C.1.5—Mass–spring period

Syllabus
First assessment 2025
Objective
Level
HL

Calculate the Period of a Mass–Spring System

Mass–spring period

For an ideal mass mm attached to a linear spring of spring constant kk,

T=2πmkT=2\pi\sqrt{\frac{m}{k}}

Use mm in kilograms and kk in Nm1\mathrm{N\,m^{-1}} to obtain TT in seconds.

Read the dependence

TmT\propto\sqrt m: more mass increases the period. T1/kT\propto1/\sqrt k: a stiffer spring decreases the period. The ideal period is independent of amplitude while Hooke's law remains valid.

Worked example from local textbook question 6

For T=1.00sT=1.00\,\mathrm s and k=84Nm1k=84\,\mathrm{N\,m^{-1}},

m=k(T2π)2=84(1.002π)2=2.13kgm=k\left(\frac{T}{2\pi}\right)^2=84\left(\frac{1.00}{2\pi}\right)^2=2.13\,\mathrm{kg}

Boundary

This model assumes a linear spring and that the stated oscillating mass includes any effective mass the question requires. Do not substitute amplitude for mm.

C.1.5 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions divide a cycle into time intervals such as 0 to T/4 and T/4 to T/2, asking which energy decreases and which increases. The evidence requires the specific stored-energy form for the oscillator.

Command terms

Describe / State

What earns marks

Name the two energy forms and state the direction of transfer over the stated time interval. From an extreme position to equilibrium, elastic/spring potential energy decreases while kinetic energy increases; from equilibrium to an extreme, the reverse occurs.

Watch for

Saying potential energy increases throughout the motion, or failing to identify elastic/spring potential energy for a spring oscillator.

Representative question

Question 1

[Maximum number: 1]

between t=0 and t=T4t=\frac{T}{4};