What are stationary waves and how do they form in the context of superposition?

Stationary waves, also known as standing waves, are waves that remain in a constant position. They form as a result of the superposition of two waves of the same frequency and amplitude traveling in opposite directions. This superposition leads to points of no displacement called nodes and points of maximum displacement called antinodes. In the Cambridge International A Levels Physics curriculum, understanding the formation of stationary waves is crucial for grasping the concept of wave interference.

How do stationary waves differ from progressive waves?

Stationary waves differ from progressive waves in that they do not transfer energy from one place to another. Instead, they oscillate in place. In a stationary wave, nodes and antinodes are fixed, whereas in progressive waves, the wave fronts move through the medium. This distinction is important in the study of wave phenomena in the Cambridge International A Levels Physics syllabus.

What are nodes and antinodes in stationary waves?

In stationary waves, nodes are points along the medium that remain stationary, experiencing no displacement. Antinodes, on the other hand, are points where the displacement is at its maximum. The distance between two consecutive nodes or antinodes is half the wavelength of the wave. Understanding nodes and antinodes is essential for analyzing wave patterns in the Cambridge International A Levels Physics course.

What are some real-world examples of stationary waves?

Real-world examples of stationary waves include the vibrations of a guitar string, the resonance in a column of air in a wind instrument, and the standing waves formed in microwave ovens. These examples illustrate the practical applications of stationary waves, a topic covered in the Cambridge International A Levels Physics curriculum.

How can the concept of stationary waves be demonstrated in a laboratory setting?

In a laboratory setting, stationary waves can be demonstrated using a string fixed at both ends or a tube with one or both ends open. By adjusting the frequency of a wave generator, students can observe the formation of nodes and antinodes. This hands-on experiment helps reinforce the theoretical concepts of stationary waves and superposition, as outlined in the Cambridge International A Levels Physics syllabus.

Why is "Saying stationary waves transfer energy" a common mistake in Stationary waves?

Why it happens: Confusing with progressive. How to avoid it: Stationary waves do NOT transfer net energy. Energy oscillates locally between KE (at antinodes) and PE (string tension). Source: 9702 Examiner Reports 2022-2024

Cambridge International A Levels Physics (9702)

Stationary waves

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Short Notes - Stationary waves

Detailed Study Notes

Detailed notes on Superposition for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Stationary Waves Study Notes — Cambridge International A Level Physics 9702 Superposition (2025-2027 syllabus)

Stationary wave formation. Nodes and antinodes. Strings and pipes.

At a glance

Stationary = two opposite progressive waves.
Node = zero amplitude.
Antinode = max amplitude.
Distance node-node = $\lambda/2$ .
Fundamental: $\lambda = 2L$ (fixed both ends).

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

8.1.1 — Describe stationary wave formation.
8.1.2 — Identify nodes and antinodes.
8.1.3 — Apply to strings and pipes.

Formation

Two opposite progressives.

Stationary (standing) wave. Formed by superposition of two progressive waves of same:

frequency,
amplitude,
speed, travelling in opposite directions.

Often: incident wave + reflected wave.

Properties.

Pattern of nodes and antinodes is stationary.
No net energy transfer.
Distance between adjacent nodes = $\lambda/2$ .
Distance between adjacent antinodes = $\lambda/2$ .
Distance from node to nearest antinode = $\lambda/4$ .

vs progressive wave.

Progressive: transfers energy; all points have same amplitude.
Stationary: no net energy transfer; amplitude varies along wave.

Cambridge tip. State the two requirements explicitly.

Figure 1 — A stationary wave formed by two opposite progressive waves

The solid and dashed red curves are the same stationary wave half a period apart. Nodes (N) never move; antinodes (A) swing between the two extremes.

Figure 2 — Node and antinode spacing, and comparison with a progressive wave

Nodes are half a wavelength apart, not a whole one — the single most common slip when converting a measured spacing into a wavelength.

Two opposite waves superpose.
Stationary pattern.
No energy transfer.

Stationary waves on strings

Both ends fixed → nodes at ends.

Both ends fixed. Nodes at each end.

Fundamental (1st harmonic): one antinode in middle. $\lambda = 2L$ .

Higher harmonics: $\lambda_n = 2L/n$ , $f_n = nv/(2L)$ for $n = 1, 2, 3, \ldots$ .

Example. Guitar string 0.65 m, fundamental 220 Hz: $\lambda = 1.30$ m, $v = 286$ m/s.

Cambridge tip. Both string and pipe with two open ends have $f_n = nv/(2L)$ .

Figure 3 — Stationary waves on a string

A stationary wave has nodes (N) where displacement is always zero and antinodes (A) where it is maximum; on a string fixed both ends fₙ = nv/(2L).

Ends fixed = nodes.
Fundamental: $\lambda = 2L$ .
$f_n = nv/(2L)$ .

See the full worked example for stationary waves →

Stationary waves in pipes

Open or closed ends.

Open end = antinode (free to vibrate). Closed end = node (cannot vibrate).

