Educerie
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Educerie · IB Diploma · Physics

Theme C Wave behaviour · C.4 Standing waves and resonance

Level
SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts)
energy, forces. A standing wave is a wave that stores energy in one place instead of carrying it away, and resonance is what happens when a periodic force feeds energy into a system at exactly the rate that suits it.
The question this unit answers
what makes a standing wave different from a travelling one, why do the ends of a string or pipe decide which notes it can play, and why does pushing at just the right frequency make something shake so hard?
Where it is examined
Paper 1A multiple choice on harmonic patterns and phase; Paper 1B, where resonance-tube or driven-oscillator data must be graphed and interpreted; Paper 2 parts of 2 to 6 marks: explain formation, sketch and label a harmonic, calculate the nth harmonic, sketch resonance curves, describe damping.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
Explain how a standing wave forms from two identical waves travelling in opposite directionsSL, HL"Explain how the standing wave forms" (2–3 marks)
Identify nodes and antinodes; describe amplitude and phase along a standing waveSL, HL"State the phase difference between P and Q" (1 mark)
Distinguish a standing wave from a travelling waveSL, HL"Outline two differences" (2 marks)
Draw standing waves on strings with fixed and free endsSL, HLSketch the 2nd harmonic, label N and A (2 marks)
Draw standing waves in open and closed pipes, as displacementSL, HLSketch the first harmonic in a closed pipe (2 marks)
Determine λ and f of the nth harmonic from length and wave speedSL, HLCalculate the third harmonic (2–3 marks); open or closed from two frequencies
Describe resonance: natural frequency, driving frequency, amplitude responseSL, HLSketch amplitude against driving frequency (2 marks)
Describe the effect of damping on peak amplitude and resonant frequencySL, HLAdd a more-damped curve to a given graph (2 marks)
Describe light, critical and heavy dampingSL, HLSketch displacement–time for each; choose one for a car suspension (2–3 marks)
Give useful and destructive effects of resonanceSL, HL"Outline one useful and one destructive example" (2–4 marks)

Before you start

You need natural frequency and phase difference from C.1, v = fλ from C.2, and the principle of superposition and reflection at a boundary from C.3. Do not expect the data booklet to hand you the harmonic formulas: you build each one from a picture, and that is the skill being tested.


1The idea in one paragraph

A wave sent along a string reflects from the far end, and the incoming and reflected waves, identical but travelling in opposite directions, superpose into a wave that goes nowhere: a standing wave. Some points, the nodes, never move; the antinodes between them swing hardest. The ends force a node or an antinode to sit there, so only certain wavelengths fit, each a harmonic with its own frequency. Those are the system's natural frequencies. Drive a system at one of them and its amplitude grows large, which is resonance; damping removes energy and lowers the peak.

2How a standing wave forms

Figure 1 draws two identical waves on one line, teal moving right and amber moving left, and their sum in black, found by adding displacements point by point.

Figure 1 · Two travelling waves add up to a standing wave Figure 1 · Two travelling waves add up to a standing wave wave travelling right → ← wave travelling left sum t = 0 t = T/8 t = T/4 t = 3T/8 t = T/2 N N N N N N = node: the two waves always cancel here, half a wavelength apart Teal travels right, amber travels left. Their sum (black) never moves along: the nodes stay put.
Figure 1 · Two travelling waves add up to a standing wave

The rows are an eighth of a period apart. At t = 0 the two waves coincide and add to twice the amplitude. A quarter-period later each has moved λ/4 in opposite directions and they cancel everywhere: the string is momentarily flat. At half a period they reinforce again, the other way up. Through all of it the black curve never slides sideways, and the dotted points stay at zero at every instant, because there the two waves are always equal and opposite. Those are the nodes.

A standing wave is the superposition of two waves of the same frequency, wavelength and amplitude travelling in opposite directions.

In practice the second wave is a reflection: a wave meets a boundary and comes back over itself, on a string between two supports, in the air of a bottle, between the metal walls of a microwave oven. If the two amplitudes differ, the cancellation is incomplete and there are no true nodes, so exam answers name all three matching properties.

3Nodes, antinodes, amplitude and phase

Figure 2 labels the parts of a standing wave. The dashed grey curves are the envelope: the furthest each point ever gets from equilibrium. The teal curve is the string at one instant.

