Physics 9702 · Paper 1

Nodes, Antinodes & Wavelength

In a stationary wave, some points stay completely still all the time. Halfway between them, points swing the widest.
The still points are nodes; the widest-swinging points are antinodes. Their spacing is fixed and even, and that spacing is exactly half a wavelength. So if you can see the pattern, you can read the wavelength from it.

Nodes and antinodes

A is a point that never moves. Here the two waves are always crest-on-trough, so they add to zero at every instant.
An is a point that swings the most. Here the two waves keep lining up, so they add to the biggest swing.
Nodes and antinodes alternate along the wave: node, antinode, node, antinode.

A standing wave at its two extremes (blue and green). Nodes (N) never move; antinodes (A) swing widest. Node to node is λ/2; node to antinode is λ/4.

The spacing rule

Two neighbouring nodes are half a wavelength apart. (So are two neighbouring antinodes.)
A node to its next antinode is a quarter of a wavelength.

\text{node to next node} = \frac{\lambda}{2}, \qquad \text{node to next antinode} = \frac{\lambda}{4}

Symbols

$\lambda$ = wavelength (m)

Measuring the wavelength

Measure the gap between two points on the pattern and you can find $\lambda$ directly. That is the use of the spacing rule.
Node spacing is $\lambda/2$ , so double the node spacing to get $\lambda$ . Then use $v = f\lambda$ to find the speed or the frequency.

Worked example

Finding λ from a row of antinodes

A stationary sound wave has 5 antinodes (5 loudest points) spread evenly over 2.0 m. Find the wavelength.

5 antinodes in a row have 4 gaps between them.
Each gap is $\lambda/2$ , so the 4 gaps total $4 \times \tfrac{\lambda}{2} = 2\lambda$ .
That total is 2.0 m, so $2\lambda = 2.0 \;\Rightarrow\; \lambda = 1.0$ m.

Answer

\lambda = 1.0 \text{ m}

Node to node (or antinode to antinode) is $\lambda/2$ . Measure that spacing, double it for $\lambda$ , then use $v = f\lambda$ . (9702/12/F/M/24 Q29)

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