Pipe open at both ends. Antinodes at both ends. $f_n = nv/(2L)$ , $n = 1, 2, 3, \ldots$ . All harmonics.

Pipe closed at one end. Node at closed, antinode at open. Fundamental: $L = \lambda/4$ → $\lambda = 4L$ .

$f_n = (2n-1)v/(4L)$ , $n = 1, 2, 3, \ldots$ . ONLY ODD harmonics ( $f, 3f, 5f$ , etc.).

Example. Closed pipe 0.34 m, $v = 340$ m/s: fundamental $\lambda = 1.36$ m → $f = 250$ Hz.

Cambridge tip. Closed pipe only has ODD harmonics — key distinction.

Figure 4 — Fundamental modes of an open pipe and a closed pipe

The closed pipe fits only a quarter of a wavelength into its length, so for the same length it sounds an octave lower than the open pipe.

Figure 5 — A closed pipe supports only odd harmonics

Every allowed mode must start with a node at the closed end and finish with an antinode at the open end, which only odd quarter-wavelengths satisfy.

Open = antinode; closed = node.
Both open: $\lambda = 2L$ fundamental.
Closed one end: $\lambda = 4L$ , odd harmonics only.

See the full worked example for stationary waves →

Quick recap

Stationary = two opposite progressive.
Node-node spacing $\lambda/2$ .
Strings/open pipes: all harmonics.
Closed pipes: odd harmonics only.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Formation conditions.
$f_n$ for string, open pipe, closed pipe.

How it’s examined

Stationary waves on every Paper 1 or 2. Most-tested: string fundamental (6 marks), pipe harmonics (6-8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Stationary waves

Step-by-step solutions to past-paper-style questions on stationary waves, written exactly the way a tutor would explain them at the board.

1Fundamental frequency on string (6 marks)
Extended• Adapted from 9702/22 May/Jun 2024• string
Question
A guitar string 0.65 m long vibrates at fundamental frequency 220 Hz. Find wave speed. (6 marks)
Step-by-step solution
1. Step 1
  Fundamental = string length is half a wavelength.
  $\lambda = 2L = 1.30 \text{ m}$
2. Step 2
  $v = f\lambda$ .
  $v = 220 \times 1.30 = 286 \text{ m/s}$
Answer
$v = 286$ m/s.
2Closed pipe (5 marks)
Extended• pipe
Question
A pipe closed at one end has length 0.34 m. Find the lowest natural frequency. ( $v = 340$ m/s.) (5 marks)
Step-by-step solution
1. Step 1
  Closed pipe. Fundamental: $\lambda/4 = L$ , so $\lambda = 4L = 1.36$ m.
2. Step 2
  Frequency.
  $f = v/\lambda = 340/1.36 = 250 \text{ Hz}$
Answer
$f = 250$ Hz.
3Identify nodes/antinodes (5 marks)
Extended• nodes
Question
A stationary wave on a string fixed at both ends shows 3 nodes (including ends) and 2 antinodes. Sketch and find the relationship between $\lambda$ and string length $L$ . (5 marks)
Step-by-step solution
1. Step 1
  3 nodes (counting ends) + 2 antinodes. This is the SECOND harmonic.
2. Step 2
  Distance between nodes = $\lambda/2$ . Two intervals → $L = \lambda$ .
Answer
$L = \lambda$ (second harmonic).

Key Formulae — Stationary waves

The formulae you need to memorise for stationary waves on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Frequencies on string (both ends fixed)
$f_n = \dfrac{nv}{2L}, \quad n = 1, 2, 3, \ldots$
$L$
string length
$n$
harmonic number
When to use
Both ends fixed = nodes. Fundamental $n = 1$ .
Closed pipe frequencies
$f_n = \dfrac{(2n-1)v}{4L}, \quad n = 1, 2, 3, \ldots$
When to use
One end closed (node), one open (antinode). Only odd harmonics.
Open pipe frequencies
$f_n = \dfrac{nv}{2L}, \quad n = 1, 2, 3, \ldots$
When to use
Both ends open (antinodes). Same as string, but harmonics generated differently.

Key Definitions and Keywords — Stationary waves

Definitions to memorise and the exact keywords mark schemes credit for stationary waves answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Stationary wave
Examiner keyword
Formed by superposition of two waves of same frequency, amplitude, travelling in opposite directions. No net energy transfer.
Node
Examiner keyword
Point of zero displacement at all times. Antinodes between.
Antinode
Examiner keyword
Point of maximum amplitude oscillation.
Fundamental frequency
Examiner keyword
Lowest natural frequency of a system. Corresponds to simplest stationary wave mode.

Common Mistakes and Misconceptions — Stationary waves

The traps other students keep falling into on stationary waves questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Saying stationary waves transfer energy
9702 Examiner Reports 2022-2024
Why it happens
Confusing with progressive.
How to avoid it
Stationary waves do NOT transfer net energy. Energy oscillates locally between KE (at antinodes) and PE (string tension).

Stationary waves — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.