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C.1.1—Conditions for SHM

Syllabus
First assessment 2025
Objective
—
Level
SL

Recognize the Conditions for SHM

Start with the restoring force

Simple harmonic motion occurs when the restoring force is proportional to displacement from equilibrium and always points back toward equilibrium: F∝−xF\propto -x. For constant mass this gives a∝−xa\propto -x.

Locate equilibrium

Measure xx from the equilibrium position, not from an arbitrary origin. At equilibrium x=0x=0, so the restoring force and acceleration are zero; away from equilibrium, the acceleration points opposite to the displacement.

Check the model

A spring–mass system is an SHM model when the spring force is linear. A simple pendulum approximates SHM only for small angular displacements, where sin⁡θ≈θ\sin\theta\approx\theta in radians.

Common trap

“Periodic” motion alone is not enough. The defining condition is the restoring acceleration a=−ω2xa=-\omega^2x, including the opposite direction and proportional dependence.

C.1.1 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions present a force–displacement relationship or an oscillator setup and ask which condition produces SHM. The evidence rewards the negative proportional relationship rather than periodicity alone.

Command terms

Identify / State

What earns marks

Identify the equilibrium position, write the restoring relationship as F ∝ −x or a = −ω²x, and check that the force reverses direction when x changes sign. For a pendulum, mention the small-angle approximation; for a spring, use the linear restoring-force region.

Watch for

Selecting any repeating motion as SHM without checking that the restoring force is proportional to displacement and opposite in direction.

Representative question

Question 1

[Maximum number: 1]

A force F acts on a particle. The displacement of the particle is x. Which variation of F with x results in simple harmonic motion?

A
B
C
D

Retrieve the Core C.1 Simple Harmonic Motion Model

Recognize SHM

SHM requires a restoring acceleration a=−ω2xa=-\omega^2x about equilibrium. Track displacement, amplitude, period, frequency and angular frequency with T=1/f=2π/ωT=1/f=2\pi/\omega.

Track one cycle

At an extreme, potential energy is maximum and kinetic energy is zero; at equilibrium, kinetic energy is maximum and potential energy is minimum. Total energy remains constant in ideal SHM.

Read the motion

The gradient of a displacement–time graph is velocity. Acceleration is opposite to displacement. Use the sign of displacement and the gradient to identify direction at any instant.

Final check

Measure displacement from equilibrium, keep amplitude distinct from peak-to-peak distance, and name the relevant potential-energy form for the oscillator.