(\lambda=\dfrac vf=\dfrac{340}{425}=\) 0.800 m ✓
(b) The displacement is zero and the graph slopes downwards (molecules on the left displaced right, on the right displaced left) at x = 0.40 m and x = 1.20 m. (At 0, 0.80 m and 1.60 m there are rarefactions.)
(c) The wave travels to the left, so the source side is to the right. A moment later, the molecule at 0.40 m has the displacement that the molecules just to its right have now, which is negative. So it is moving to the left — at its maximum speed, since \(y=0\).
(d) The nearest compression to the right of \(x=0\) is at 0.40 m and the pattern moves left at 340 m s−1: \(t=\dfrac{0.40}{340}=\) 1.2 × 10−3 s (1.18 ms).
Check: \(T=\dfrac{1}{425}=2.35\ \text{ms}\), and a compression replaces a rarefaction after half a period, 1.18 ms. ✓
(e) Wrong. The compression (the pattern) moves 0.40 m; the molecule does not. It oscillates about \(x=0.40\) m with an amplitude of only 3.0 μm. The wave transfers energy, not air.
What is the nature of sound waves?
- Sound is produced by a vibrating object (a loudspeaker cone, a string, vocal folds). As the object moves forward it pushes the neighbouring layer of air together (compression); as it moves back it leaves a region of lower pressure (rarefaction).
- The pressure and density of the medium vary along the wave: higher than normal at compressions, lower at rarefactions.
- Sound needs a medium. It cannot travel through a vacuum: in the classic bell-in-a-jar demonstration the ringing fades as the air is pumped out, although the bell can still be seen (light needs no medium).
- The speed of sound depends on the medium: roughly 340 m s−1 in air at room temperature (larger in warmer air), about 1500 m s−1 in water and several thousand m s−1 in solids such as steel. Values are given in questions; they are not in the data booklet.
- Humans hear frequencies from about 20 Hz to 20 kHz; in air this is wavelengths from about 17 m down to 1.7 cm. Sound above 20 kHz is ultrasound — still a longitudinal mechanical wave.
- Qualitatively, a higher frequency is heard as a higher pitch, and a larger amplitude as a louder sound. Neither changes the speed of sound in a given medium.
To measure the speed of sound, a student stands 102 m in front of a tall wall and strikes two wooden blocks together. A microphone next to her, connected to a data logger, records 0.598 s between the sound of the strike and the arrival of the echo.
(a) Show that the speed of sound in air is about 340 m s−1. (b) A tuning fork emits a note of frequency 440 Hz. Calculate the wavelength of this sound in air. (c) Suggest why the student stands far from the wall rather than close to it.
(a) The sound travels to the wall and back: distance \(=2\times102=204\ \text{m}\).
\(v=\dfrac{204}{0.598}=\) 341 m s−1 ≈ 340 m s−1
(b) \(\lambda=\dfrac vf=\dfrac{340}{440}=\) 0.77 m (using 341 m s−1 gives 0.78 m; either is accepted)
(c) A longer distance gives a longer time interval, so the same absolute uncertainty in the time (for example ±0.01 s) is a smaller percentage uncertainty: here about 2 %. Close to the wall the echo would also overlap with the original sound.
Forgetting the factor of 2 for the return journey is the classic error: it would give 171 m s−1.
What is the nature of electromagnetic waves?
- Electromagnetic waves are produced by accelerating charges (for example electrons oscillating in a radio aerial). Nothing material oscillates — only the fields — so no medium is needed.
- All electromagnetic waves travel at the same speed \(c\) in a vacuum, whatever their frequency. In air the speed is almost exactly \(c\); in glass or water it is smaller (C.3).
- They are transverse, they carry energy (sunlight warming the Earth, B.2) and they obey \(c=f\lambda\).
- Visible light is one small part of a continuous electromagnetic spectrum. The data booklet chart gives a rough order of magnitude for the wavelength range of each region — learn to use it, not the exact boundaries, which overlap.
| Region | Order of magnitude of λ / m | Typical source or use |
|---|---|---|
| γ rays | 10−16 – 10−11 | radioactive nuclei (E.3); sterilising, cancer treatment |
| X-rays | 10−12 – 10−8 | fast electrons hitting a metal target; medical imaging |
| Ultraviolet | 10−8 – 4 × 10−7 | the Sun; causes sunburn, makes some materials fluoresce |
| Visible | 4 × 10−7 – 7 × 10−7 (400–700 nm) | hot objects such as the Sun and lamps; sight, optical fibres |
| Infrared | 7 × 10−7 – 10−3 | all warm bodies (B.1, B.2); remote controls, thermal imaging |
| Microwaves | 10−3 – 10−1 | radar, mobile phone and wireless networks, microwave ovens |
| Radio waves | 10−1 and longer | oscillating currents in aerials; broadcasting, communication |
An airport radar emits electromagnetic waves of wavelength 3.2 cm. Take the speed of the waves in air to be \(c\).