Figure 2 · Nodes, antinodes, amplitude and phase Figure 2 · Nodes, antinodes, amplitude and phase N N N N A A A P Q R λ/2 node to node λ/4 node to antinode λ (two loops) dashed: the envelope teal: one instant Every point between two neighbouring nodes moves in phase; across a node the motion flips (antiphase).
Figure 2 · Nodes, antinodes, amplitude and phase

A node is a point where the displacement is always zero. An antinode is a point where the amplitude is a maximum. The spacings come straight from Figure 1:

  • adjacent nodes are λ/2 apart, and so are adjacent antinodes;
  • a node and the next antinode are λ/4 apart;
  • one full wavelength spans two loops.

Here λ is the wavelength of the two travelling waves that make the pattern, so in a picture you count loops and multiply by λ/2.

Amplitude depends on position. In a travelling wave every point has the same amplitude. In a standing wave the amplitude runs from zero at a node to a maximum at an antinode. In Figure 2, P and Q are in the same loop but P is nearer the node, so P has the smaller amplitude.

Phase is simpler than you expect. Every point between two neighbouring nodes rises, peaks and passes through zero together, so P and Q are in phase even though their amplitudes differ. Cross a node and the motion flips: R, in the next loop, is in antiphase with P and Q, a phase difference of π rad (180°). In a standing wave those are the only two possibilities.

Travelling (progressive) waveStanding wave
Energytransferred along the wavenot transferred; stored in the pattern
Amplitudethe same at every point (no losses)varies with position: zero at nodes, maximum at antinodes
Phasechanges continuously along the wave; points λ apart are in phaseall points between adjacent nodes in phase; antiphase across a node
Wave profilemoves along at speed vstays in place; only its size changes
Wavelengthdistance between neighbouring points in phasetwice the distance between adjacent nodes

4Boundary conditions: what the ends do

The ends of a string or an air column are its boundary conditions, and they decide which standing waves can exist.

On a string:

  • A fixed end cannot move, so it is always a node.
  • A free end (a string tied to a light ring sliding on a smooth rod) moves most, so it is always an antinode.

In a pipe, the wave is sound, which is longitudinal, and we describe it by the displacement of the air along the pipe.

  • At a closed end the air is up against a wall and cannot move along the pipe, so it is always a displacement node.
  • At an open end the air is free to move in and out, so it is always a displacement antinode.

Pressure nodes and antinodes are not required, and the end correction (the antinode sitting slightly beyond an open end) is ignored: put the antinode exactly at the open end. After that, finding harmonics is a drawing problem. Put the right thing at each end, fit the fewest loops that obey both ends, then add loops one at a time.

5Standing waves on strings

Figure 3 shows the first three possible patterns for each of the three kinds of string. In each sketch the two teal curves are the extreme positions a quarter-cycle apart; the dashed line is equilibrium.

Figure 3 · Harmonics on a string: the ends decide Figure 3 · Harmonics on a string: the ends decide (a) Fixed – fixed 1st harmonic λ = 2L 2nd harmonic λ = L 3rd harmonic λ = 2L/3 f = n v / 2L, every n (b) Fixed – free 1st harmonic λ = 4L 3rd harmonic λ = 4L/3 5th harmonic λ = 4L/5 f = n v / 4L, odd n only (c) Free – free 1st harmonic λ = 2L 2nd harmonic λ = L 3rd harmonic λ = 2L/3 f = n v / 2L, every n A fixed end is always a node; a free end is always an antinode. Only wavelengths that fit are allowed.
Figure 3 · Harmonics on a string: the ends decide

The lowest-frequency standing wave is the first harmonic (the guide bans fundamental and overtone), and the nth harmonic has frequency fₙ = n f₁.

Fixed at both ends (panel a), and free at both ends (panel c). Both ends are the same kind, nodes or antinodes, so the length must hold a whole number of half-wavelengths.

L = n λn / 2 so λn = 2L / nn = 1, 2, 3, …
fn = v / λn = n v / 2Levery harmonic is possible

One end fixed, one free (panel b). One end is a node and the other an antinode, so the length holds an odd number of quarter-wavelengths.

