An object in SHM has amplitude 0.040\ m and angular frequency 12\ rad s^-1. Find its maximum speed. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A mass of 0.20\ kg on a spring oscillates with simple harmonic motion of amplitude 0.030\ m and period 0.75\ s. Calculate the maximum speed of the mass and its maximum acceleration.

Angular frequency: ω = 2πT = 2π0.75 = 8.38\ rad s^-1. Maximum speed: v_max = ω A = 8.38 × 0.030 = 0.25\ m s^-1. Maximum acceleration: a_max = ω^2 A = (8.38)^2 × 0.030 = 70.2 × 0.030 = 2.1\ m s^-2. Markers reward ω = (2π)/(T), the maximum speed about 0.25\ m s^-1, and the maximum acceleration about 2.1\ m s^-2.

EnglandPhysicsSyllabus dot point

What defines simple harmonic motion, and how do displacement, velocity, acceleration and energy vary through a cycle?

Vibrations: the defining condition for simple harmonic motion, displacement, velocity and acceleration in SHM, the period of mass-spring and pendulum systems, and the interchange of kinetic and potential energy.

A focused answer to the Eduqas A-Level Physics Component 1 vibrations content, covering the defining condition for simple harmonic motion, the displacement, velocity and acceleration equations, the period of a mass-spring system and a simple pendulum, and the interchange of kinetic and potential energy.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Quick answer

Simple harmonic motion (SHM) occurs when the acceleration is proportional to the displacement from equilibrium and always directed back towards it: $a = -\omega^2 x$ . The displacement varies as $x = A\cos(\omega t)$ (starting from maximum) or $A\sin(\omega t)$ , with maximum speed $v_\text{max} = \omega A$ at the centre and maximum acceleration $a_\text{max} = \omega^2 A$ at the extremes. The angular frequency is $\omega = 2\pi f = \frac{2\pi}{T}$ . A mass on a spring has period $T = 2\pi\sqrt{\frac{m}{k}}$ ; a simple pendulum has $T = 2\pi\sqrt{\frac{l}{g}}$ . Energy interchanges continuously between kinetic (greatest at the centre) and potential (greatest at the extremes), the total constant if undamped.

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to state the defining condition for simple harmonic motion, use the displacement, velocity and acceleration relations, apply the period equations for a mass-spring system and a simple pendulum, and describe the interchange of kinetic and potential energy during an oscillation.

The answer

The defining condition

Displacement, velocity and acceleration

Mass-spring and pendulum periods

Energy in SHM

Examples in context

Simple harmonic motion models the swing of a pendulum clock, the bounce of a mass on a spring, the vibration of atoms in a solid lattice, and the motion of a tuning fork. The isochronous property (period independent of amplitude) is why a pendulum keeps good time even as its swing slowly decays, and why a mass-spring system is used as the timing element in mechanical watches and vehicle suspension models.

Try this

Q1. State the defining condition for simple harmonic motion. [2 marks]

Cue. The acceleration is proportional to the displacement from equilibrium and directed towards it, $a = -\omega^2 x$ .

Q2. An object in SHM has amplitude $0.040\ \text{m}$ and angular frequency $12\ \text{rad s}^{-1}$ . Find its maximum speed. [2 marks]

Cue. $v_\text{max} = \omega A = 12 \times 0.040 = 0.48\ \text{m s}^{-1}$ .

Q3. State where in the oscillation the kinetic energy is greatest. [1 mark]

Cue. At the centre (equilibrium position), where the speed is greatest.

Exam-style practice questions

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

Eduqas 20194 marksA mass of

0.20\ \text{kg}

on a spring oscillates with simple harmonic motion of amplitude

0.030\ \text{m}

and period

0.75\ \text{s}

. Calculate the maximum speed of the mass and its maximum acceleration.

Show worked answer →

Angular frequency: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.75} = 8.38\ \text{rad s}^{-1}$ .

Maximum speed: $v_\text{max} = \omega A = 8.38 \times 0.030 = 0.25\ \text{m s}^{-1}$ .

Maximum acceleration: $a_\text{max} = \omega^2 A = (8.38)^2 \times 0.030 = 70.2 \times 0.030 = 2.1\ \text{m s}^{-2}$ .

Markers reward $\omega = \frac{2\pi}{T}$ , the maximum speed about $0.25\ \text{m s}^{-1}$ , and the maximum acceleration about $2.1\ \text{m s}^{-2}$ .

Eduqas 20224 marksA simple pendulum has a period of

1.6\ \text{s}

on Earth. Calculate its length, and state with a reason what happens to its period if the mass of the bob is doubled. Take

g = 9.81\ \text{m s}^{-2}

Show worked answer →

Length from $T = 2\pi\sqrt{\dfrac{l}{g}}$ : square and rearrange, $l = \dfrac{gT^2}{4\pi^2} = \dfrac{9.81 \times 1.6^2}{4\pi^2} = \dfrac{25.1}{39.5} = 0.64\ \text{m}$ .

The period of a simple pendulum depends only on its length and $g$ , not on the mass of the bob, so doubling the mass leaves the period unchanged at $1.6\ \text{s}$ .

Markers reward rearranging for the length about $0.64\ \text{m}$ , and stating the period is independent of the bob's mass so it is unchanged.

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)