C.1 Simple harmonic motion

Lesson C.1 — Simple Harmonic Motion (IB Physics)

These notes develop a clear, self-study friendly understanding of oscillations and simple harmonic motion (SHM), including definitions, conditions for SHM, time period relationships, pendulums and springs, energy transfers, and (HL) phase relationships.

Syllabus map

C.1.1 Simple harmonic motion
C.1.2 Simple pendulum and mass–spring system
C.1.3 Position and velocity
C.1.4 (HL) Energy in SHM

C.1.1 Simple Harmonic Motion (SL)

Learning objectives

Identify examples of SHM and describe the conditions required for SHM.
Use the terms: equilibrium position, displacement $x$ , amplitude $A$ , time period $T$ , frequency $f$ , angular frequency $\omega$ .
Use the defining equation of SHM: $a=-\omega^2 x$ .
Use $f=\frac{1}{T}$ and $\omega=2\pi f$ .

Equilibrium position

Imagine pushing a friend on a swing. If you stop pushing and your friend does not pump the swing,
the oscillations gradually get smaller and the swing comes to rest directly underneath the pivot.

The position where the object would naturally rest if it were not moving is called the equilibrium position.

At equilibrium:

$F_{\text{net}}=0$

Key idea: SHM requires that if the object is displaced from equilibrium, there is a restoring force (and therefore acceleration)
that pulls it back toward equilibrium.

Displacement and amplitude

Definitions

Displacement, $x$ , is the distance from equilibrium at a particular instant (can be positive or negative).
Amplitude, $A$ , is the maximum displacement from equilibrium.

$|x| \leq A$

Scaffolding (how to think about it):

$x$ changes throughout the motion.
$A$ is fixed for that oscillation unless energy is added/removed.
Amplitude is linked to energy: bigger $A$ means more total energy.

Time period and frequency

Definitions

The time period, $T$ , is the time taken for one complete oscillation.
The frequency, $f$ , is the number of oscillations per second.

$f=\frac{1}{T}$

Units: $T$ in seconds (s), $f$ in hertz (Hz).

Angular frequency

Angular frequency $\omega$ measures how fast the oscillator moves through its cycle in radians per second.
One complete oscillation corresponds to $2\pi$ radians.

$\omega = 2\pi f$

$\omega = \frac{2\pi}{T}$

Common misconception

$f$ and $\omega$ are related but not the same.
$f$ counts cycles per second; $\omega$ counts radians per second.

Defining condition for SHM

A system performs simple harmonic motion if the acceleration is:

proportional to displacement from equilibrium, and
directed toward equilibrium.

This is written as:

$a=-\omega^2 x$

Why the negative sign matters:

If $x$ is positive (object is to the right), acceleration is negative (pulls left).
If $x$ is negative, acceleration is positive (pulls right).
This restoring acceleration is what produces oscillation.

Worked example 1 — Angular frequency ratio

Oscillator A has time period $2.5\,\text{s}$ .
Oscillator B has frequency $0.20\,\text{Hz}$ .
Determine the ratio $\omega_A : \omega_B$ .

Step 1: Find $f_A$ .

$f_A=\frac{1}{T_A}=\frac{1}{2.5}=0.40\,\text{Hz}$

Step 2: Use $\omega=2\pi f$ .

$\omega_A=2\pi(0.40)=0.80\pi$

$\omega_B=2\pi(0.20)=0.40\pi$

Step 3: Ratio

$\frac{\omega_A}{\omega_B}=\frac{0.80\pi}{0.40\pi}=2$

Answer: $\omega_A:\omega_B=2:1$

IB exam tip

If you are given a graph of $a$ against $x$ for SHM, it is a straight line through the origin with gradient $-\omega^2$ .
If you are given a graph of $x$ against $t$ , the gradient is not constant (because $x$ changes sinusoidally).