L = n λn / 4 so λn = 4L / nn = 1, 3, 5, … only
fn = n v / 4L

An even number of quarter-wavelengths would put the same kind of point at both ends, so even harmonics are missing and the second pattern is the third harmonic (3f₁). In every case v, set by the string's tension and mass per unit length, is the same for all harmonics.

Worked example: a guitar string. A string of vibrating length 0.650 m, fixed at both ends, has a first harmonic of 196 Hz. Determine the wave speed on the string and the wavelength and frequency of its third harmonic.

first harmonic: λ1 = 2L = 2 × 0.650 = 1.30 m
v = f1 λ1 = 196 × 1.30 = 255 m s-1254.8, same for every harmonic
third harmonic: λ3 = 2L / 3 = 1.30 / 3 = 0.433 m
f3 = 3 f1 = 3 × 196 = 588 Hzcheck: v / λ3 = 254.8 / 0.4333 = 588 Hz

A classic slip: the sound you hear has the string's frequency, 196 Hz, but not its wavelength. In air at 340 m s⁻¹ the sound's wavelength is 340 ÷ 196 = 1.73 m.

6Standing waves in pipes

Figure 4 does the same for pipes. The curves are a graph of displacement amplitude along the pipe; the air itself moves left and right, not up and down.

Figure 4 · Harmonics in a pipe, drawn as displacement of the air Figure 4 · Harmonics in a pipe, drawn as displacement of the air (a) Open – open 1st harmonic λ = 2L 2nd harmonic λ = L 3rd harmonic λ = 2L/3 f = n v / 2L, every n (b) Closed – closed 1st harmonic λ = 2L 2nd harmonic λ = L 3rd harmonic λ = 2L/3 f = n v / 2L, every n (c) Closed – open 1st harmonic λ = 4L 3rd harmonic λ = 4L/3 5th harmonic λ = 4L/5 f = n v / 4L, odd n only A closed end is a displacement node, an open end a displacement antinode. The curves show how far the air moves along the pipe, not air moving sideways.
Figure 4 · Harmonics in a pipe, drawn as displacement of the air

Closed ends behave like fixed ends, open ends like free ends.

PipeλₙfₙHarmonics
Open at both ends (A, A)2L/nnv/2Lall
Closed at both ends (N, N)2L/nnv/2Lall
Closed at one end (N, A)4L/nnv/4Lodd only

So a pipe closed at one end has a first harmonic half the frequency of an open pipe of the same length (λ₁ = 4L, not 2L), and only odd harmonics.

Worked example: open or closed? A pipe resonates at 450 Hz and at 750 Hz, and at no frequency in between. The speed of sound is 340 m s⁻¹. Determine whether the pipe is open at both ends or closed at one end, and find its length.

gap between consecutive resonances = 750 − 450 = 300 Hz
if open at both ends: consecutive harmonics differ by f1, so f1 = 300 Hz
450 / 300 = 1.5not a whole number: rejected
if closed at one end: consecutive harmonics are odd, so they differ by 2 f1
f1 = 300 / 2 = 150 Hz
450 / 150 = 3, 750 / 150 = 53rd and 5th harmonics: consistent
closed at one end: L = v / 4 f1 = 340 / (4 × 150) = 0.567 m

Reading it from data. A Paper 1B resonance tube is closed at one end, with its length changed while a speaker sounds above it. For the first harmonic f₁ = (v/4) × (1/L), so f₁ against 1/L is a straight line through the origin with gradient v/4, as in the invented results of Figure 5.

Figure 5 · First harmonic against 1/L for a tube closed at one end Figure 5 · First harmonic against 1/L for a tube closed at one end f₁ (Hz) 1/L (m⁻¹) 1 2 3 4 5 100 200 300 400 Δ(1/L) = 4.0 m⁻¹ Δf = 344 Hz Invented data. A straight line through the origin: f₁ = (v/4) × (1/L), so the gradient is v/4.
Figure 5 · First harmonic against 1/L for a tube closed at one end
gradient = Δf / Δ(1/L) = 344 / 4.0 = 86 Hz mlarge triangle, read from the line
v = 4 × gradient = 4 × 86 = 344 m s-1

Choosing axes that make a straight line, then interpreting the gradient, is the Paper 1B skill.

