Understanding Progressive Waves Revision Notes | A Level AQA Physics

1. What Is a Progressive Wave?

Drop a stone into still water and watch the ripples spread outward. Energy travels across the surface, yet the water itself does not travel with it — the water molecules bob up and down as the wave passes through. That is the defining feature of a progressive wave: it transfers energy through the oscillation of particles, without transferring the medium itself.

Every progressive wave has a source that disturbs the medium, causing neighbouring particles to oscillate in turn. The pattern of disturbance — the wave profile — moves outward from the source in the direction of energy transfer. This is what distinguishes a progressive wave from a stationary wave, which you will meet later in the course.

Progressive waves fall into two families depending on the relationship between the direction of particle oscillation and the direction of wave travel. In a transverse wave — such as light or a wave on a string — particles oscillate perpendicular to the direction of energy transfer. In a longitudinal wave — such as sound — particles oscillate parallel to the direction of energy transfer, creating alternating compressions and rarefactions.

2. The Language of Waves: Key Quantities

To describe any progressive wave precisely, you need six quantities. Examiners test these definitions directly, so the exact wording matters. Work through each one carefully — they will reappear in every wave calculation you do at A Level.

Amplitude and wavelength are both read from a displacement–distance graph of the wave. Amplitude is the height from the equilibrium line to a crest (or trough); wavelength is the horizontal distance from one crest to the next — or more precisely, between any two adjacent points that are in phase.

Frequency and period are linked by a simple but important relationship. If a wave completes 50 full oscillations every second, its frequency is 50 Hz and each oscillation takes $\frac{1}{50}$ of a second. That reciprocal relationship is captured in the equation $f = \frac{1}{T}$.

3. The Wave Equation: Linking Speed, Frequency, and Wavelength

Here is where the quantities connect. Imagine watching a wave on a rope: in one second, $f$ complete wave cycles pass any fixed point. Each cycle occupies a length of $\lambda$ metres along the rope. So in one second, the wave profile advances a total distance of $f \times \lambda$ metres — and that distance per second is the wave speed.

A critical point: for a wave travelling through a fixed medium, the wave speed $c$ is determined by the properties of that medium — not by the source. If you increase the frequency of a wave in the same medium, the wavelength decreases proportionally so that $c$ stays constant. Frequency and wavelength are inversely proportional at constant wave speed.

Worked example: Calculating Wave Speed from Frequency and Wavelength

A water wave has a frequency of 3.0 Hz and a wavelength of 0.40 m. Calculate the wave speed. [3 marks]

1. $c = f\lambda$ — [M1] Selecting the correct equation earns the method mark.
2. $f = 3.0 \text{ Hz}, \quad \lambda = 0.40 \text{ m}$ — Always list your known quantities before substituting — this prevents unit errors.
3. $c = 3.0 \times 0.40$ — [M1] Correct substitution with consistent units.
4. $c = 1.2 \text{ m s}^{-1}$ — [A1] Correct answer with units. Note: $3.0 \times 0.40 = 1.2$ exactly.

Answer: $c = 1.2 \text{ m s}^{-1}$

Worked example: Finding Wavelength from Wave Speed and Period

A radio wave travels at $3.0 \times 10^8 \text{ m s}^{-1}$ and has a period of $2.0 \times 10^{-7}$ s. Calculate its wavelength. [4 marks]

1. $f = \frac{1}{T} = \frac{1}{2.0 \times 10^{-7}}$ — [M1] Must convert period to frequency before using the wave equation.
2. $f = 5.0 \times 10^{6} \text{ Hz}$ — Dividing 1 by $2.0 \times 10^{-7}$: move the power, $\frac{1}{2.0} = 0.5$, and $10^{0} \div 10^{-7} = 10^{7}$, giving $0.5 \times 10^{7} = 5.0 \times 10^{6}$ Hz. [A1]
3. $\lambda = \frac{c}{f}$ — [M1] Rearranging $c = f\lambda$ correctly.
4. $\lambda = \frac{3.0 \times 10^{8}}{5.0 \times 10^{6}} = \frac{3.0}{5.0} \times 10^{8-6} = 0.60 \times 10^{2} = 60 \text{ m}$ — [A1] Correct answer with units. Dividing the coefficients: $3.0 \div 5.0 = 0.60$. Subtracting exponents: $10^{8} \div 10^{6} = 10^{2}$.

