# Stationary Waves

> CIE A-Level Physics · Unit 8: Superposition
> Source: https://www.owlsprep.com/study/cie-9702-u8-stationary-waves/

This module explains how stationary (standing) waves form from the superposition of two identical progressive waves travelling in opposite directions. You will learn to identify key features and solve common exam problems for strings and air columns.

**Prerequisites:** [Progressive wave properties and superposition principle](https://www.owlsprep.com/study/cie-9702-u8-progressive-waves/)

## Learning objectives

- Explain how stationary waves form via superposition
- Identify nodes and antinodes for different media
- Calculate wavelength and frequency from harmonic data
- Compare properties of stationary and progressive waves
- Solve problems for strings and open/closed air columns

## Formation and Key Features

**Stationary (Standing) Wave** — A wave that forms when two identical progressive waves of the same frequency and amplitude travel in opposite directions along the same medium, superpose, and produce no net propagation of energy

*Example:* Vibrations on a guitar string held fixed at both ends

When the two waves superpose, points of permanent destructive interference are called nodes (zero amplitude), and points of permanent constructive interference are called antinodes (maximum amplitude). All points between two adjacent nodes vibrate in phase with each other.

**Worked example:** A stationary wave forms on a string of length 1.2 m fixed at both ends, with 3 antinodes between the fixed ends. How many nodes are present in total on the string?

1. Fixed ends of a string are always nodes, so we start with 2 nodes at each end.
2. Between 3 antinodes, there are 2 additional internal nodes, between each pair of adjacent antinodes.
3. Total nodes = 2 (ends) + 2 (internal) = 4 nodes total

> **Exam tip:** Always remember: fixed ends of strings and closed ends of air tubes are always nodes; open ends of air tubes are always antinodes.

## Wavelength and Harmonic Calculations

A core relationship for all stationary waves is that the distance between two consecutive nodes (or two consecutive antinodes) is always $\frac{\lambda}{2}$. The distance between a node and its adjacent antinode is $\frac{\lambda}{4}$. For a string fixed at both ends, the length of the string $L$ relates to wavelength by:

$$L = \frac{n\lambda}{2}, \quad n = 1, 2, 3,...$$

$n$ is the harmonic number, where $n=1$ is the first (fundamental) harmonic.

**Worked example:** A string of length 0.8 m fixed at both ends vibrates in its 3rd harmonic. Calculate the wavelength of the stationary wave.

1. For a string fixed at both ends, use the relation: $L = \frac{n\lambda}{2}$
2. Substitute $L=0.8$ m and $n=3$:
3. $$\lambda = \frac{2L}{n} = \frac{2 \times 0.8}{3} = \frac{1.6}{3} \approx 0.53 \, \text{m}$$

**Check your understanding**

Test your understanding:

1. What is the distance between a node and the next adjacent antinode?

   - $\lambda/4$
   - $\lambda/2$
   - $\lambda$
   - $3\lambda/4$

   *Answer:* $\lambda/4$

   *Why:* Consecutive nodes are $\lambda/2$ apart, so half that distance between a node and adjacent antinode is $\lambda/4$.

## Stationary Waves in Air Columns

Sound waves reflect off the ends of air tubes to form stationary waves. There are two common cases: open-ended tubes (open at both ends) and closed-ended tubes (closed at one end, open at the other). The end conditions and wavelength relations are different for each case.

- Open at both ends: Antinodes at both ends, $L = \frac{n\lambda}{2}$, $n=1,2,3...$ (all integers)
- Closed at one end: Node at closed end, antinode at open end, $L = \frac{n\lambda}{4}$, $n=1,3,5...$ (only odd harmonics)

**Worked example:** A 25 cm long closed-end air tube produces its 1st harmonic. The speed of sound is 340 m/s. Calculate the frequency of the harmonic.

