An object in SHM has amplitude 0.050\ m and angular frequency 10\ rad s^-1. Find its maximum speed. [2 marks]

OCR OCR 2019-style practice: A mass of 0.25\ kg on a spring oscillates with simple harmonic motion of amplitude 0.040\ m and period 0.80\ s. Calculate the maximum speed of the mass and its maximum acceleration.

Angular frequency: ω = (2π)/(T) = (2π)/(0.80) = 7.85\ rad s^-1. Maximum speed: v_ = ω A = 7.85 × 0.040 = 0.31\ m s^-1. Maximum acceleration: a_ = ω^2 A = (7.85)^2 × 0.040 = 61.6 × 0.040 = 2.5\ m s^-2. Markers reward ω = (2π)/(T), the maximum speed about 0.31\ m s^-1, and the maximum acceleration about 2.5\ m s^-2.

EnglandPhysicsSyllabus dot point

What defines simple harmonic motion, and how do damping and resonance affect oscillations?

Simple harmonic motion: the defining condition, displacement, velocity and acceleration in SHM, the energy interchange, the period of mass-spring and pendulum systems, and free and forced oscillations with damping and resonance.

A focused answer to the OCR H556 oscillations content, covering the defining condition for simple harmonic motion, the displacement, velocity and acceleration equations, the interchange of kinetic and potential energy, the period of a mass-spring system and a simple pendulum, and free and forced oscillations with damping and resonance.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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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), with maximum speed $v_{\max} = \omega A$ at the centre and maximum acceleration $a_{\max} = \omega^2 A$ at the extremes. Energy interchanges between kinetic (greatest at the centre) and potential (greatest at the extremes), with the total constant if undamped. A mass on a spring has $T = 2\pi\sqrt{\frac{m}{k}}$ ; a simple pendulum has $T = 2\pi\sqrt{\frac{l}{g}}$ . Damping reduces amplitude over time; resonance is the large-amplitude response when a driving frequency matches the natural frequency.

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What this dot point is asking

OCR wants you to state the defining condition for simple harmonic motion, use the displacement, velocity and acceleration equations, describe the interchange of kinetic and potential energy, use the period equations for a mass-spring system and a simple pendulum, and describe free and forced oscillations, damping and resonance.

The answer

The defining condition

Displacement, velocity and acceleration

Energy in SHM

Mass-spring and pendulum periods

Free oscillations, damping and resonance

Examples in context

Resonance is exploited in musical instruments, radio tuning (matching a circuit's natural frequency to a station) and magnetic resonance imaging. It can also be destructive: marching soldiers break step on bridges, and engineers add dampers to skyscrapers and bridges to prevent wind or earthquakes driving dangerous oscillations. Car suspensions use near-critical damping so the vehicle settles quickly after a bump without bouncing. A pendulum clock relies on the isochronous property of SHM to keep time.

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 $0.050\ \text{m}$ and angular frequency $10\ \text{rad s}^{-1}$ . Find its maximum speed. [2 marks]

Cue. $v_{\max} = \omega A = 10 \times 0.050 = 0.50\ \text{m s}^{-1}$ .

Q3. Explain what is meant by resonance. [2 marks]

Cue. When the driving frequency of a forced oscillation equals the system's natural frequency, energy is transferred most efficiently and the amplitude becomes very large.

Exam-style practice questions

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

OCR 20194 marksA mass of

0.25\ \text{kg}

on a spring oscillates with simple harmonic motion of amplitude

0.040\ \text{m}

and period

0.80\ \text{s}

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

Show worked answer →

Angular frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.80} = 7.85\ \text{rad s}^{-1}$ .

Maximum speed: $v_{\max} = \omega A = 7.85 \times 0.040 = 0.31\ \text{m s}^{-1}$ .

Maximum acceleration: $a_{\max} = \omega^2 A = (7.85)^2 \times 0.040 = 61.6 \times 0.040 = 2.5\ \text{m s}^{-2}$ .

Markers reward $\omega = \frac{2\pi}{T}$ , the maximum speed about $0.31\ \text{m s}^{-1}$ , and the maximum acceleration about $2.5\ \text{m s}^{-2}$ .

OCR 20224 marksA simple pendulum has a period of

2.0\ \text{s}

on Earth. Calculate its length. The pendulum is then taken to the Moon where

g = 1.6\ \text{m s}^{-2}

. State and explain what happens to its period. Take

g = 9.8\ \text{m s}^{-2}

on Earth.

Show worked answer →

Length: from $T = 2\pi\sqrt{\frac{l}{g}}$ , square and rearrange: $l = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 2.0^2}{4\pi^2} = \frac{39.2}{39.5} = 0.99\ \text{m}$ , about $1.0\ \text{m}$ .

On the Moon $g$ is smaller, so since $T \propto \frac{1}{\sqrt{g}}$ the period increases. Specifically $T_{\text{Moon}} = 2\pi\sqrt{\frac{0.99}{1.6}} = 4.9\ \text{s}$ , so the pendulum swings more slowly.

Markers reward rearranging for the length about $1.0\ \text{m}$ , recognising $T \propto \frac{1}{\sqrt{g}}$ , and a longer period on the Moon.

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)