Graphical Representations of SHM Revision Notes | A Level AQA Physics

1. Why Three Graphs? The Big Picture

A mass on a spring bobs up and down. You already know it undergoes SHM — but how do you read its story from a graph? The answer is that displacement $x$, velocity $v$, and acceleration $a$ each tell a different chapter of the same story, and the three graphs are mathematically locked together.

Here is the key insight that drives the whole lesson: velocity is the rate of change of displacement, so the $v$–$t$ graph is the gradient function of the $x$–$t$ graph. Equally, acceleration is the rate of change of velocity, so the $a$–$t$ graph is the gradient function of the $v$–$t$ graph. Once you see this chain, the shapes of all three graphs follow automatically.

2. The Displacement-Time Graph

Start with the oscillator released from maximum positive displacement $+A$ at $t = 0$. The displacement varies as a cosine: $x = A\cos(\omega t)$. Picture the graph: it begins at its peak, sweeps down through zero, reaches $-A$, returns through zero, and completes one full cycle back at $+A$ after one period $T$.

Notice the key features: the graph crosses zero at $t = \frac{T}{4}$ and $t = \frac{3T}{4}$, and it reaches its negative peak at $t = \frac{T}{2}$. These crossing points are where the gradient — and therefore the velocity — is at its maximum magnitude.

3. Deriving the Velocity-Time Graph from the Gradient

To build the $v$–$t$ graph, ask yourself: at every moment, how steep is the $x$–$t$ graph? At $t = 0$ the cosine curve is at its peak — it is momentarily flat, so the gradient is zero and $v = 0$. As the curve sweeps downward towards $t = \frac{T}{4}$, the gradient becomes increasingly negative, reaching its most negative value right at the zero-crossing. After that, the gradient becomes less steep again as the curve approaches $-A$.

Tracking this gradient point by point produces a negative sine curve: $v = -A\omega\sin(\omega t)$. The amplitude of the velocity graph is $A\omega$, not $A$ — the faster the oscillation (larger $\omega$), the greater the peak speed. This is a crucial distinction that examiners test directly.

Compare the two graphs side by side. The $v$–$t$ graph is shifted a quarter period to the right relative to the $x$–$t$ graph — velocity lags displacement by a quarter period ($T/4$), or equivalently, displacement leads velocity by $T/4$. Wherever $x$ is at a maximum or minimum, $v$ is zero; wherever $x$ is zero, $v$ is at its maximum magnitude. These two facts are the graphical signature of SHM.

4. Deriving the Acceleration-Time Graph from the Gradient

Apply exactly the same logic one more time: the gradient of the $v$–$t$ graph gives the acceleration. At $t = 0$, the $v$–$t$ graph (a negative sine) has its steepest negative gradient — so acceleration is at its most negative value. At $t = \frac{T}{4}$, the $v$–$t$ graph is at its trough ($-A\omega$), so its gradient is zero and $a = 0$.

Differentiating $v = -A\omega\sin(\omega t)$ gives $a = -A\omega^2\cos(\omega t)$. The acceleration graph is therefore a cosine with amplitude $A\omega^2$ — but it is inverted relative to the $x$–$t$ graph. This inversion is the graphical expression of the defining condition of SHM: $a = -\omega^2 x$. When displacement is positive, acceleration is negative (directed back towards equilibrium), and vice versa.

The $a$–$t$ graph is in antiphase with the $x$–$t$ graph — they are mirror images of each other about the time axis. This is the most important phase relationship to state in an exam answer. The $a$–$t$ graph is also a quarter period out of phase with the $v$–$t$ graph, for the same gradient-chain reason.

5. The Three Graphs Together — Phase Relationships

Placing all three graphs on the same time axis reveals the full phase structure of SHM. Starting from $x = +A$ at $t = 0$: displacement follows $\cos(\omega t)$, velocity follows $-\sin(\omega t)$ (a quarter period behind the cosine peak), and acceleration follows $-\cos(\omega t)$ (half a period behind the displacement peak, i.e. in antiphase).