1. Convert length to meters: $25 \, \text{cm} = 0.25 \, \text{m}$
2. 1st harmonic for closed-end tube: $n=1$, so $L = \frac{\lambda}{4}$
3. Rearrange to get $\lambda = 4L = 4 \times 0.25 = 1 \, \text{m}$
4. Use wave equation $v = f\lambda$, rearrange for $f$:
5. $$f = \frac{v}{\lambda} = \frac{340}{1} = 340 \, \text{Hz}$$

> **Exam tip:** Always confirm if the air column is closed at one end or open at both before starting calculations, this is a common source of lost marks.

*Calculator:* allowed

## Stationary vs Progressive Waves

| Property | Stationary Wave | Progressive Wave |
| --- | --- | --- |
| Net energy transfer | No net transfer | Transfers energy |
| Amplitude | Varies from 0 at nodes to max at antinodes | Same for all points |
| Phase | All points between 2 nodes are in phase | All points have different phase |
| Wavelength relation | $\lambda = 2 \times$ inter-nodal distance | $\lambda = $ distance between consecutive in-phase points |

**Worked example:** State two differences between a stationary wave and a progressive wave of the same frequency, and state the net speed of energy transfer for a stationary wave.

1. Two valid differences are: (1) Stationary waves have no net energy transfer, while progressive waves transfer energy along the medium. (2) Amplitude varies along a stationary wave, but is constant for a progressive wave.
2. The net speed of energy transfer for a stationary wave is 0 m/s, since energy is trapped in the oscillations of each segment.

## Common pitfalls

- **Wrong:** Assuming open ends of air columns are nodes
  - Why it fails: Confusion with fixed ends of strings, which are always nodes
  - Correct: Open ends of air columns allow maximum air movement, so they are always antinodes
- **Wrong:** Using $L = n\lambda/2$ for the fundamental harmonic of a closed-end air column
  - Why it fails: Ignoring the different boundary condition for a closed end
  - Correct: Use $L = n\lambda/4$ for one-closed-end air columns, with only odd values of n
- **Wrong:** Taking inter-nodal distance equal to a full wavelength $\lambda$
  - Why it fails: Mixing up wavelength definitions for progressive and stationary waves
  - Correct: Distance between two consecutive nodes is $\lambda/2$, so $\lambda = 2 \times$ inter-nodal distance
- **Wrong:** Claiming stationary waves have no energy at all
  - Why it fails: Misinterpretation of the 'no net transfer' rule
  - Correct: Energy is stored in the oscillations of the wave; there is just no net propagation of energy along the medium
- **Wrong:** Stating all points on a stationary wave have the same amplitude
  - Why it fails: Confusing the property of stationary waves with progressive waves
  - Correct: Amplitude ranges from zero at nodes to maximum at antinodes along a stationary wave

## Cheatsheet

| Medium Type | End Conditions | Wavelength Relation | Allowed Harmonics |
| --- | --- | --- | --- |
| String fixed both ends | Nodes at both ends | $L = \frac{n\lambda}{2}$ | $n = 1, 2, 3...$ |
| Air column open both ends | Antinodes at both ends | $L = \frac{n\lambda}{2}$ | $n = 1, 2, 3...$ |
| Air column one closed end | Node at closed, antinode at open | $L = \frac{n\lambda}{4}$ | $n = 1, 3, 5...$ (only odd) |
| Consecutive nodes/antinodes | - | distance = $\frac{\lambda}{2}$ | - |
| Node to adjacent antinode | - | distance = $\frac{\lambda}{4}$ | - |

## What's next

Stationary waves are a core application of the superposition principle, and underpin the behavior of all musical instruments, resonance, and acoustic systems. They are one of the most commonly tested topics in CIE A-Level Physics, appearing in both multiple choice and structured questions, so mastering boundary conditions and harmonic relations is critical for full marks. Next, we will extend the superposition principle to two dimensions to cover interference and diffraction, which build directly on the wave superposition concepts you learned here.

- [Interference](https://www.owlsprep.com/study/cie-9702-u8-interference/)
- [Diffraction](https://www.owlsprep.com/study/cie-9702-u8-diffraction/)
- [Young's double-slit experiment](https://www.owlsprep.com/study/cie-9702-u8-young-s-double-slit-experiment/)

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