Simple Harmonic Motion — Defining Equation

Spec mapping: OCR H556 Module 5.3 — Oscillations (displacement, amplitude, period, frequency, angular frequency; the defining equation $a = -\omega^2 x$a=−ω2x; conditions for simple harmonic motion; period independence from amplitude; mass-spring and simple pendulum as canonical SHM systems). Refer to the official OCR H556 specification document for exact wording.

Simple harmonic motion (SHM) is one of the most important ideas in physics. It describes the way anything wobbles: a mass on a spring, the pendulum of a clock, a guitar string, the atoms in a crystal, even molecules of carbon dioxide absorbing infrared radiation in the atmosphere (the absorption mechanism behind the greenhouse effect). Whenever a system is slightly disturbed from a stable equilibrium and then released, its motion — at least for small displacements — is almost always simple harmonic. This is a remarkably universal claim, and the reason for it is a piece of mathematics so general that we shall return to it in the "Going further" section: any smooth potential-energy well looks parabolic at its bottom.

This lesson introduces the defining equation of SHM, explains the minus sign that distinguishes a stable oscillator from an unstable runaway, states the two conditions that OCR mark schemes demand, and sets up the mathematical machinery we will use in the four SHM lessons that follow. Get the minus sign and the two conditions correct, and the rest of the module unfolds cleanly. Forget the minus sign — by far the single most common A-Level SHM mistake — and you have an equation that describes an exploding exponential, not an oscillation.

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What is an Oscillation?

An oscillation is any repeating to-and-fro motion about an equilibrium position. The object returns through the same sequence of positions at regular intervals. Familiar examples:

• A child on a swing (a pendulum, near-SHM at small amplitudes).
• A ruler twanged and held flat against the edge of a desk (a cantilever, SHM in the small-deflection limit).
• The piston of a car engine (sinusoidal to a good approximation, driven by the crankshaft).
• The beating of a heart (not SHM — a relaxation oscillator with non-linear restoring "force" — but periodic).
• A guitar string after a pluck (a superposition of many SHM normal modes).
• The vibration of an atom in a lattice (SHM in the harmonic approximation, source of solid-state heat capacity).

Each full repetition is a cycle. The time for one complete cycle is the period $T$T (seconds), and the number of cycles per second is the frequency $f = 1/T$f=1/T (hertz, Hz). Both quantities have exactly the same meaning as in circular motion — indeed there is a deep mathematical link between circular motion and SHM which we will exploit in the next lesson: the projection of uniform circular motion onto a diameter is exactly SHM with the same $\omega$ω.

The amplitude $A$A is the maximum displacement from equilibrium. Phase $\varphi$φ (the next lesson) records where in the cycle the oscillator is at $t = 0$t=0. Together, these are the four canonical "SHM coordinates" — $A$A, $\omega$ω, $T$T, $\varphi$φ — and almost every quantitative SHM problem boils down to extracting one of them from given data.

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The Restoring Force

What makes something oscillate rather than just sit still? It needs a restoring force — a force that always pushes or pulls the object back towards its equilibrium position.

Consider a mass on a spring. If you pull the mass to the right, the spring stretches and pulls it back to the left. If you push the mass to the left, the spring compresses and pushes it back to the right. Either way, the force is always directed toward the equilibrium point. The restoring force is the mechanical reason the body cannot just keep moving in one direction — it is constantly being decelerated and reversed by the spring.

flowchart TD
  S[System disturbed from equilibrium] --> R{Is there a restoring force?}
  R -- "No (no force; or force away from equilibrium)" --> N[No oscillation: drift or runaway]
  R -- "Yes, but F not proportional to x" --> A[Anharmonic oscillation — period depends on amplitude]
  R -- "Yes, F = −kx (linear in displacement)" --> H[SHM: a = −ω²x with ω² = k/m, period independent of amplitude]

If the restoring force is proportional to the displacement (and only then), something magical happens: the motion turns out to be simple harmonic, with sinusoidal solutions and a fixed period that does not depend on the amplitude. This is the defining condition of SHM. If the restoring force has any other form — proportional to $x^3$x3, say, or to $\sin x$sinx instead of $x$x, or some patchwork combination — the period will depend on amplitude and the motion will be anharmonic.