Reading across each row of this table is a powerful exam technique. If a question gives you the displacement at a particular instant, you can immediately state the sign and magnitude of the acceleration using $a = -\omega^2 x$, and you can reason about whether velocity is increasing or decreasing from the gradient of the $x$–$t$ graph at that point.

6. Worked Example — Reading Graphs and Calculating Values

Exam questions often give you one graph and ask you to deduce features of the others, or to calculate $\omega$, $v_{max}$, or $a_{max}$ from the graph. The step-by-step below models the full method.

Worked example: Extracting $\omega$, $v_{max}$, and $a_{max}$ from a Displacement-Time Graph

A displacement-time graph for a mass oscillating on a spring shows a cosine curve with amplitude $A = 0.040$ m and period $T = 0.50$ s. Calculate: (a) the angular frequency $\omega$, (b) the maximum speed $v_{max}$, and (c) the maximum acceleration $a_{max}$.

1. $A = 0.040$ m, $T = 0.50$ s — These are read directly from the $x$–$t$ graph: $A$ is the peak value, $T$ is the time for one complete cycle. [B1]
2. $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.50}$ — Angular frequency links the period to the sinusoidal equations. [M1]
3. $\omega = \frac{2 \times 3.1416}{0.50} = \frac{6.2832}{0.50} = 12.6$ rad s$^{-1}$ (3 s.f.) — Keep at least 3 significant figures for intermediate values to avoid rounding errors in later steps. [A1]
4. $v_{max} = A\omega = 0.040 \times 12.6 = 0.504$ m s$^{-1}$ — The amplitude of the $v$–$t$ graph is $A\omega$. This is the speed at the equilibrium position. [M1, A1]
5. $a_{max} = A\omega^2 = 0.040 \times (12.6)^2$ $12.6 \times 12.6 = 158.76$ $a_{max} = 0.040 \times 158.76 = 6.35$ m s$^{-2}$ (3 s.f.) — The amplitude of the $a$–$t$ graph is $A\omega^2$. Note: $\omega^2$ is computed as $12.6 \times 12.6 = 158.76$, not estimated. [M1, A1]
6. At $x = +A = +0.040$ m: $a = -158.76 \times 0.040 = -6.35$ m s$^{-2}$ — The magnitude matches $a_{max}$ and the negative sign confirms the acceleration opposes the displacement — consistent with SHM. This cross-check is worth doing in an exam.

Answer: (a) $\omega = 12.6$ rad s$^{-1}$, (b) $v_{max} = 0.504$ m s$^{-1}$, (c) $a_{max} = 6.35$ m s$^{-2}$

7. The Gradient Chain — A Deeper Look

The relationship between the three graphs is not just a pattern to memorise — it is a direct consequence of calculus. Differentiating $x = A\cos(\omega t)$ with respect to time gives $v = -A\omega\sin(\omega t)$, and differentiating again gives $a = -A\omega^2\cos(\omega t)$. You are not expected to perform this differentiation in the AQA exam, but understanding the logic is what allows you to reconstruct any graph from any other.

A practical consequence: if you are given a $v$–$t$ graph and asked to sketch the $x$–$t$ graph, you reverse the process. Where the $v$–$t$ graph is positive, the $x$–$t$ graph has a positive gradient (displacement is increasing). Where $v = 0$, the $x$–$t$ graph has a turning point. This is the graphical equivalent of integration, and it is a common exam task.

8. Exam Technique and Common Mistakes

AQA questions on SHM graphs tend to fall into three categories: sketch a named graph, describe the phase relationship between two quantities, or calculate a maximum value from graph data. Each demands a slightly different approach, and the marks are often lost through imprecise language rather than wrong physics.

For a 'describe and explain' question worth 3–4 marks, a full-mark response typically includes: the name of the graph shape (sinusoidal), the phase relationship stated precisely (e.g. 'velocity leads displacement by a quarter period'), and the physical reason linked to the gradient chain ('because velocity is the rate of change of displacement'). Hitting all three points is what separates a 2-mark answer from a 4-mark one.

The three graphs of SHM — displacement, velocity, and acceleration against time — are connected by the gradient chain: each graph is the gradient function of the one before it, and this single principle generates all the phase relationships and amplitude scalings that AQA tests.