ScotlandMathematics of MechanicsSyllabus dot point

What is simple harmonic motion, and how do you find the period, speed and displacement of an oscillating body, including a mass on a spring obeying Hooke's law?

Define simple harmonic motion by the equation a equals minus omega squared x; derive and use the displacement, velocity and period results; apply Hooke's law to springs and strings; and analyse the energy of an oscillation.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on simple harmonic motion, covering the defining equation, the displacement and velocity solutions, the period and amplitude, Hooke's law for springs and elastic strings, and the interchange of kinetic and potential energy in an oscillation.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The defining equation and its solution
Hooke's law and the spring oscillator
Energy in an oscillation
The phase and starting conditions
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What this dot point is asking

Simple harmonic motion (SHM) is the standard model for any oscillation about an equilibrium - a mass bobbing on a spring, a pendulum at small angle, a particle on an elastic string. The SQA wants you to recognise SHM from its defining equation, use the displacement, velocity and period results, apply Hooke's law to set up the equation for a spring or string, and track the exchange of kinetic and potential energy.

The defining equation and its solution

What makes motion "simple harmonic" is a single relationship between acceleration and displacement. Everything else follows from it.

The velocity relation is the most used result: it lets you find the speed at any displacement without involving time. The fastest point is the centre ( $x = 0$ , $v = \omega a$ ) and the slowest are the turning points ( $x = \pm a$ , $v = 0$ ).

Speed and period for a given SHM

A particle in SHM has $\omega = 3$ rad s $^{-1}$ and amplitude $0.05$ m. Find its speed at $x = 0.03$ m and its period.

Step 1: Apply the velocity relation

The time-free relation $v^2 = \omega^2(a^2 - x^2)$ gives the speed at any displacement without needing to know the time. Substituting $\omega = 3$ , $a = 0.05$ , $x = 0.03$ :

v^2 = 9(0.05^2 - 0.03^2) = 9(0.0025 - 0.0009) = 9(0.0016) = 0.0144.

Step 2: Take the square root for the speed

Speed is positive, so:

v = 0.12\ \text{m s}^{-1}.

Step 3: Find the period

The period depends only on $\omega$ , not on the amplitude, so pulling a spring further does not change how long each oscillation takes:

T = \frac{2\pi}{\omega} = \frac{2\pi}{3} \approx 2.09\ \text{s.}

Final answer: speed at $x = 0.03\ \text{m}$ is $0.12\ \text{m s}^{-1}$ ; period is approximately $2.09\ \text{s}$ .

Hooke's law and the spring oscillator

Hooke's law states that the tension (or thrust) in a spring or elastic string is proportional to its extension. It is what supplies the restoring force that makes a spring-mass system oscillate harmonically.

The neat point is that the steady weight does not affect the oscillation: at the equilibrium position the spring tension already balances gravity, so the extra displacement feels only the extra spring force $-kx$ , which is exactly the SHM condition.

Energy in an oscillation

Because the only force doing work is the conservative spring force, the total mechanical energy stays constant. Energy continuously transfers between kinetic and elastic potential.

The total energy follows directly from the velocity relation. The kinetic energy at displacement $x$ is $\tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2(a^2 - x^2)$ , and the potential energy stored in the restoring force is $\tfrac{1}{2}m\omega^2 x^2$ . Adding them, the $x^2$ terms cancel and the constant total $\tfrac{1}{2}m\omega^2 a^2$ remains, which is a clean way to check a piece of SHM work: if your kinetic and potential energies do not sum to a constant, there is a slip somewhere.

Energy in an SHM oscillation

A $0.2$ kg mass performs SHM with $\omega = 4$ rad s $^{-1}$ and amplitude $0.3$ m. Find its total energy and its kinetic energy at $x = 0.15$ m.