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The Defining Equation of SHM

An object undergoes simple harmonic motion if, and only if, its acceleration satisfies

$a = -\omega^2 x$a=−ω2x

where $x$x is the displacement from equilibrium and $\omega$ω is a positive constant called the angular frequency (units: rad s$^{-1}$−1). Read the equation in English:

"The acceleration is proportional to the displacement and directed towards the equilibrium position."

The minus sign is essential — it encodes the restoring nature of the force. Without it you would have an unstable system that accelerates away from equilibrium, like a marble balanced on top of an inverted bowl. With the minus sign, you have an oscillator; without, a runaway exponential. This is the single most critical algebraic feature of the entire SHM topic.

Conditions for SHM (memorise verbatim)

For a system to execute SHM, two conditions must be satisfied:

1. The acceleration must be proportional to the displacement from a fixed equilibrium point: $|a| \propto |x|$∣a∣∝∣x∣.
2. The acceleration must be directed towards the equilibrium point (opposite in sign to $x$x).

Both conditions together are encoded in the single equation $a = -\omega^2 x$a=−ω2x. The proportionality constant is $\omega^2$ω2, with $\omega > 0$ω>0.

Exam Tip: OCR very often asks "State the two conditions for a body to execute simple harmonic motion." Memorise the two bullet points above; mark schemes typically allow some paraphrase but always demand both the proportionality and the directional condition. A reliable full-mark answer: "(i) The acceleration is proportional to the displacement from a fixed equilibrium position; (ii) the acceleration is always directed towards this equilibrium position."

Why "Angular Frequency"?

The constant $\omega$ω is called the angular frequency of the oscillation, by analogy with the angular velocity of uniform circular motion. Both have units of rad s$^{-1}$−1, both relate to period via $T = 2\pi/\omega$T=2π/ω, both relate to frequency via $\omega = 2\pi f$ω=2πf. The deeper reason — uncovered in lesson 4 — is that the projection of a point in uniform circular motion onto a diameter executes SHM with exactly the same $\omega$ω. Circular motion and SHM are mathematically the same object, viewed from two angles.

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Why It Is Called "Simple"

There are plenty of periodic motions that are not simple harmonic. A ball bouncing elastically between two walls has a restoring "force" only at the walls — zero everywhere else; its $a$a-$x$x graph is two spikes, not a straight line, and its motion is triangular, not sinusoidal. A pendulum swung through a huge amplitude has a restoring torque proportional to $\sin\theta$sinθ, not to $\theta$θ, so its period lengthens with amplitude — only the small-angle approximation $\sin\theta \approx \theta$sinθ≈θ (valid up to about $\theta = 10°$θ=10° with $<1\%$<1% error) reduces it to SHM.

"Simple" in SHM refers to the linear relationship $a \propto x$a∝x. Any more complicated relationship is called anharmonic, and its period generally depends on amplitude. The "harmonic" in "harmonic motion" comes from the deep connection between SHM, sinusoidal functions and the harmonics of musical notes (a topic explored further in undergraduate Fourier analysis).

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Connecting $\omega$ω to Period and Frequency

The constant $\omega$ω in the defining equation plays exactly the same role as the angular velocity in circular motion. Once we solve the differential equation $a = -\omega^2 x$a=−ω2x (next lesson) we will find that one complete oscillation takes a time

$T = \frac{2\pi}{\omega} \qquad \text{and} \qquad \omega = 2\pi f$T=ω2π​andω=2πf

These are identical to the circular-motion relationships. You should commit them to memory — they appear in virtually every SHM calculation, alongside the defining equation itself.

Exam Tip: OCR does not require you to memorise the period formulae for specific systems (e.g. $T = 2\pi\sqrt{m/k}$T=2πm/k​ for a mass-spring or $T = 2\pi\sqrt{L/g}$T=2πL/g​ for a small-angle pendulum). If such a formula is needed, the examiner will either provide it or expect you to derive it within the question. You must, however, know $a = -\omega^2 x$a=−ω2x and $\omega = 2\pi/T = 2\pi f$ω=2π/T=2πf by heart. The mass-spring derivation (worked example 3 below) is so frequently asked that it should be muscle memory.