(a) Calculate the frequency of the waves. (b) State, with a reason, the region of the electromagnetic spectrum to which they belong. (c) A pulse reflected from an aircraft returns to the radar 0.12 ms after it was emitted. Calculate the distance to the aircraft.
(a) \(f=\dfrac c\lambda=\dfrac{3.00\times10^8}{3.2\times10^{-2}}=\) 9.4 × 109 Hz (9.4 GHz)
(b) \(\lambda=3.2\times10^{-2}\ \text{m}\) lies between about 10−3 m and 10−1 m, so the waves are microwaves.
(c) The pulse travels there and back: \(2d=ct\Rightarrow d=\dfrac{3.00\times10^8\times0.12\times10^{-3}}{2}=\) 1.8 × 104 m (18 km)
What happens to the wavelength when the frequency or the medium changes?
- Change the source frequency, same medium: the speed is unchanged, so \(\lambda\propto\dfrac1f\). Doubling the frequency halves the wavelength.
- Same source, new medium: the particles at the boundary are driven at the rate at which the waves arrive, so the number of oscillations per second cannot change. If the speed falls, the wavelength falls in the same ratio.
- The speed depends on the properties of the medium — for a rope, how taut and how heavy it is; for sound, the material and its temperature. In the situations met in IB questions it does not depend on the frequency or the amplitude.
- Light entering glass from air slows down, so its wavelength decreases while its frequency (and colour) stays the same. Sound entering water from air speeds up, so its wavelength increases. The change of direction that can accompany this, refraction, is studied in C.3.
C.2 Refraction & Snell's law simulation: send waves into a slower medium and watch the wavefronts bunch up — the wavelength shrinks while the frequency stays fixed. (The angles and Snell's law come in C.3.)
A medical ultrasound probe emits pulses of frequency 3.0 MHz into soft tissue, where the speed of sound is 1540 m s−1. The pulses then pass into bone, where the speed of sound is 3500 m s−1.
(a) Calculate the wavelength in soft tissue. (b) Explain what happens to the frequency, the speed and the wavelength as the ultrasound passes from tissue into bone, and determine the wavelength in bone. (c) Each pulse lasts for four complete oscillations. Determine the duration of a pulse and its length in soft tissue. (d) The probe is switched to 6.0 MHz. State the effect on the speed and on the wavelength in soft tissue.
(a) \(\lambda_t=\dfrac vf=\dfrac{1540}{3.0\times10^6}=\) 5.1 × 10−4 m (0.51 mm)
(b) The frequency is set by the probe and is unchanged at 3.0 MHz: the bone at the boundary is made to oscillate at the rate the tissue drives it. The speed increases because bone is a different medium. Since \(\lambda=v/f\) with \(f\) constant, the wavelength increases in the same ratio (3500/1540 = 2.27):
\(\lambda_b=\dfrac{3500}{3.0\times10^6}=\) 1.2 × 10−3 m (1.17 mm)
(c) Duration \(=4T=\dfrac{4}{3.0\times10^6}=\) 1.3 × 10−6 s; length in tissue \(=4\lambda_t=4\times0.513\ \text{mm}=\) 2.1 mm
(d) The speed is unchanged (same medium, 1540 m s−1); the wavelength halves to 0.26 mm.
How do mechanical waves differ from electromagnetic waves?
