Recall and apply the wave equation $v = f \lambda$ to determine the speed, frequency or wavelength of a wave, including across media in which the wave speed changes

What this dot point is asking

QCAA wants you to use the wave equation $v = f \lambda$ to relate the three fundamental wave quantities, including the case where a wave crosses from one medium to another and the speed (and therefore wavelength) changes while frequency stays fixed.

The wave equation

The speed at which a wave propagates equals the frequency times the wavelength:

Each variable can be made the subject:

Units: speed in m s$^{-1}$, frequency in hertz ($1$ Hz $= 1$ s$^{-1}$), wavelength in metres. Always check unit consistency before substituting.

Why $v = f \lambda$ works

In one period $T = 1/f$, a wave moves one wavelength $\lambda$. So speed equals $\lambda / T = f \lambda$. The same derivation works for transverse and longitudinal waves, mechanical waves and electromagnetic waves.

Speed in different media

Wave speed depends only on the medium, not on the source.

Crossing a boundary

When a wave passes from one medium to another, three things happen.

1. Frequency $f$ is unchanged (set by the source; cycles cannot pile up at the boundary).
2. Speed $v$ changes to the value in the new medium.
3. Wavelength $\lambda$ adjusts to satisfy $v = f \lambda$. Slower medium = shorter wavelength. Faster medium = longer wavelength.

This is the principle behind refraction (the path bends because the wavelength changes at the boundary).

Worked example

A radio station broadcasts at $96.5$ MHz. Find the wavelength of the radio waves in air.

Radio waves are electromagnetic, so $v = c = 3.00 \times 10^8$ m s$^{-1}$ in air to a good approximation.

$\lambda = c / f = (3.00 \times 10^8) / (96.5 \times 10^6) = 3.11$ m.

This is why FM antennas are roughly $1.5$ m long (a quarter wavelength at typical broadcast frequencies).

Common traps

Mixing up MHz and Hz. $96.5$ MHz $= 96.5 \times 10^6$ Hz. Plugging in $96.5$ on its own gives an answer that is off by a factor of one million.

Forgetting frequency invariance at a boundary. If a question asks for the wavelength in glass given the wavelength in air, you usually need to compute frequency from the air values first, then divide the glass speed by that frequency.

Using $c$ for sound. $c = 3.00 \times 10^8$ m s$^{-1}$ is the speed of light. Sound is far slower; use the medium-specific value given in the question.

Substituting before identifying units. Frequency in $1/$ms gives wavelength in mm, not m. Convert before substituting.

How this appears in IA1 and EA

IA1 data test. Often a slinky or oscilloscope stimulus, with a measured period or frequency and a wavelength read off a diagram. Compute speed.

EA Paper 1. Standard multiple choice: which quantity changes when a wave crosses media (answer: speed and wavelength, not frequency).

EA Paper 2. Used as a setup for refraction calculations and standing-wave problems in Unit 4 quantum context (where photon wavelength sets ionising potential).

In one sentence

The wave equation $v = f \lambda$ connects wave speed, frequency and wavelength for any wave, with frequency fixed by the source and speed fixed by the medium; when a wave crosses a boundary the frequency is unchanged and the wavelength scales with the new speed.