Answer: $\lambda = 60 \text{ m}$

4. Phase and Phase Difference

Two particles on the same progressive wave are rarely doing exactly the same thing at the same moment. The concept of phase describes where a particle is in its oscillation cycle at a given instant. Phase difference then tells you how far apart two particles are in their cycles — and this is where the maths gets precise.

Picture a displacement–distance snapshot of a wave. Two particles separated by exactly one full wavelength are perfectly in step: one crest, one crest; one trough, one trough. They are said to be in phase. Two particles separated by half a wavelength are perfectly out of step — when one is at a crest, the other is at a trough. They are in antiphase.

The AQA specification requires you to be fluent in all three ways of expressing phase difference. The conversion is straightforward: one complete cycle corresponds to 360°, which corresponds to $2\pi$ radians. So a phase difference of half a cycle is 180° or $\pi$ rad; a quarter cycle is 90° or $\frac{\pi}{2}$ rad.

To calculate phase difference from a wave diagram, measure the path difference — the distance between the two points along the direction of wave travel — and express it as a fraction of the wavelength. Then multiply by $2\pi$ to get the phase difference in radians, or by 360° to get it in degrees.

Worked example: Calculating Phase Difference Between Two Points on a Wave

A progressive wave has a wavelength of 0.80 m. Two points P and Q on the wave are separated by a distance of 0.20 m along the direction of travel. Calculate the phase difference between P and Q in (a) radians and (b) degrees.

1. $\frac{\Delta x}{\lambda} = \frac{0.20}{0.80} = 0.25$ — The two points are one quarter of a wavelength apart. This fraction is the key intermediate result.
2. $\phi = 0.25 \times 2\pi = 0.50\pi = \frac{\pi}{2} \approx 1.57 \text{ rad}$ — $0.25 \times 2\pi$: multiply $0.25 \times 2 = 0.50$, so $\phi = 0.50\pi$ rad. This is $\frac{\pi}{2}$ rad exactly.
3. $\phi = 0.25 \times 360° = 90°$ — $0.25 \times 360 = 90$. Consistent with $\frac{\pi}{2}$ rad — a quarter cycle is always 90°.
4. $\phi = \frac{\pi}{2} \text{ rad} = 90°$ — Both forms are acceptable in AQA answers. Stating both shows full understanding.

Answer: Phase difference = $\frac{\pi}{2}$ rad (90°). The points are a quarter of a cycle apart.

5. Reading Wave Diagrams

AQA examiners frequently present wave data as graphs rather than numbers. You need to extract amplitude, wavelength, and phase difference directly from a displacement–distance or displacement–time graph. The approach is the same each time: identify the equilibrium line, read off the relevant distances, and apply the definitions precisely.

On a displacement–distance graph, the amplitude is the maximum displacement from the equilibrium line — read it as the vertical distance from the centre to a crest. The wavelength is the horizontal distance between two successive crests (or any two adjacent in-phase points). On a displacement–time graph, the same vertical reading gives amplitude, but the horizontal axis now gives the period $T$ rather than wavelength.

To find phase difference from a graph, measure the horizontal shift between two wave profiles (or two marked points) and divide by the wavelength. Multiply by $2\pi$ for radians or 360° for degrees. This is the same formula as before — the graph just gives you $\Delta x$ visually.

6. Exam Technique and Common Mistakes

Wave questions are reliable mark-earners if you are disciplined about units and definitions. The errors that cost students marks are almost always the same ones — so knowing them in advance puts you ahead.

For extended-response questions asking you to 'describe' a progressive wave, structure your answer around the particle oscillation: state that particles oscillate about their equilibrium positions, that the wave profile moves in the direction of energy transfer, and that energy is transferred without net movement of matter. Each of those three points is a potential mark.

Bringing it all together: a progressive wave is characterised by oscillating particles, a travelling wave profile, and a set of measurable quantities — amplitude, frequency, wavelength, speed, period, and phase difference — all linked by $c = f\lambda$ and $f = \frac{1}{T}$. Mastery of these definitions and equations, combined with the ability to read wave diagrams and express phase difference in any required form, covers the full AQA specification requirement for progressive waves.

To reach A-level standard, do more than define a progressive wave: analyse how the fixed phase relationship between neighbouring particles explains energy transfer without bulk transport of matter, and justify why a snapshot of the wave profile alone is not enough unless you also consider particle motion over time.