Step 1: Calculate the total energy

The total mechanical energy in SHM is $\tfrac{1}{2}m\omega^2 a^2$ , the value the kinetic energy takes when the body passes through the centre at maximum speed. It stays constant throughout the motion:

E = \tfrac{1}{2}(0.2)(16)(0.09) = 0.144\ \text{J.}

Step 2: Find the kinetic energy at $x = 0.15\ \text{m}$

First use the velocity relation to get $v^2$ , then halve and multiply by mass:

v^2 = \omega^2(a^2 - x^2) = 16(0.09 - 0.0225) = 16(0.0675) = 1.08,

\text{KE} = \tfrac{1}{2}(0.2)(1.08) = 0.108\ \text{J.}

Step 3: Verify with the potential energy

At the same displacement the potential energy stored in the restoring force is $\tfrac{1}{2}m\omega^2 x^2$ . Checking that KE and PE sum to the total is a useful error-catcher:

\text{PE} = \tfrac{1}{2}(0.2)(16)(0.0225) = 0.036\ \text{J}, \quad 0.108 + 0.036 = 0.144\ \text{J.}

Final answer: total energy is $0.144\ \text{J}$ ; kinetic energy at $x = 0.15\ \text{m}$ is $0.108\ \text{J}$ .

The phase and starting conditions

The arbitrary phase $\varepsilon$ in $x = a\sin(\omega t + \varepsilon)$ is fixed by where the body starts. A body released from rest at the extreme $x = a$ has $x = a\cos\omega t$ ; a body passing through the centre at $t = 0$ has $x = a\sin\omega t$ . Choosing the form that matches the starting state, rather than carrying $\varepsilon$ through the algebra, usually keeps the working short.

Common traps

Thinking the period depends on amplitude: In SHM the period $T = \dfrac{2\pi}{\omega}$ is set only by $\omega$ , so pulling a spring further does not change how long an oscillation takes. The amplitude changes the maximum speed, not the period.
Putting maximum speed at the extremes: The speed is greatest at the centre ( $v_{\max} = \omega a$ ) and zero at the turning points. Reversing this is a common slip; the velocity relation $v^2 = \omega^2(a^2 - x^2)$ makes it clear that $v$ is largest when $x = 0$ .
Forgetting the equilibrium weight cancels for a spring: When setting up SHM for a hanging mass, measure the displacement from the equilibrium position. The steady weight is already balanced by the equilibrium tension, so only the extra spring force $-kx$ drives the motion. Including the weight again gives the wrong equation.

Try this

Q1. An SHM has period $4$ s. Find its angular frequency. [2 marks]

Cue. $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$ rad s $^{-1}$ .

Q2. A particle in SHM has amplitude $0.1$ m and maximum speed $0.6$ m s $^{-1}$ . Find $\omega$ . [2 marks]

Cue. $v_{\max} = \omega a$ , so $\omega = \dfrac{0.6}{0.1} = 6$ rad s $^{-1}$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: SHM speed5 marksA particle moves with simple harmonic motion of amplitude

0.2

m and period

\pi

seconds. Find the maximum speed and the speed when the particle is

0.1

m from the centre.

Show worked answer →

The angular frequency is $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{\pi} = 2$ rad s $^{-1}$ (1 mark).

Maximum speed occurs at the centre, $v_{\max} = \omega a = 2\times 0.2 = 0.4$ m s $^{-1}$ (1 mark).

Using $v^2 = \omega^2(a^2 - x^2)$ with $x = 0.1$ : $v^2 = 4(0.04 - 0.01) = 4(0.03) = 0.12$ (2 marks).

So $v = \sqrt{0.12} \approx 0.346$ m s $^{-1}$ (1 mark). Markers reward finding $\omega$ from the period, the maximum speed $\omega a$ , and the velocity relation $v^2 = \omega^2(a^2 - x^2)$ .

AH style: Hooke's law SHM6 marksA particle of mass

0.4

kg hangs in equilibrium on a light spring of stiffness

k = 16

N m

^{-1}

. It is pulled down a small distance and released. Show that the motion is simple harmonic and find the period. Take

g = 9.8

m s

^{-2}

Show worked answer →

At equilibrium the spring extension $e$ satisfies $ke = mg$ , so $e = \dfrac{0.4\times 9.8}{16} = 0.245$ m (1 mark).

Displace the mass a further $x$ below equilibrium. The net restoring force is $-(k(e + x) - mg) = -kx$ , since $ke = mg$ cancels (2 marks).

Newton's second law: $m\ddot{x} = -kx$ , so $\ddot{x} = -\dfrac{k}{m}x$ , which is SHM with $\omega^2 = \dfrac{k}{m} = \dfrac{16}{0.4} = 40$ (2 marks).

Period $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt{40}} \approx 0.993$ s (1 mark). Markers reward showing the restoring force is proportional to displacement (the equilibrium terms cancel) and the period from $\omega = \sqrt{k/m}$ .

Related dot points

Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)