C.1.2 Simple Pendulum and Mass–Spring System (SL + HL derivation)

Learning objectives

Use $T=2\pi\sqrt{\frac{m}{k}}$ for a mass–spring system.
Use $T=2\pi\sqrt{\frac{L}{g}}$ for a simple pendulum (small angles).
(HL) Derive the pendulum period using the small-angle approximation.

Mass–spring system

A spring exerts a restoring force that is proportional to displacement (Hooke’s law):

$F=-kx$

Using Newton’s second law $F=ma$ :

$ma=-kx$

$a=-\frac{k}{m}x$

Comparing with the SHM definition $a=-\omega^2 x$ :

$\omega^2=\frac{k}{m} \quad\Rightarrow\quad \omega=\sqrt{\frac{k}{m}}$

Therefore the period is:

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}}$

Scaffolding (what this means)

Larger mass $\Rightarrow$ more inertia $\Rightarrow$ slower oscillation $\Rightarrow$ larger $T$ .
Stiffer spring (larger $k$ ) $\Rightarrow$ stronger restoring force $\Rightarrow$ faster oscillation $\Rightarrow$ smaller $T$ .

Simple pendulum — full derivation (HL rigour)

Consider a pendulum of length $L$ displaced by a small angle $\theta$ from the vertical.
The bob has weight $mg$ acting downward.

The restoring effect is due to the component of weight along the arc:

$F_{\text{tangential}}=-mg\sin\theta$

For motion along an arc, Newton’s second law becomes:

$F_{\text{tangential}} = ma_{\text{tangential}}$

The tangential acceleration is related to angular acceleration:

$a_{\text{tangential}} = L\frac{d^2\theta}{dt^2}$

Substitute into Newton’s second law:

$-mg\sin\theta = mL\frac{d^2\theta}{dt^2}$

Cancel $m$ :

$-g\sin\theta = L\frac{d^2\theta}{dt^2}$

Rearrange:

$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta$

Small-angle approximation: for $\theta$ in radians and $\theta<10^\circ$ ,

$\sin\theta \approx \theta$

So the equation becomes:

$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta$

This has the same form as SHM:

$\frac{d^2(\text{displacement})}{dt^2} = -\omega^2(\text{displacement})$

Therefore:

$\omega^2=\frac{g}{L} \quad\Rightarrow\quad \omega=\sqrt{\frac{g}{L}}$

Finally, the period is:

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{g}}$

IB exam tip

Always state the assumption: “small-angle approximation, $\sin\theta\approx\theta$ (radians)”.
If the angle is large, the motion is still oscillatory but not perfectly SHM and the formula becomes less accurate.

Worked example 2 — Pendulum length

A simple pendulum completes 4 full oscillations in 2.0 s.
Determine the length of the pendulum. Take $g=9.81\,\text{m s}^{-2}$ .

Step 1: Find frequency.

$f=\frac{4}{2.0}=2.0\,\text{Hz}$

Step 2: Find period.

$T=\frac{1}{f}=\frac{1}{2.0}=0.50\,\text{s}$

Step 3: Use $T=2\pi\sqrt{\frac{L}{g}}$ and rearrange.

$T^2=4\pi^2\frac{L}{g} \Rightarrow L=\frac{gT^2}{4\pi^2}$

$L=\frac{(9.81)(0.50)^2}{4\pi^2}=0.062\,\text{m}$

Answer: $L=6.2\,\text{cm}$

Worked example 3 — Spring constant

A mass of $230\,\text{g}$ attached to a spring undergoes SHM with frequency $16\,\text{Hz}$ .
Calculate the spring constant $k$ .

Use $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}} \Rightarrow k=m(2\pi f)^2$ .

