Standing waves are where everything in IB Physics HL wave theory comes together: superposition (C.3), boundary conditions, harmonics, and the link between geometry and frequency. The same harmonic formulas govern guitar strings, organ pipes, microwave ovens, atomic orbitals, and the resonance modes of bridges. Topic C.4 also introduces resonance and damping, which appear again in electromagnetism (Topic D) and atomic physics (Topic E).
This cheatsheet condenses every Topic C.4 formula and exam trap into one revision sheet — node and antinode separations, the four boundary cases (N–N, N–A, A–A) for both strings and pipes, harmonic-frequency formulas, the effect of damping on a resonance curve, and the three damping regimes (light, critical, heavy). The most common HL pitfalls — saying energy travels along a standing wave, claiming adjacent antinodes are in phase, confusing critical with heavy damping — are flagged in red.
§1 — Formation of Standing Waves C.4 SL + HL
A standing wave is the superposition of two identical travelling waves moving in opposite directions:
$$y(x, t) = 2A \sin(kx) \cos(\omega t)$$
The amplitude at position $x$ is $|2A \sin(kx)|$ — fixed in space, varying with position. The wave does not travel; the entire pattern oscillates in place at frequency $\omega/(2\pi)$.
§2 — Nodes, Antinodes & Phase Rules C.4 SL + HL
Phase rules
- Points in the same segment (between two adjacent nodes): in phase, $\Delta\varphi = 0$.
- Points in adjacent segments (separated by one node): in antiphase, $\Delta\varphi = \pi$.
- Crossing $n$ nodes: $\Delta\varphi = n\pi$ rad.
- Adjacent antinodes are in antiphase ($\Delta\varphi = \pi$) — they sit in adjacent segments.
§3 — Standing Waves on Strings C.4 SL + HL
| Boundary | Harmonics present | Frequency formula |
|---|---|---|
| Both fixed (N–N) | All: $1, 2, 3, \ldots$ | $f_n = \dfrac{n v}{2L}$ |
| Fixed + free (N–A) | Odd only: $1, 3, 5, \ldots$ | $f_n = \dfrac{(2n-1) v}{4L}$ |
| Both free (A–A) | All: $1, 2, 3, \ldots$ | $f_n = \dfrac{n v}{2L}$ |
§4 — Standing Waves in Pipes C.4 SL + HL
| Boundary | Harmonics present | Frequency formula |
|---|---|---|
| Both open (A–A) | All: $1, 2, 3, \ldots$ | $f_n = \dfrac{n v}{2L}$ |
| Closed + open (N–A) | Odd only: $1, 3, 5, \ldots$ | $f_n = \dfrac{(2n-1) v}{4L}$ |
| Both closed (N–N) | All: $1, 2, 3, \ldots$ | $f_n = \dfrac{n v}{2L}$ |
§5 — Resonance C.4 SL + HL
Resonance occurs when the driving frequency $f_\text{drive}$ matches (or is close to) the natural frequency $f_0$ of the system. At resonance, the amplitude of oscillation is maximum and energy transfer from driver to oscillator is most efficient.
Effect of damping on the resonance curve
| As damping increases | Effect on resonance curve |
|---|---|
| Peak amplitude | Decreases |
| Peak width | Increases (broader) |
| Resonant frequency | Shifts slightly below $f_0$ |
§6 — Types of Damping C.4 SL + HL
| Type | Motion | Example |
|---|---|---|
| Light damping | Oscillatory; amplitude decays exponentially | Pendulum in air |
| Critical damping | Fastest return, no oscillation | Car shock absorbers |
| Heavy (over-)damping | Slow return, no oscillation | Hydraulic door closer |
§7 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Find the first harmonic frequency" | Identify boundary type (N–N, N–A, A–A); apply $f_n$ formula with $n = 1$. |
| "Sketch the third harmonic" | Place 3 antinodes (or appropriate number for boundary type). |
| "Phase difference between two points" | Count the nodes between them; multiply by $\pi$. |
| "Distance between adjacent nodes" | $\lambda/2$. |
| "Wavelength on the string / in the pipe" | From boundary diagram; for N–N, $\lambda_n = 2L/n$. |
| "Closed-open vs open-open of same length" | Closed-open has half the first-harmonic frequency. |
| "What happens at resonance?" | Maximum amplitude; energy transfer most efficient. |
| "Effect of damping on resonance" | Lower peak, broader curve, slight downward shift in peak frequency. |
| "Which damping returns fastest without oscillating?" | Critical damping. |
| "Does a standing wave transfer energy?" | No — energy oscillates within segments, no net transfer. |
Worked Example — IB-Style Standing Wave Problem
Question (HL Paper 2 style — 7 marks)
An organ pipe is open at one end and closed at the other. Its length is $L = 0.85$ m. The speed of sound in air is 340 m s$^{-1}$. (a) State which end is a displacement node. (b) Calculate the frequency of the first harmonic. (c) Calculate the frequency of the third harmonic. (d) Could a standing wave at exactly twice the first-harmonic frequency exist in this pipe? Explain.
Solution
- The closed end is a displacement node (air molecules cannot move there). The open end is a displacement antinode. (A1)
- For a closed-open (N–A) pipe, $f_n = (2n - 1) v / (4L)$. With $n = 1$: $f_1 = 340 / (4 \times 0.85) = 340 / 3.40 = 100\ \mathrm{Hz}$. (M1)(A1)
- The "third harmonic" in a closed-open pipe corresponds to $n = 2$ in the formula (odd harmonics only): $f = (2 \times 2 - 1) \times 340 / (4 \times 0.85) = 3 \times 100 = 300\ \mathrm{Hz}$. (M1)(A1)
- No — a frequency of $2 f_1 = 200\ \mathrm{Hz}$ corresponds to the second harmonic, which is forbidden in a closed-open pipe. (R1)
- Reason: only odd harmonics ($f_1, 3 f_1, 5 f_1, \ldots$) satisfy both boundary conditions (node at the closed end, antinode at the open end). (R1)
Examiner's note: The trap is that students conflate "third harmonic" with "$n = 3$" in the closed-open formula. In a closed-open pipe, the $n$-th allowed mode in the formula gives the $(2n-1)$-th harmonic. The third harmonic is the second allowed mode, with frequency $3 f_1$.
Common Student Questions
Do standing waves transfer energy along the medium?
What is the phase difference between adjacent antinodes?
Why does a closed-open pipe only have odd harmonics?
What is the difference between critical damping and heavy damping?
What happens to the resonance curve as damping increases?
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