7Damping: light, critical and heavy

A real oscillator left alone loses energy to friction, air resistance or forces inside the material. That is damping: a force opposing the velocity, so it always takes energy out. Figure 6 shows the three degrees the guide names, each for an oscillator pulled aside and released from rest.

Figure 6 · Light, critical and heavy damping Figure 6 · Light, critical and heavy damping (a) Light damping x t amplitude decays gradually oscillates, period ≈ unchanged (b) Critical damping x t no oscillation, quickest return (c) Heavy damping x t no oscillation, slow creep back All three start from the same displacement at rest. Critical damping reaches equilibrium soonest without overshooting.
Figure 6 · Light, critical and heavy damping

Light damping. The system still oscillates, but the amplitude decreases gradually, cycle by cycle, with the period almost unchanged. A plucked guitar string is lightly damped.

Critical damping. The system returns to equilibrium in the shortest possible time without oscillating, with no overshoot. Car suspension is designed close to this, so the body settles after a bump without bouncing; so is a self-closing door.

Heavy damping. No oscillation, but the return to equilibrium is slower than with critical damping: it creeps back, like a pendulum in thick oil. The large resistive force prevents overshoot but also slows the return, so heavy damping is not the fastest.

8Driven oscillations and resonance

Every oscillating system has a natural frequency, f₀, at which it oscillates if displaced and released; a string or pipe has a whole set, its harmonics. Push the system with a periodic force at a driving frequency fd, and after a short settling time it oscillates at fd, not at f₀. This is a forced oscillation, and what depends on fd is the amplitude.

Resonance is the large increase in amplitude when the driving frequency equals, or is very close to, the natural frequency. Figure 7 plots steady amplitude against driving frequency for three levels of damping.

Figure 7 · Amplitude of a driven oscillator against driving frequency Figure 7 · Amplitude of a driven oscillator against driving frequency amplitude of oscillation driving frequency light damping more damping heavy damping f₀ well above f₀: amplitude falls away More damping: a lower peak, a broader curve, and a peak slightly below f₀.
Figure 7 · Amplitude of a driven oscillator against driving frequency

Why the peak happens, in terms of energy. At f₀ the driving force stays in step with the velocity, so it does positive work throughout every cycle. The amplitude grows until the energy lost to damping per cycle equals the energy supplied per cycle. Away from f₀ the force is out of step for part of each cycle and takes energy back, so the steady amplitude is smaller.

What damping does to the curve (qualitative only, as in Figure 7):

  1. The maximum amplitude is lower, because losses balance the input at a smaller amplitude.
  2. The resonant frequency is slightly lower. The peak moves a little below f₀; with light damping the shift is too small to see.
  3. The peak is broader: a moderate response over a wide range of frequencies rather than a violent one at a single frequency.

With no damping the amplitude at resonance would grow without limit; real peaks are high but finite.

Standing waves are resonance. A string or pipe has a natural frequency for each harmonic, and energy fed in at one of them builds a large standing wave. That is why the resonance tube in section 6 goes loud at particular lengths: there its natural frequency matches the fork.

9Resonance: useful and destructive

Useful resonance.

  • Musical instruments. An organ pipe's air column or a guitar string resonates at its harmonics, turning a small input into a loud, clear note.
  • Tuning a radio. The receiving circuit's natural frequency is adjusted until it matches one station's frequency, which it then picks out from all the others.
  • Quartz clocks. A quartz crystal driven at its natural frequency oscillates very steadily, and counting the oscillations keeps time.
  • Molecules and infrared. Carbon dioxide and water vapour molecules have natural vibration frequencies in the infrared, so they absorb the Earth's infrared emission at those frequencies: the link to the greenhouse effect in B.2.

Destructive resonance.

  • Bridges and buildings. Wind, footsteps or earthquakes can drive a structure near its natural frequency. London's Millennium Bridge swayed sideways under crowds when it opened in 2000 and was closed until dampers were fitted; tall buildings in earthquake zones carry damping systems for the same reason.
  • Machinery and vehicles. An engine speed that matches the natural frequency of a panel makes it rattle, and repeated large oscillation can crack metal through fatigue.

The cure is always one of two: move the natural frequency away from the driving frequency, or add damping to cut the peak.

10Where marks are lost

"A wave that doesn't move", or "two waves meet". The particles do oscillate; the profile stays put and no energy is transferred. And formation needs two waves of the same frequency, wavelength and amplitude travelling in opposite directions and superposing.