$m=0.230\,\text{kg}$

$k=0.230(2\pi \times 16)^2$

$k=0.230(100.53)^2$

$k=2.32\times 10^3\,\text{N m}^{-1}$

Answer: $k=2.32\times 10^3\,\text{N m}^{-1}$

Energy of SHM

In SHM, energy continuously transfers between:

Potential energy (gravitational for pendulum, elastic for spring)
Kinetic energy (due to motion)

At maximum displacement ( $x=\pm A$ ):

$v=0$
energy is entirely potential

At equilibrium ( $x=0$ ):

$v$ is maximum
energy is entirely kinetic

C.1.3 Position and Velocity (HL)

For an oscillator with amplitude $A$ :

Displacement:

$x=A\sin(\omega t)$

Velocity:

$v=A\omega\cos(\omega t)$

Acceleration:

$a=-\omega^2x$

Scaffolding (reading the graphs)

$x$ is maximum at turning points.
$v$ is maximum at equilibrium.
$a$ is maximum at turning points (because $|x|$ is maximum).
$v$ is zero when $x=\pm A$ .

C.1.4 Energy (HL)

Total energy is constant (no damping). For a mass–spring system:

$E=\frac{1}{2}kA^2$

Kinetic energy and potential energy vary with time:

$KE=\frac{1}{2}kA^2\cos^2(\omega t)$

$PE=\frac{1}{2}kA^2\sin^2(\omega t)$

Kinetic energy has two maxima per cycle because it depends on $v^2$ (always positive).

IB-Style Exam Practice (SL + HL)

Instructions

Show full working and include units.
Use $g=9.81\,\text{m s}^{-2}$ unless otherwise stated.
State assumptions (e.g. small-angle approximation) where needed.

Question 1 (SL) — Period, frequency and angular frequency

A particle oscillates with time period $T=0.80\,\text{s}$ .

(a) Calculate the frequency $f$ .
(b) Calculate the angular frequency $\omega$ .
(c) State the units of $f$ and $\omega$ .

Solution

(a)

$f=\frac{1}{T}=\frac{1}{0.80}=1.25\,\text{Hz}$

(b)

$\omega=2\pi f=2\pi(1.25)=7.85\,\text{rad s}^{-1}$

Question 2 (SL) — Identifying SHM from an $a$ – $x$ graph

A graph of acceleration $a$ against displacement $x$ for an oscillator is a straight line through the origin with gradient $-36\,\text{s}^{-2}$ .

(a) Explain why this indicates SHM.
(b) Determine $\omega$ .
(c) Determine the period $T$ .

Solution

(a) SHM requires $a\propto -x$ , so an $a$ – $x$ straight line through the origin with negative gradient matches $a=-\omega^2 x$ .
(b) Gradient $=-\omega^2$ , so $\omega^2=36 \Rightarrow \omega=6.0\,\text{rad s}^{-1}$ .
(c)

$T=\frac{2\pi}{\omega}=\frac{2\pi}{6.0}=1.05\,\text{s}$

Question 3 (SL) — Mass–spring period and change in parameters

A mass $m$ on a spring oscillates with period $T=1.20\,\text{s}$ .
The mass is changed to $4m$ while the spring is unchanged.

(a) Determine the new period.
(b) Explain physically why the period changes in this way.

Solution

(a)

$T=2\pi\sqrt{\frac{m}{k}} \Rightarrow T\propto \sqrt{m}$

So:

$T_{\text{new}}=T\sqrt{\frac{4m}{m}}=T\sqrt{4}=2T=2(1.20)=2.40\,\text{s}$

(b) Increasing mass increases inertia; the system accelerates less for the same restoring force, so oscillations take longer.

Question 4 (SL) — Simple pendulum and ratios

Pendulum A has twice the mass and half the length of pendulum B.
Both oscillate with small amplitude.

Determine the ratio $T_A:T_B$ .

Solution

For a simple pendulum:

$T=2\pi\sqrt{\frac{L}{g}}$

So mass does not matter and $T\propto\sqrt{L}$ .
Given $L_A=\frac{1}{2}L_B$ :

$\frac{T_A}{T_B}=\sqrt{\frac{L_A}{L_B}}=\sqrt{\frac{1/2}{1}}=\frac{1}{\sqrt{2}}$

Answer:

$T_A:T_B = 1:\sqrt{2}$

Question 5 (HL) — Displacement and velocity at a given time

An oscillator has amplitude $A=0.060\,\text{m}$ and frequency $f=5.0\,\text{Hz}$ .
It starts at equilibrium at $t=0$ moving in the positive direction, so $x=A\sin(\omega t)$ .