The wave model — oscillation, wavelength, frequency, speed, \(v=f\lambda\), energy transfer — applies to both families. What differs is what oscillates and what the wave needs in order to travel.
| Mechanical waves | Electromagnetic waves | |
|---|---|---|
| What oscillates | particles of a material medium | electric and magnetic fields |
| Medium | needed — cannot cross a vacuum | not needed — travel through a vacuum |
| Type | transverse (rope, string) or longitudinal (sound) | always transverse |
| Speed | depends on the medium; from about 1 m s−1 (water waves) to several thousand m s−1 (sound in solids) | \(c=3.00\times10^8\ \text{m s}^{-1}\) in a vacuum for every frequency; less in matter |
| Produced by | a vibrating object in contact with the medium | accelerating (e.g. oscillating) charges |
| Examples | sound, ultrasound, water waves, seismic waves, waves on strings | radio, microwaves, IR, light, UV, X-rays, γ rays |
Exam tips for C.2
On a displacement–position graph for a sound wave, the centre of a compression is where the displacement is zero and neighbouring particles are displaced towards it. With “positive = displaced in the +x direction”, that is where the graph crosses the axis going down. Choosing a crest is the most common wrong answer in multiple-choice questions on this.
When asked how the wavelength changes as a wave enters a new medium, the marks go to two ideas: the frequency stays the same and the speed changes, so by \(v=f\lambda\) the wavelength changes in the same direction as the speed. Stating only “the wavelength decreases” earns at most one mark.
A repeat length on a \(y\)–\(x\) graph is λ; on a \(y\)–\(t\) graph it is T. Students who read “0.40” off a time axis and call it a wavelength lose every subsequent mark. Write down which graph you are using next to each value you read.
When asked to describe how the particles move, say that they oscillate about fixed equilibrium positions, parallel (longitudinal) or perpendicular (transverse) to the direction of energy transfer, and that there is no net movement of the medium. “The particles move with the wave” is a contradiction of the wave model and scores zero.
Most lost marks in electromagnetic-spectrum calculations are prefix errors: nm is 10−9 m, MHz is 106 Hz. After calculating a wavelength, check its region against the data booklet chart — a “radio wave” of wavelength 10−7 m tells you a power of ten has gone wrong.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The particles of the medium travel along with the wave. | Each particle returns to its equilibrium position after every oscillation. | Energy and the wave pattern travel; the medium only oscillates. |
| A higher frequency makes a wave travel faster. | In a given medium \(v\) is fixed; increasing \(f\) reduces \(\lambda\) instead. | The medium sets \(v\); the source sets \(f\); \(\lambda=v/f\). |
| A larger amplitude makes a wave travel faster. | Amplitude affects the energy carried, not the speed. | Louder sound and quieter sound reach you at the same time. |
| The frequency changes when a wave enters a new medium. | The boundary oscillates at the rate the waves arrive; oscillations cannot pile up or disappear there. | Frequency is constant; speed and wavelength change together. |
| Sound can travel through space, and radio waves are a kind of sound. | Sound needs particles to oscillate; radio waves are electromagnetic. | Sound is mechanical and longitudinal; radio is electromagnetic and transverse. |
| A displacement–time graph shows the shape of the wave. | It shows one particle's motion over time, not the wave in space. | Only a displacement–distance graph is a “snapshot”, and only for a transverse wave does it look like the wave. |
How this topic connects
Wavelength from distance = speed \(\times\) time
A wave pattern moves through a uniform medium at constant speed, so distance = speed × time applies (A.1). In one period \(T\) the source completes one oscillation and the disturbance it started has moved forward by \(vT\). That distance is, by definition, one wavelength: \(\lambda=vT\), which is \(v=f\lambda\). The same kinematic idea measures \(v\) itself: time an echo or a pulse over a known distance, as in worked examples 4 and 5.
No medium needed: how electromagnetic waves cross empty space
A mechanical wave is passed on by forces between neighbouring particles, so it needs a medium. In an electromagnetic wave the oscillating quantities are fields: a changing electric field is accompanied by a changing magnetic field and vice versa (Maxwell's theory, which is not required in detail; the link between changing magnetic fields and induced emf is met in D.4, HL only). The two fields sustain each other as they travel, so no particles are needed. Light from the Sun and from distant stars crosses the vacuum of space this way.
Spreading power over a sphere: the inverse square law for radiation
A point source radiates energy equally in all directions. At distance \(d\) the power \(L\) is spread over a sphere of area \(4\pi d^2\), so the power per unit area is \(b=\dfrac{L}{4\pi d^2}\) (data booklet, B.1). Doubling the distance spreads the same power over four times the area, so the intensity falls to one quarter. This assumes no energy is absorbed on the way — true in a vacuum, approximately true in clear air.