Taking λ as the node-to-node distance. Adjacent nodes are λ/2 apart. Count loops, multiply by λ/2.

Giving a phase difference like π/2 between two points on a standing wave. Only 0 (same loop) or π (adjacent loops, across a node) is possible. Amplitude varies within a loop; phase does not.

Calling the second pattern of a closed pipe the "second harmonic". A pipe closed at one end has odd harmonics only. The second pattern is the third harmonic, frequency 3f₁.

Using "fundamental" and "overtone". The guide bans them. Use first harmonic, second harmonic, and so on.

Swapping the ends of pipes. A closed end is a displacement node (the wall stops the air); an open end is a displacement antinode.

Saying heavy damping returns fastest, or that the frequency of a forced oscillation is the natural frequency. Critical damping is fastest without overshoot. A driven system oscillates at the driving frequency; only its amplitude depends on how close that is to f₀.

11Draw it right

  1. Standing waves: both extreme positions drawn, nodes labelled N and antinodes A, ends matching the boundary (fixed or closed = node; free or open = antinode).
  2. Equal loops: each λ/2 long, with a quarter-loop (λ/4) at a free or open end.
  3. Name harmonics by frequency ratio: a closed–open pipe runs 1st, 3rd, 5th.
  4. Pipes: mark the closed end and label the envelope "displacement".
  5. Resonance curve: amplitude against driving frequency, starting non-zero at low frequency, peaking at f₀, falling towards zero above it.
  6. More damping: a lower, broader peak, below the old curve near f₀ and slightly to its left.
  7. Damping sketches: light oscillates in a decaying envelope at constant period; critical reaches zero soonest with no overshoot; heavy approaches zero more slowly and never crosses.

12Try it

Marks in brackets. Answers and marker's notes are at the end. Take the speed of sound in air as 340 m s⁻¹.

Q1. A pipe of length L is closed at one end and open at the other. Which of these is not a possible wavelength of a standing wave in the pipe? 1 mark

A. 4L

B. 2L

C. 4L/3

D. 4L/5

Q2. A string is stretched between two fixed supports and vibrated at one end.

(a) Explain how a standing wave forms on the string. 2 marks

(b) State two differences between the standing wave and a travelling wave on the same string. 2 marks

Q3. An organ pipe is 0.85 m long and open at both ends.

(a) Calculate the frequency of its first harmonic. 2 marks

(b) Sketch the displacement pattern of the second harmonic, labelling the nodes and antinodes. 2 marks

(c) One end of the pipe is now closed. Determine the two lowest frequencies at which it resonates. 2 marks

Q4. A string 0.90 m long, fixed at both ends, vibrates in its third harmonic at 240 Hz. Points X and Y on the string are 0.20 m and 0.40 m from one end.

(a) Determine the speed of the waves on the string. 2 marks

(b) State and explain the phase difference between X and Y. 2 marks

Q5. A student hangs a 0.20 kg mass from a spring and drives the top of the spring up and down with a vibration generator. She records the amplitude of the mass's oscillation at different driving frequencies (invented data).

Driving frequency / Hz1.01.52.02.22.42.62.83.03.5
Amplitude / mm471630523117115

(a) Estimate the natural frequency of the mass–spring system. 1 mark

(b) Hence estimate the spring constant, using T = 2π√(m/k) from C.1. 2 marks

(c) The student attaches a large card to the mass to increase the air resistance. Describe how the graph of amplitude against driving frequency changes. 2 marks

(d) Explain, in terms of energy, why the amplitude is largest near the natural frequency. 2 marks

Q6. Car suspensions are designed to be close to critically damped.

(a) Describe what is meant by critical damping. 2 marks

(b) Explain why neither light damping nor heavy damping would be suitable. 2 marks

(c) Outline one situation in which resonance is useful. 1 mark

13In one breath

Two identical waves travelling in opposite directions, usually a wave and its reflection, superpose into a standing wave. Nodes never move; antinodes have the largest amplitude; nodes are λ/2 apart and a node to the next antinode is λ/4. No energy is transferred, amplitude varies with position, points in one loop are in phase and adjacent loops are in antiphase. Fixed and closed ends are displacement nodes, free and open ends antinodes. Like ends give λₙ = 2L/n, fₙ = nv/2L, every harmonic; unlike ends give λₙ = 4L/n, fₙ = nv/4L, odd harmonics only. The lowest is the first harmonic. Driving at the natural frequency gives resonance, the amplitude growing until damping losses match the input; more damping lowers and broadens the peak and moves it slightly below f₀. Light damping oscillates and decays, critical returns fastest without overshoot, heavy creeps back. Resonance makes music and tunes radios; it also shakes bridges.