(a) Determine $\omega$ .
(b) Determine the displacement $x$ at $t=0.050\,\text{s}$ .
(c) Determine the velocity $v$ at $t=0.050\,\text{s}$ .

Solution

(a)

$\omega=2\pi f=2\pi(5.0)=31.4\,\text{rad s}^{-1}$

(b)

$x=A\sin(\omega t)=0.060\sin(31.4\times 0.050)$

$x=0.060\sin(1.57)\approx 0.060(1.00)=0.060\,\text{m}$

(c)

$v=A\omega\cos(\omega t)=0.060(31.4)\cos(1.57)$

$v\approx 1.88\times 0 \approx 0\,\text{m s}^{-1}$

This makes sense because at $t=0.050\,\text{s}$ the oscillator is at (approximately) maximum displacement, so $v\approx 0$ .

Question 6 (HL) — Energy at a displacement

A mass–spring system has spring constant $k=200\,\text{N m}^{-1}$ and amplitude $A=0.080\,\text{m}$ .

(a) Determine the total energy of the oscillator.
(b) Determine the elastic potential energy when $x=0.050\,\text{m}$ .
(c) Determine the kinetic energy when $x=0.050\,\text{m}$ .

Solution

(a)

$E=\frac{1}{2}kA^2=\frac{1}{2}(200)(0.080)^2=100(0.0064)=0.64\,\text{J}$

(b)

$EPE=\frac{1}{2}kx^2=\frac{1}{2}(200)(0.050)^2=100(0.0025)=0.25\,\text{J}$

(c)

$KE=E-EPE=0.64-0.25=0.39\,\text{J}$

Key ideas to take away

SHM is defined by $a=-\omega^2x$ (restoring and proportional).
$f=\frac{1}{T}$ and $\omega=2\pi f$ link the time and angular descriptions.
Mass–spring: $T=2\pi\sqrt{\frac{m}{k}}$ ; pendulum: $T=2\pi\sqrt{\frac{L}{g}}$ (small angles).
Energy transfers between kinetic and potential while total energy stays constant (no damping).
(HL) $x$ , $v$ , and $a$ are sinusoidal and phase-shifted relative to each other.

IB Physics SL Book 2025; Papers 1 & 2 Practice 5 Sections39 Lessons19 Quizzes12 Weeks

Curriculum

C.1 Simple harmonic motion

Lesson C.1 — Simple Harmonic Motion (IB Physics)

Syllabus map

C.1.1 Simple Harmonic Motion (SL)

Learning objectives

Equilibrium position

Displacement and amplitude

Definitions

Time period and frequency

Definitions

Angular frequency

Common misconception

Defining condition for SHM

Worked example 1 — Angular frequency ratio

IB exam tip

C.1.2 Simple Pendulum and Mass–Spring System (SL + HL derivation)

Learning objectives

Mass–spring system

Scaffolding (what this means)

Simple pendulum — full derivation (HL rigour)

IB exam tip

Worked example 2 — Pendulum length

Worked example 3 — Spring constant

Energy of SHM

C.1.3 Position and Velocity (HL)

Scaffolding (reading the graphs)

C.1.4 Energy (HL)

IB-Style Exam Practice (SL + HL)

Instructions

Question 1 (SL) — Period, frequency and angular frequency

Question 2 (SL) — Identifying SHM from an – graph

Question 3 (SL) — Mass–spring period and change in parameters

Question 4 (SL) — Simple pendulum and ratios

Question 5 (HL) — Displacement and velocity at a given time

Question 6 (HL) — Energy at a displacement

Key ideas to take away

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Question 2 (SL) — Identifying SHM from an $a$ – $x$ graph