Tools & inquiry — measuring waves (NOS: models, evidence and serendipity)
- Speed of sound by echo or two microphones: time a sharp sound over a measured path with a data logger. Two microphones a metre or more apart, triggered by the same clap, remove the human reaction time. Reduce background noise — work in a quiet space, use sound-absorbing material around the microphones and set a trigger threshold above the noise level.
- Waves in a ripple tank or on a string: freeze the pattern with a stroboscope or slow-motion video and measure several wavelengths at once; read \(f\) from the signal generator; compare \(f\lambda\) with a direct timing of a pulse.
- Sound on an oscilloscope: a microphone turns the pressure variations into a voltage–time trace. The trace is a displacement–time graph, so the horizontal repeat gives \(T\), not \(\lambda\).
- NOS — how were X-rays discovered? In 1895 Wilhelm Röntgen noticed that a fluorescent screen glowed near a covered cathode-ray tube. He did not ignore the anomaly: he tested it systematically and showed that an unknown, penetrating radiation was responsible. Only later were X-rays shown to be electromagnetic waves of very short wavelength.
Key equations
| Equation | Use | In data booklet? |
|---|---|---|
| \(v=f\lambda=\dfrac{\lambda}{T}\) | wave speed, for every type of wave | yes (C.2) |
| \(T=\dfrac1f\) | period and frequency | yes (C.1) |
| \(c=f\lambda\), \(c=3.00\times10^8\ \text{m s}^{-1}\) | electromagnetic waves in a vacuum | \(c\) is in the constants; the equation is \(v=f\lambda\) with \(v=c\) |
| distance = speed × time; echo: \(2d=vt\) | pulse and echo timing | no — kinematics; remember the factor of 2 |
| \(b=\dfrac{L}{4\pi d^2}\) | inverse square fall-off of intensity from a point source | yes (B.1) |
| electromagnetic spectrum chart | orders of magnitude of λ for each region | yes |
Check your understanding
1. Water-wave crests 1.2 m apart pass a floating buoy once every 2.0 s. Calculate the wave speed and describe the motion of the buoy.
\(v=\dfrac{\lambda}{T}=\dfrac{1.2}{2.0}=0.60\ \text{m s}^{-1}\). The buoy bobs up and down about the same position with a period of 2.0 s; it is not carried along by the wave (ignoring any current or wind).
2. Classify each as transverse or longitudinal, and as mechanical or electromagnetic: sound along a steel rail; starlight; a wave on a guitar string; ultrasound in tissue; X-rays.
Sound in steel: longitudinal, mechanical. Starlight: transverse, electromagnetic. Guitar string: transverse, mechanical. Ultrasound: longitudinal, mechanical. X-rays: transverse, electromagnetic.
3. Visible light has wavelengths from 400 nm to 700 nm in a vacuum. Calculate the corresponding range of frequencies.
\(f=\dfrac c\lambda\): \(\dfrac{3.00\times10^8}{400\times10^{-9}}=7.5\times10^{14}\ \text{Hz}\) (violet) to \(\dfrac{3.00\times10^8}{700\times10^{-9}}=4.3\times10^{14}\ \text{Hz}\) (red).
4. A displacement–time graph for one point on a string shows 15 complete oscillations in 6.0 s. The wave speed on the string is 2.4 m s−1. Determine the wavelength.
\(f=\dfrac{15}{6.0}=2.5\ \text{Hz}\), so \(\lambda=\dfrac vf=\dfrac{2.4}{2.5}=0.96\ \text{m}\). The time graph alone gives only \(f\); you need \(v\) to get \(\lambda\).
5. Thunder is heard 4.5 s after the lightning flash is seen. The speed of sound is 340 m s−1. Estimate the distance to the lightning and explain why the travel time of the light can be ignored.
\(d=vt=340\times4.5=1.5\times10^3\ \text{m}\) (1530 m). The light takes \(\dfrac{1530}{3.00\times10^8}=5.1\times10^{-6}\ \text{s}\) — about a millionth of the sound's travel time, far smaller than the uncertainty in timing 4.5 s.
Practise C.2 with exam-style questions
Original Paper 1A, 1B and 2 questions on wave types, wave graphs, \(v=f\lambda\), sound and electromagnetic waves, with full mark schemes. Tick the questions you want and print them as your own test.
Revise and explore
- Flashcards — choose IB Physics and filter to C.2 Wave model. Free, no sign-up.
- C.2 Refraction & Snell's law simulation — see the wavelength change as waves enter a new medium
- Full IB Physics question bank (all sub-topics, SL & HL)
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