NSWMaths Extension 2Syllabus dot point

How does the defining equation of simple harmonic motion lead to its displacement, velocity and period, and how do we model an oscillating particle?

Derive and apply the equations of simple harmonic motion, relating acceleration, velocity, displacement, amplitude and period for an oscillating particle

A focused answer to the HSC Maths Extension 2 dot point on simple harmonic motion. The defining equation, displacement and velocity functions, amplitude and period, and the velocity-displacement relation, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking
The defining equation
Displacement as a function of time
Velocity as a function of time
The velocity-displacement relation

What this dot point is asking

NESA wants you to model simple harmonic motion (SHM), the oscillation of a particle whose acceleration is proportional to its displacement and directed towards a fixed centre. You must use the defining equation, write displacement and velocity as functions of time, and find amplitude, period and the velocity-displacement relation.

The defining equation

A particle moves in simple harmonic motion if its acceleration satisfies

\ddot{x} = -n^2 x,

where $x$ is displacement from a fixed centre and $n > 0$ is a positive constant. The minus sign means acceleration always points back towards the centre: the further the particle strays, the harder it is pulled back. This is the mathematical signature of an oscillation about equilibrium.

Displacement as a function of time

The general solution of $\ddot{x} = -n^2 x$ is

x = a\cos(nt + \alpha),

where $a$ is the amplitude (maximum displacement) and $\alpha$ is a phase constant fixed by initial conditions. Equivalently it can be written $x = A\cos nt + B\sin nt$ . The motion repeats every time $nt$ increases by $2\pi$ , so the period is

T = \frac{2\pi}{n}.

The frequency (oscillations per unit time) is $\dfrac{1}{T} = \dfrac{n}{2\pi}$ .

Velocity as a function of time

Differentiating $x = a\cos(nt + \alpha)$ gives

v = \dot{x} = -an\sin(nt + \alpha).

The speed reaches its maximum $an$ when $\sin(nt + \alpha) = \pm 1$ , which occurs as the particle passes through the centre $x = 0$ . The velocity is zero at the extremes $x = \pm a$ , where the particle momentarily stops before reversing.

The velocity-displacement relation

It is often more useful to relate speed directly to position, without time. Using $\ddot{x} = \dfrac{d}{dx}\left(\tfrac{1}{2}v^2\right)$ and integrating $\dfrac{d}{dx}\left(\tfrac{1}{2}v^2\right) = -n^2 x$ gives $\tfrac{1}{2}v^2 = -\tfrac{1}{2}n^2 x^2 + c$ . Applying $v = 0$ at $x = a$ fixes the constant, yielding

v^2 = n^2(a^2 - x^2).

This confirms maximum speed $v = an$ at the centre $x = 0$ and zero speed at $x = \pm a$ .

Exam-style practice questions

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

2023 HSC2 marksA particle moves in simple harmonic motion described by the equation x'' = -9(x - 4). Find the period and the central point of motion.

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Compare with the standard SHM equation x'' = -n^2 (x - c), where c is the centre of motion and n is the angular frequency.

Here -n^2 = -9, so n^2 = 9 and n = 3. The bracket (x - 4) shows the centre is c = 4.

Period: T = 2 pi/n = 2 pi/3 seconds.

So the period is 2 pi/3 and the central point of motion is x = 4.

Mark notes: 1 mark for the period 2 pi/3 (from n = 3), 1 mark for the centre of motion x = 4.

2024 HSC3 marksA particle is moving in simple harmonic motion, described by x'' = -4(x + 1). When the particle passes through the origin, the speed of the particle is 4 m s^-1. What distance does the particle travel during a full period of its motion?

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From x'' = -4(x + 1), the form x'' = -n^2 (x - c) gives n^2 = 4 (so n = 2) and centre c = -1.

The velocity-displacement relation for SHM is v^2 = n^2 (A^2 - (x - c)^2), where A is the amplitude. Here v^2 = 4(A^2 - (x + 1)^2).

At the origin x = 0 the speed is 4, so 4^2 = 4(A^2 - (0 + 1)^2):
16 = 4(A^2 - 1), so A^2 - 1 = 4, giving A^2 = 5 and A = sqrt(5).

In one full period the particle travels from one extreme to the other and back, covering 4 amplitudes:

distance = 4A = 4 sqrt(5) metres.

Mark notes: 1 mark for identifying n = 2 and centre c = -1, 1 mark for using v^2 = n^2(A^2 - (x - c)^2) to find A = sqrt(5), 1 mark for the distance 4A = 4 sqrt(5) metres.

2021 HSC4 marksAn object moves in simple harmonic motion along the x-axis with x'' = -4(x - 3), x in metres after t seconds. Initially the object is 5.5 metres to the right of the origin and moving towards the origin. It has a speed of 8 m s^-1 as it passes through the origin. (i) Between which two values of x is the particle oscillating? (ii) Find the first value of t for which x = 0, correct to 2 decimal places.

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From x'' = -4(x - 3): n^2 = 4 so n = 2, and the centre is c = 3.

Part (i). Use v^2 = n^2(A^2 - (x - c)^2) = 4(A^2 - (x - 3)^2). At the origin x = 0 the speed is 8:
64 = 4(A^2 - 9), so A^2 - 9 = 16, giving A^2 = 25 and A = 5.
The particle oscillates between c - A = 3 - 5 = -2 and c + A = 3 + 5 = 8. So between x = -2 and x = 8.

Part (ii). Write x = c + A cos(n t + phi) = 3 + 5 cos(2t + phi). At t = 0, x = 5.5: cos(phi) = 2.5/5 = 0.5, so phi = pi/3 or -pi/3. Since the object is moving towards the origin (x decreasing), x'(0) = -10 sin(phi) < 0, so sin(phi) > 0 and phi = pi/3.

Thus x = 3 + 5 cos(2t + pi/3). Setting x = 0: cos(2t + pi/3) = -0.6, so the first relevant solution is 2t + pi/3 = arccos(-0.6) = 2.2143...
2t = 2.2143 - 1.0472 = 1.1671, so t = 0.58 seconds (to 2 decimal places).

Mark notes: parts were 2 marks each. 1 mark for n = 2 and centre 3, 1 mark for A = 5 giving the range -2 to 8; then 1 mark for the displacement equation with phi = pi/3, 1 mark for t = 0.58.