C.1.6—Simple pendulum period

Syllabus
First assessment 2025
Objective
Level
SL

Calculate the Period of a Simple Pendulum

Simple-pendulum period

For a pendulum of length ll undergoing small-angle oscillations,

T=2πlgT=2\pi\sqrt{\frac{l}{g}}

Measure ll from the pivot to the bob's centre of mass and use gg in ms2\mathrm{m\,s^{-2}}.

Read the dependence

TlT\propto\sqrt l and T1/gT\propto1/\sqrt g. Bob mass does not appear, so changing mass alone does not change the ideal period.

Worked example from local practice question 9

Changing MM to 4M4M has no effect. Changing ll to 0.25l0.25l gives

T=2π0.25lg=0.5TT'=2\pi\sqrt{\frac{0.25l}{g}}=0.5T

Boundary

The equation is the small-angle approximation, where sinθθ\sin\theta\approx\theta with θ\theta in radians. Large amplitudes do not follow this period exactly.

C.1.6 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

Questions give a velocity or displacement graph and ask you to identify the other graphs or the direction of motion at a time. The evidence rewards gradient reasoning and the correct restoring direction.

Command terms

What is / State / Explain

What earns marks

Use the gradient of a displacement–time graph for velocity, then use a = −ω²x for acceleration. Check the point’s displacement sign and gradient separately; at an extreme, v = 0 but |a| is maximum.

Watch for

Reading velocity from the height of a displacement graph instead of its gradient.

Representative question

Question 1

[Maximum number: 1]

An object performs simple harmonic motion (shm). The graph shows how the velocity v of the object varies with time t.

The displacement of the object is x and its acceleration is a. What is the variation of x with t and the variation of a with t ?

A
B
C
D