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Worked Example 1 — Recognising SHM from a Graph

A student measures the acceleration of a trolley attached to a pair of springs at various displacements from its equilibrium position:

$x$x / m	$a$a / m s$^{-2}$−2
$+0.20$+0.20	$-8.0$−8.0
$+0.10$+0.10	$-4.0$−4.0
$0.00$0.00	$0.0$0.0
$-0.10$−0.10	$+4.0$+4.0
$-0.20$−0.20	$+8.0$+8.0

Does the trolley execute SHM? If so, find $\omega$ω and the period $T$T.

Analysis. The data show $a = -40x$a=−40x (in SI units). Compare with $a = -\omega^2 x$a=−ω2x:

$\begin{aligned} \omega^2 &= 40 \text{ s}^{-2} \\ \omega &= \sqrt{40} = 6.32 \text{ rad s}^{-1} \\ T &= \frac{2\pi}{\omega} = \frac{2\pi}{6.32} = 0.993 \text{ s} \\ f &= \frac{1}{T} = 1.01 \text{ Hz} \end{aligned}$ω2ωTf​=40 s−2=40​=6.32 rad s−1=ω2π​=6.322π​=0.993 s=T1​=1.01 Hz​

The acceleration is strictly proportional to $x$x and opposite in sign, so both conditions for SHM are satisfied. The trolley executes SHM with angular frequency $6.32$6.32 rad s$^{-1}$−1 and period almost exactly $1$1 s.

Graph check. Plot $a$a on the vertical axis against $x$x on the horizontal. The data lie on a perfect straight line through the origin, with negative gradient $-40$−40 s$^{-2}$−2. The line passes through the second and fourth quadrants (positive $x$x, negative $a$a and vice versa) — the hallmark visual signature of SHM.

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Worked Example 2 — Maximum Acceleration of a Loudspeaker

A loudspeaker cone oscillates at $500$500 Hz with amplitude $0.50$0.50 mm. Assuming the motion is simple harmonic, find the maximum acceleration of the cone.

$\omega = 2\pi f = 2\pi \times 500 = 3\,141 \text{ rad s}^{-1}$ω=2πf=2π×500=3141 rad s−1

The maximum displacement is the amplitude, $x_\text{max} = A = 5.0 \times 10^{-4}$xmax​=A=5.0×10−4 m. By the defining equation $a = -\omega^2 x$a=−ω2x, the maximum magnitude of acceleration occurs when $|x| = A$∣x∣=A:

$|a_\text{max}| = \omega^2 A = (3\,141)^2 \times 5.0 \times 10^{-4} = 9.87 \times 10^6 \times 5.0 \times 10^{-4} = 4.93 \times 10^3 \text{ m s}^{-2}$∣amax​∣=ω2A=(3141)2×5.0×10−4=9.87×106×5.0×10−4=4.93×103 m s−2

That is about $500\,g$500g — remarkable for a speaker cone, but it is only a peak value reached twice per cycle (at the two extremes of the motion). The motion is gentle (sub-millimetre amplitude); it is the frequency that drives the acceleration up, via the $\omega^2$ω2 factor. This is a general feature of SHM at high frequency: even tiny amplitudes produce huge accelerations.

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Worked Example 3 — Mass on a Spring (Canonical Derivation)

A $0.50$0.50 kg mass is attached to a spring of stiffness $k = 200$k=200 N m$^{-1}$−1 on a smooth horizontal surface. The mass is pulled $0.10$0.10 m from equilibrium and released. Show that the motion is simple harmonic and find the angular frequency and period.

Step 1 — Restoring force. By Hooke's law, the spring exerts a force on the mass given by

$F_\text{spring} = -kx$Fspring​=−kx

where $x$x is the extension of the spring (= displacement of the mass from equilibrium) and the minus sign indicates the force is opposite to the displacement (restoring). This is the Hooke-1660 result.

Step 2 — Newton's second law. With the mass on a smooth surface, the only horizontal force is the spring force. Newton 2 gives

$\begin{aligned} ma &= -kx \\ a &= -\frac{k}{m}\,x \end{aligned}$maa​=−kx=−mk​x​

Step 3 — Compare with the SHM defining equation. $a = -\omega^2 x$a=−ω2x matches our derived equation if and only if

$\omega^2 = \frac{k}{m}$ω2=mk​

Both required conditions for SHM are satisfied: (i) acceleration is proportional to displacement ($a \propto x$a∝x); (ii) acceleration is opposite in sign to displacement ($a$a and $x$x have opposite signs). The motion is therefore simple harmonic.

