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The wave equation

Covers AS91170

The relationship between speed, frequency and wavelength

In one period $T$ , a wave advances exactly one wavelength $\lambda$ . Speed is distance over time, so:

$v = f\lambda$

$v$ — wave speed (m s⁻¹)
$f$ — frequency (Hz)
$\lambda$ — wavelength (m)
Two supporting relationships appear on the same resource sheet:

$f = \frac{1}{T} \qquad v = \frac{d}{t}$

Use $v = d/t$ when a question gives a distance travelled and a time (an echo, a pulse down a rope, light crossing a gap).
Use $v = f\lambda$ when it gives any two of speed, frequency and wavelength.

What stays the same and what changes

This is the idea the rest of the standard is built on:

Quantity	Set by	Changes when the wave enters a new medium?
Frequency	the source	Never
Speed	the medium	Yes
Wavelength	both, via $v = f\lambda$	Yes

The frequency cannot change at a boundary. Waves arrive at the boundary at a certain rate, and they must leave at the same rate — otherwise waves would pile up or vanish there.
The speed is a property of the medium. Light slows in glass; water waves slow in shallow water.
Because $f$ is fixed and $v$ changes, the wavelength must change in proportion:

$\frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}$

Slower wave → shorter wavelength. Faster wave → longer wavelength.

Rearranging and working in the right units

$f = \dfrac{v}{\lambda}$ and $\lambda = \dfrac{v}{f}$ .
Convert everything to metres, seconds and hertz before substituting:
- $1$ cm $= 10^{-2}$ m, $1$ mm $= 10^{-3}$ m, $1$ nm $= 10^{-9}$ m,
- $1$ kHz $= 10^{3}$ Hz, $1$ MHz $= 10^{6}$ Hz.
The speed of light in a vacuum, $c = 3.0 \times 10^{8}$ m s⁻¹, is on the resource sheet.

Worked ExampleSpeed of a water wave

Water waves of wavelength $0.45$ m pass a post at a rate of $12$ waves every $4.0$ s. Find the frequency and the speed of the waves.

Step 1 — Frequency from the count and the time

$f = \frac{\text{number of waves}}{\text{time}} = \frac{12}{4.0} = 3.0 \text{ Hz}$

Step 2 — Speed from the wave equation

$v = f\lambda = (3.0)(0.45) = 1.35 \approx 1.4 \text{ m s}^{-1}$

$f = 3.0 \text{ Hz}, \qquad v = 1.4 \text{ m s}^{-1}$

Worked ExampleA wave crossing a boundary

Water waves travelling at $0.36$ m s⁻¹ in deep water have a wavelength of $0.12$ m. They cross into shallow water, where they slow to $0.24$ m s⁻¹. Find (a) the frequency in the deep water, (b) the frequency in the shallow water, and (c) the new wavelength.

(a) Frequency in the deep water

$f = \frac{v}{\lambda} = \frac{0.36}{0.12} = 3.0 \text{ Hz}$

(b) Frequency in the shallow water

The frequency is set by the source and does not change at a boundary:

$f = 3.0 \text{ Hz}$

(c) New wavelength

Use the wave equation again, with the new speed and the unchanged frequency:

$\lambda = \frac{v}{f} = \frac{0.24}{3.0} = 0.080 \text{ m}$

$f = 3.0 \text{ Hz in both}, \qquad \lambda_{\text{shallow}} = 0.080 \text{ m}$

Worked ExampleLight slowing in glass

Light of wavelength $600$ nm in air travels at $3.0 \times 10^{8}$ m s⁻¹. In a glass block it slows to $2.0 \times 10^{8}$ m s⁻¹. Find the frequency of the light and its wavelength inside the glass.

Step 1 — Convert the wavelength

$\lambda = 600 \text{ nm} = 600 \times 10^{-9} = 6.0 \times 10^{-7} \text{ m}$

Step 2 — Frequency, from the values in air

$f = \frac{v}{\lambda} = \frac{3.0 \times 10^{8}}{6.0 \times 10^{-7}} = 5.0 \times 10^{14} \text{ Hz}$

Step 3 — Wavelength in the glass

The frequency is unchanged at $5.0 \times 10^{14}$ Hz, so:

$\lambda = \frac{v}{f} = \frac{2.0 \times 10^{8}}{5.0 \times 10^{14}} = 4.0 \times 10^{-7} \text{ m} = 400 \text{ nm}$

$f = 5.0 \times 10^{14} \text{ Hz}, \qquad \lambda_{\text{glass}} = 400 \text{ nm}$