The wave equation $v = f\lambda$ and applications (QCE Physics Unit 2)

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).

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).

Examples in context

Example 1. A Cairns light-rail track inspection ultrasonic gauge transmits at $5.0 \text{ MHz}$. In steel ($v = 5900 \text{ m s}^{-1}$), $\lambda = v/f = 1.18 \text{ mm}$, fine enough to resolve weld-line defects of comparable size. In a polymer pad ($v = 2400 \text{ m s}^{-1}$) the wavelength shrinks to $0.48 \text{ mm}$ but the frequency is unchanged - exactly the QCAA-style multi-medium application of $v = f\lambda$.

Example 2. ANSTO Mt Cotton satellite tracking uses $2.2 \text{ GHz}$ S-band telemetry. In vacuum $\lambda = c/f = 0.136 \text{ m}$, sets dish-aperture sizing for a wanted beam-width. The same signal passing through the moist tropical Cairns troposphere slows by under $0.03$ per cent, a deviation engineers correct in Doppler-tracking software. QCAA EA Unit 2 thematic stems often pair a domestic application with such a propagation-medium twist.

Try this

Q1. State the wave equation and define each symbol. [2 marks]

• Cue. $v = f\lambda$; $v$ wave speed, $f$ frequency, $\lambda$ wavelength.

Q2. A sound wave of frequency $440 \text{ Hz}$ moves from air ($v = 340 \text{ m s}^{-1}$) into water ($v = 1480 \text{ m s}^{-1}$). Calculate the wavelength in each medium. [3 marks]

• Cue. Air: $\lambda = 0.773 \text{ m}$; water: $\lambda = 3.36 \text{ m}$; frequency unchanged.

Q3. An ultrasonic gauge at $5.0 \text{ MHz}$ examines a Cairns light-rail track. (a) Calculate $\lambda$ in steel ($v = 5900 \text{ m s}^{-1}$). (b) Calculate $\lambda$ in a $2400 \text{ m s}^{-1}$ polymer pad coupling. (c) Explain how the unchanged frequency and changing wavelength relate to resolution. [2+2+3 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

• Cue. (a) $1.18 \text{ mm}$; (b) $0.48 \text{ mm}$; (c) frequency fixed at source, wavelength sets resolvable defect size, so polymer couples but reduces in-steel resolution.