Finding the wavelength from nodes and antinodes

By the end of this lesson

By the end you can find a wavelength from the positions of nodes or antinodes, choose measurements that reduce uncertainty, and combine the result with $v = f\lambda$ .

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$L_2$	second length
$L_1$	first length

The pattern is a ruler

A stationary wave gives you a ruler printed on the medium. Its marks are the nodes, and they are always in the same places.

Neighbouring nodes are half a wavelength apart:

$\text{node spacing} = \frac{\lambda}{2}$

So one measurement is enough in principle:

$\lambda = 2 \times (\text{distance between neighbouring nodes})$

The same applies to antinodes: neighbouring antinodes are also $\dfrac{\lambda}{2}$ apart. A node and the nearest antinode are $\dfrac{\lambda}{4}$ apart.

Count intervals, not points

Measuring one small gap gives a large percentage uncertainty. It is much better to measure across many gaps and divide.

If you measure from the first node to the last node and there are $N$ intervals between them, then

$d = N\frac{\lambda}{2} \quad\Rightarrow\quad \lambda = \frac{2d}{N}$

The trap: counting the nodes gives one more than the number of intervals.

nodes marked	intervals	relation
2	1	$d = \dfrac{\lambda}{2}$
3	2	$d = \lambda$
5	4	$d = 2\lambda$

So five nodes spread over $0.48\ \text{m}$ give $\lambda = \dfrac{2 \times 0.48}{4} = 0.24\ \text{m}$ , not $\dfrac{0.48}{5}$ .

Two more routes to the wavelength

From two resonances in a tube. Successive resonances differ in length by half a wavelength:

$\lambda = 2(L_2 - L_1)$

This is more accurate than using $4L_1$ , because any end-correction error cancels in the difference (and in this course end corrections are negligible anyway).

From the wave equation. Once you know $\lambda$ and the frequency $f$ , the speed is

$v = f\lambda$

For a string, $f$ is the vibrator frequency; for sound, it is the tuning-fork frequency; for microwaves, it is the transmitter frequency.

Worked example: String: wavelength and speed

A string of length $1.50\ \text{m}$ is fixed at both ends and driven at $120\ \text{Hz}$ . It vibrates in a steady pattern with three loops.

Calculate (a) the wavelength and (b) the speed of the wave on the string.

Have a go first?

Worked example: Air column: frequency from a resonance length

A tube closed by a water surface resonates first when the air column is $0.20\ \text{m}$ long. The speed of sound is $340\ \text{m s}^{-1}$ .

Calculate (a) the wavelength and (b) the frequency of the sound.

Have a go first?

Common mistakes

Five nodes means five half-wavelengths.

Five nodes make only four intervals, so there are four half-wavelengths between the first and last node.

The distance between a node and an antinode is a half-wavelength.

Node to nearest antinode is a quarter wavelength. Half a wavelength takes you from a node to the next node.

You can read the wavelength straight off the length of the string.

Only if you also know how many half-wavelengths fit on it: $L = n\dfrac{\lambda}{2}$ .

In the exam

When a question gives node positions, find the spacing between neighbours and double it — or count intervals across the whole span.

Show the interval count explicitly, e.g. " $5$ nodes $\Rightarrow 4$ intervals $\Rightarrow 2\lambda = 0.48\ \text{m}$ ". Examiners award that.

If a length and a frequency are given, finish with $v = f\lambda$ and quote the unit $\text{m s}^{-1}$ .

Check yourself

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.