Equations for Simple Harmonic Motion (SHM) (HL) | HL IB Physics Revision Notes 2025

Equations for Simple Harmonic Motion

There are many equations to learn in the topic of simple harmonic motion
The key equations are summarised below:

3-1-8-new-equations-for-shm-1

3-1-8-new-equations-for-shm-2

3-1-8-new-equations-for-shm-3

A summary of the equations related to simple harmonic motion. The green stars indicate equations which are not included in the IB data booklet and the highlighted equations are for HL students only

Worked example

The graph shows the potential energy, E_P, for a particle oscillating with SHM. The particle has mass 45 g.

9-1-3-worked-example-1

(a)

Use the graph to determine the amplitude and the period of the oscillation.

(b)

Calculate the maximum speed which the particle achieves.

Answer:

(a)

Step 1: Use the graph to determine the maximum potential energy of the particle

9-1-3-worked-example-1-solution

Maximum potential energy, E_Pmax = 60 mJ = 60 × 10⁻³ J

Step 2: Determine the amplitude of the oscillation

The amplitude of the motion $x_{0}$ is the maximum displacement
At the maximum displacement, the particle is at its highest point, hence this is the position of maximum potential energy
From the graph, when E_P = 60 mJ:

${x_{0}}^{2}$ = 8.0 cm = 0.08 m

Amplitude: $x_{0} = \sqrt{0.08} = 0.28 m$

Step 3: Write down the equation for the potential energy of an oscillator and rearrange for angular velocity ω

$E_{P} = \frac{1}{2} m ω^{2} {x_{0}}^{2}$

$ω^{2} = \frac{2 E_{P}}{m {x_{0}}^{2}} \Rightarrow ω = \sqrt{\frac{2 E_{P}}{m {x_{0}}^{2}}}$

Step 4: Substitute the known values and calculate ω

Mass of the particle, m = 45 g = 45 × 10⁻³ kg

$ω = \sqrt{\frac{2 \times (60 \times 10^{- 3})}{(45 \times 10^{- 3}) \times 0.08}} = 5.77 rad s^{- 1}$

Step 5: Determine the time period of the oscillation

$T = \frac{2 π}{ω} = \frac{2 π}{5.77}$ = 1.1 s

(b)

The maximum speed of the oscillator is equal to

$v_{m a x} = ω x_{0}$

$v_{m a x}$ = 5.77 × 0.28 = 1.6 m s⁻¹

Worked example

A student investigated the behaviour of a 200 g mass oscillating on a spring, and produced the graph shown.

9-1-3-worked-example-2

(a)

Determine the values for amplitude and time period

(b)

Hence find the maximum kinetic energy of the oscillating mass.

Answer:

(a)

Read the values of amplitude and time period from the graph

Amplitude, $x_{0}$ = 0.3 m
Time period, T = 2.0 s

(b)

Step 1: Write down the equation for the maximum speed of an oscillator

$v_{m a x} = ω x_{0}$

Step 2: Write down the equation relating angular speed and time period

$ω = \frac{2 π}{T}$

Step 3: Combine the two equations and calculate the maximum speed

$v_{m a x} = \frac{2 π x_{0}}{T} = \frac{2 π \times 0.3}{2.0}$

$v_{m a x}$ = 0.942 m s⁻¹

Step 4: Use the maximum speed to calculate the maximum kinetic energy of the oscillating mass

Mass of the oscillator, m = 200 g = 0.2 kg

E_{K m a x} = \frac{1}{2} m {v_{m a x}}^{2}

E_Kmax = 0.5 × 0.2 × 0.942² = 0.1 J

Exam Tip

There are a large number of equations associated with SHM. Most of them are given in the data booklet which you will be given to use in the exam

Make sure you are familiar with the equations, as you will probably need to use several different ones to solve the longer questions.

Equations for Simple Harmonic Motion (SHM) (HL) (HL IB Physics)

Revision Note