SHM Equations & Graphs
Knowing acceleration is proportional to displacement is only half the story — the practical power of SHM comes from the worked-out equations for displacement, velocity and acceleration at any moment, and from two neat period formulae that predict exactly how fast a mass-spring system or pendulum will oscillate before you ever set it swinging.
The equations of motion for SHM
The defining equation $a=-\omega^2x$ is a starting condition, not a complete description — to know where an oscillating object actually is at a given time, you need the solved equations for $x$, $v$ and $a$ as functions of $t$. For an object released from rest at maximum displacement $x_0$ (the amplitude) at $t=0$:
Velocity and acceleration follow from displacement by successive differentiation with respect to time, though for A-level purposes it's the equations themselves that matter most.
If instead the object starts at $t=0$ passing through equilibrium (moving at maximum speed), sine and cosine swap:
Which pair to use depends on the starting condition in the question — cosine for "released from maximum displacement", sine for "starts at the centre moving at maximum speed". Both describe the same SHM, just with a different choice of $t=0$.
A displacement-independent relationship for velocity
It's often more useful to know speed at a given displacement rather than a given time. Combining the $x$ and $v$ equations (using $\sin^2+\cos^2=1$) eliminates $t$ entirely:
This applies regardless of starting condition, and confirms two intuitive facts: speed is greatest ($v=\pm\omega x_0$) at $x=0$, and zero at $x=\pm x_0$, where the object momentarily stops before reversing.
The $x$, $v$ and $a$ graphs share the same period $T$ but are offset: $v$ is a quarter-cycle (90°) ahead of $x$ — velocity is zero when displacement is maximum, and maximum when displacement is zero. $a$ is a half-cycle (180°) out of phase with $x$ — acceleration is always the exact opposite sign to displacement, the SHM condition $a=-\omega^2x$ restated graphically.
Period of a mass-spring system and a simple pendulum
Since $\omega = 2\pi/T$, and each SHM system has its own characteristic $\omega^2$ set by its physical properties, each has its own period formula.
Mass-spring system
A mass $m$ on a spring of spring constant $k$ has $\omega^2 = k/m$. Substituting into $\omega = 2\pi/T$ and rearranging:
A homogeneity check: $\sqrt{\dfrac{\text{kg}}{\text{N}\,\text{m}^{-1}}} = \sqrt{\dfrac{\text{kg}}{\text{kg}\,\text{s}^{-2}}} = \sqrt{\text{s}^2} = \text{s}$, a time, as expected.
Simple pendulum
For a simple pendulum of length $l$ swinging through small angles:
where $g$ is gravitational field strength. Same check: $\sqrt{\dfrac{\text{m}}{\text{m}\,\text{s}^{-2}}} = \sqrt{\text{s}^2} = \text{s}$.
Neither formula contains amplitude $x_0$. A mass-spring system or small-angle pendulum takes exactly the same time per oscillation whether swinging through a large amplitude or a tiny one — why pendulum clocks keep time even as their swing shrinks due to friction. It's a direct consequence of $a\propto x$: a bigger displacement comes with proportionally bigger restoring acceleration, so extra distance and extra acceleration scale together, leaving the cycle time unchanged.
Required Practical: Investigating Simple Harmonic Motion
This is AQA's Required Practical 7: investigating the period of an oscillating system.
Equipment
| Equipment | Purpose |
|---|---|
| Spring + mass set, or pendulum bob and string | The oscillating system under investigation |
| Retort stand and clamp | Suspends the spring or pendulum securely |
| Split cork / clamp | Fixes the pendulum's point of suspension precisely, for a consistent length measurement |
| Stopwatch | Times multiple oscillations |
| Metre ruler | Measures pendulum length $l$ (or spring extension, if needed) |
| Fiducial marker | Marks the equilibrium position, giving a fixed, high-speed point to time from |
| Protractor | Checks the pendulum's starting angle stays small (below about 10°) |
Method
Sources of error and how to minimise them
A fixed reaction-time uncertainty (perhaps ±0.2 s) becomes a much smaller percentage of the total time when timing 20 oscillations rather than 1 — always time as many oscillations as practical and divide, rather than timing a single swing.
A pendulum swinging through too large an angle deviates from true SHM, systematically distorting the measured period away from $T=2\pi\sqrt{l/g}$. Keep the starting displacement small (below about 10°) throughout, checked with a protractor.
Measuring $l$ to the top or bottom of the bob instead of consistently to its centre introduces a small but avoidable systematic error into every reading, and hence into the gradient (and $g$) extracted from the graph. Always measure to the same point — the centre of the bob.
Damping gradually shrinks the amplitude and can slightly affect timing over many oscillations. Use a dense, compact bob to minimise air resistance, and minimise friction at the pivot.
Key takeaways
- SHM solutions: $x=x_0\cos(\omega t)$, $v=-x_0\omega\sin(\omega t)$, $a=-x_0\omega^2\cos(\omega t)$ (released from rest at max displacement); sine and cosine swap if released from the centre at maximum speed instead.
- $v = \pm\omega\sqrt{x_0^2-x^2}$ links speed directly to displacement, without needing time — maximum speed at $x=0$, zero speed at $x=\pm x_0$.
- On x/v/a-against-time graphs: $v$ leads $x$ by a quarter cycle (90°); $a$ is exactly out of phase with $x$ (180°), consistent with $a=-\omega^2x$.
- Period formulae: mass-spring $T=2\pi\sqrt{m/k}$; simple pendulum (small angle) $T=2\pi\sqrt{l/g}$. Neither depends on amplitude.
- AQA Required Practical 7: time 10–20 oscillations and divide to reduce percentage timing error; plot $T^2$ against $l$ (or $m$) for a straight line through the origin, and use the gradient to find $g$ (or $k$).