Free Cheatsheet · Topic C.4 · SL + HL

IB Physics HL Standing Waves & Resonance — Complete Cheatsheet

Every formula, boundary-condition diagram, trick, and trap for IB Physics HL Topic C.4 — formation of standing waves, strings and pipes, harmonics, resonance, and damping.

Topic: C.4 Standing Waves & Resonance Syllabus: SL + HL (common content) Read time: ~13 minutes Last updated: Apr 2026

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 $\omega/(2\pi)$.

TrickBoth component travelling waves must have the same $f$, $\lambda$ and $A$. If $A$ differs, the nodes are not perfectly zero-amplitude — but IB always assumes perfect reflection unless stated otherwise.
TrapStanding waves do not transfer energy along the medium. Marks are lost if you write "energy travels from N to A" or "energy flows from one end to the other".
NoteNode-to-node separation $= \lambda/2$. Antinode-to-antinode separation $= \lambda/2$. Node-to-antinode separation $= \lambda/4$.

§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.
TrapAdjacent antinodes have $\Delta\varphi = \pi$, NOT zero. They oscillate in antiphase even though both are antinodes. Many students write "in phase" because the two points look similar — IB examiners hammer this in markschemes.
TrickTo find the phase difference between two points on a standing wave, count the nodes between them. Each node adds $\pi$ to the phase difference.

§3 — Standing Waves on Strings C.4 SL + HL

BoundaryHarmonics presentFrequency 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}$
TrickFor an N–A or A–N system, only odd harmonics exist. The frequencies are in the ratio $1:3:5:7\ldots$ — never $1:2:3$.
TrapThe lowest frequency mode is called the first harmonic, not the "fundamental" or "first overtone". IB markschemes only accept "first harmonic" — older textbooks use different conventions.
NoteFor a fixed-fixed string, the number of nodes equals harmonic number $+ 1$, and the number of antinodes equals the harmonic number.

§4 — Standing Waves in Pipes C.4 SL + HL

BoundaryHarmonics presentFrequency 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}$
TrapA closed end is a displacement node (air can't move). An open end is a displacement antinode. This is for displacement of air molecules — pressure has the opposite pattern (closed end = pressure antinode). IB usually asks about displacement.
TrickFor the same length pipe, $f_{1,\text{open-open}} = 2 \times f_{1,\text{closed-open}}$. The closed-open pipe has half the first-harmonic frequency because the fundamental wavelength is $4L$ instead of $2L$.
NoteNo end corrections are needed in IB Physics Topic C.4. Use the pipe length $L$ directly in the harmonic formulas.

§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 increasesEffect on resonance curve
Peak amplitudeDecreases
Peak widthIncreases (broader)
Resonant frequencyShifts slightly below $f_0$
TrickSketch: light damping = tall narrow peak; heavy damping = short broad peak. Both curves rise from a common low value at very low driving frequency and cross the axis well below $f_0$.
TrapResonant frequency (the peak of the curve) $\ne f_0$ for heavily damped systems — it sits slightly below. For lightly damped systems they are approximately equal. Be careful which "resonant frequency" the question is asking about.

§6 — Types of Damping C.4 SL + HL

TypeMotionExample
Light dampingOscillatory; amplitude decays exponentiallyPendulum in air
Critical dampingFastest return, no oscillationCar shock absorbers
Heavy (over-)dampingSlow return, no oscillationHydraulic door closer
NoteCritical damping is the minimum damping needed to prevent oscillation and return the system to equilibrium as fast as possible.
TrapHeavy damping is not faster than critical — it is slower. Critical damping is the fastest non-oscillatory return. Many students assume "more damping = faster decay" and lose the mark.

§7 — Exam Attack Plan All sections

When you see this in the question — reach for that:

Question triggerReach 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

  1. The closed end is a displacement node (air molecules cannot move there). The open end is a displacement antinode.  (A1)
  2. 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)
  3. 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)
  4. No — a frequency of $2 f_1 = 200\ \mathrm{Hz}$ corresponds to the second harmonic, which is forbidden in a closed-open pipe.  (R1)
  5. 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?
No. A standing wave is the superposition of two travelling waves of equal amplitude, frequency and wavelength moving in opposite directions. The two energy flows cancel, so the net energy transfer along the medium is zero. Energy is still oscillating between kinetic and potential within each segment, but it does not travel from one node to another. Saying "energy travels from N to A" is a classic mark loss in IB Topic C.4.
What is the phase difference between adjacent antinodes?
Adjacent antinodes are in antiphase — phase difference $\pi$ (180°) — because they sit in adjacent segments separated by one node. Each node adds $\pi$ to the phase difference. Points within the same segment (between two nodes) are in phase. This is heavily examined and many students incorrectly say adjacent antinodes are in phase because they look similar.
Why does a closed-open pipe only have odd harmonics?
At a closed end, air molecules cannot move, so there must be a displacement node. At the open end, air is free to move, so there is a displacement antinode. The boundary conditions (N–A) only fit when the pipe length $L$ equals $(2n-1)\lambda/4$, giving $f_n = (2n-1) v / (4L)$ for $n = 1, 2, 3, \ldots$ — that is, harmonics in the ratio $1:3:5:7$. Even harmonics simply do not satisfy both boundary conditions.
What is the difference between critical damping and heavy damping?
Critical damping is the minimum damping needed to prevent oscillation, so the system returns to equilibrium as fast as possible without overshooting. Heavy (over-)damping returns the system to equilibrium more slowly than critical damping — the extra friction slows everything down. Light damping allows the system to oscillate but with exponentially decreasing amplitude. Common engineering example: car shock absorbers use approximately critical damping for the smoothest ride.
What happens to the resonance curve as damping increases?
The peak amplitude decreases (less efficient energy transfer) and the peak becomes broader (the system responds over a wider range of driving frequencies). The peak frequency also shifts slightly below the natural frequency $f_0$ for heavily damped systems, though for lightly damped systems the peak is approximately at $f_0$. A useful sketch: light damping = tall narrow peak; heavy damping = short broad peak.

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