Step 4 — Compute $\omega$ω and $T$T.

$\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20 \text{ rad s}^{-1}$ω=k/m​=200/0.50​=400​=20 rad s−1

$T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314 \text{ s}$T=ω2π​=202π​=0.314 s

Notice that the amplitude $0.10$0.10 m does not appear anywhere in the final answer — the period is independent of amplitude, which is the fundamental and beautiful feature of SHM. Pull the mass $0.05$0.05 m, $0.10$0.10 m, or $0.20$0.20 m from equilibrium and it will return through equilibrium with the same period $0.314$0.314 s every time (as long as Hooke's law continues to hold — i.e. you don't stretch the spring beyond its elastic limit).

Note: Even though this derivation produces $T = 2\pi\sqrt{m/k}$T=2πm/k​, OCR does not require you to memorise it as a standalone formula. You derive it on the spot, as we did, from $a = -\omega^2 x$a=−ω2x and Hooke's law whenever you need it. The marks are typically awarded for the derivation, not just the numerical answer.

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Worked Example 4 — The Simple Pendulum (Small-Angle SHM)

A simple pendulum consists of a bob of mass $m$m on a light inextensible string of length $L$L. The bob is displaced by a small angle $\theta$θ from the vertical. Show that for small $\theta$θ the motion is approximately simple harmonic, and find $\omega^2$ω2.

Step 1 — Restoring force. When the bob is at angle $\theta$θ to the vertical, the component of gravity tangential to the arc (perpendicular to the string) is the restoring force. Its magnitude is

$F_\text{restoring} = mg\sin\theta$Frestoring​=mgsinθ

directed back toward equilibrium ($\theta = 0$θ=0).

Step 2 — Newton's second law along the arc. The bob's arc-length displacement is $s = L\theta$s=Lθ, and its acceleration along the arc is $a = L\ddot\theta$a=Lθ¨. Newton 2 gives

$mL\ddot\theta = -mg\sin\theta \quad \Rightarrow \quad \ddot\theta = -\frac{g}{L}\sin\theta$mLθ¨=−mgsinθ⇒θ¨=−Lg​sinθ

Step 3 — Small-angle approximation. For small $\theta$θ (in radians), $\sin\theta \approx \theta$sinθ≈θ to within $<1\%$<1% accuracy for $\theta < 14°$θ<14°. Then

$\ddot\theta \approx -\frac{g}{L}\theta$θ¨≈−Lg​θ

Comparing with the SHM defining equation written in terms of $\theta$θ as the angular displacement, we identify

$\omega^2 = \frac{g}{L} \quad \Rightarrow \quad T = 2\pi\sqrt{L/g}$ω2=Lg​⇒T=2πL/g​

For a $1.00$1.00 m pendulum on Earth, $T = 2\pi\sqrt{1.00/9.81} = 2.01$T=2π1.00/9.81​=2.01 s — slightly longer than the often-quoted "$2$2-second pendulum" only because $g$g at Britain's latitude is a touch under $9.81$9.81 m s$^{-2}$−2. Notice mass cancels — the pendulum period depends only on length and $g$g. This is why Galileo's pendulum experiments worked: he could verify the result using bobs of different mass and the period was identical.

Beyond small angles: at $\theta = 30°$θ=30° the actual period is about $1.7\%$1.7% longer than the SHM prediction; at $\theta = 90°$θ=90° it is about $18\%$18% longer; at $\theta = 179°$θ=179° (almost inverted) it diverges to infinity. So the pendulum is not a perfect SHM oscillator — it is the harmonic approximation that holds for small swings. This is a beautiful example of the more general truth that all potential wells look harmonic at the bottom.

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Sketching $a$a vs $x$x Graphs for SHM

Because $a = -\omega^2 x$a=−ω2x is linear in $x$x, a plot of $a$a against $x$x is a straight line through the origin with negative gradient $-\omega^2$−ω2. This is a favourite OCR graph-sketching question.