Answers

Q1. B. Node at one end and antinode at the other, so L is an odd number of quarter-wavelengths: λ = 4L/n, n odd. 2L would need an even number. A, C and D are the 1st, 3rd and 5th harmonics.

Q2. (a) The wave reflects at the far support. The reflected and incoming waves have the same frequency, wavelength and amplitude but travel in opposite directions, and they superpose, cancelling permanently at the nodes. (b) Any two, each stated for both waves: energy is transferred by a travelling wave but not by a standing wave; amplitude is the same everywhere on a travelling wave but varies with position on a standing wave; phase changes continuously along a travelling wave, while points in one loop of a standing wave are in phase. (a) 1 for reflection giving a second wave travelling the opposite way, 1 for same f, λ and amplitude superposing. (b) 1 for each correct difference, each stated for both waves. "One moves and one doesn't" scores 0.

Q3. (a) Open at both ends: λ₁ = 2L = 1.70 m, so f₁ = v/λ₁ = 340 ÷ 1.70 = 200 Hz. (b) Antinodes at both ends and the centre, nodes at L/4 and 3L/4. (c) Closed at one end: f₁ = v/4L = 340 ÷ 3.40 = 100 Hz; the next harmonic is the third, 3 × 100 = 300 Hz. (a) M1 for λ = 2L, A1 for 200 Hz. (b) 1 for antinodes at both ends, 1 for two nodes correctly placed. (c) A1 for 100 Hz, A1 for 300 Hz. Giving 200 Hz as the second frequency in (c) ignores the missing even harmonics and scores A1 for 100 Hz only.

Q4. (a) Third harmonic, fixed ends: λ = 2L/3 = 2 × 0.90 ÷ 3 = 0.60 m. v = fλ = 240 × 0.60 = 144 m s⁻¹. (b) Nodes are λ/2 = 0.30 m apart, at 0, 0.30, 0.60 and 0.90 m. The node at 0.30 m lies between X and Y, so they are in adjacent loops and in antiphase: π rad (180°). (a) M1 for λ = 0.60 m, A1 for 144 m s⁻¹. (b) A1 for π or 180°, R1 for a node between them or adjacent loops. A phase difference found from path difference (0.20 ÷ 0.60 × 2π) treats it as a travelling wave and scores 0.

Q5. (a) About 2.4 Hz, where the amplitude is greatest. (b) T = 1/f = 1 ÷ 2.4 = 0.417 s. From T = 2π√(m/k), k = 4π²m/T² = 4π²mf² = 4π² × 0.20 × 2.4² = 45 N m⁻¹. (c) The peak is lower and broader, and occurs at a slightly lower frequency. (d) At the natural frequency the driving force is in step with the velocity and does positive work throughout each cycle, so energy is added every cycle; the amplitude grows until the energy lost to damping per cycle equals the energy supplied. (a) A1. (b) M1 for rearranging with f = 2.4 Hz, A1 for 45 N m⁻¹ (accept 43–48 for f between 2.35 and 2.45 Hz). (c) 1 for lower (and broader) peak, 1 for peak shifted to lower frequency. (d) 1 for energy transferred most efficiently / force in phase with velocity, 1 for amplitude limited when input equals losses.

Q6. (a) The system returns to its equilibrium position in the shortest possible time without oscillating (without overshooting). (b) Light damping would let the car bounce several times after each bump; heavy damping would return so slowly that the suspension would still be compressed at the next bump. (c) For example, a radio receiver, whose circuit is adjusted until its natural frequency matches one station's frequency. (a) 1 for shortest time, 1 for no oscillation. (b) 1 for light: continued oscillation; 1 for heavy: slow return. (c) 1 for any correct example with its mechanism in a few words; "a swing" with no explanation scores 0.


Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section C.4 Standing waves and resonance. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!