C.1 Simple harmonic motion

IB Physics HL· IB Physics 2025 Syllabus: Theme C.1· 15 min read

1. Defining Simple Harmonic Motion★★☆☆☆⏱ 4 min

📘 Definition

Simple Harmonic Motion (SHM)

SHM

A type of periodic motion where the acceleration of the object is directly proportional to its displacement from equilibrium, and always directed opposite to the displacement.

Example:

Mass on a frictionless horizontal spring, simple pendulum oscillating at small angles

This definition is the core test for SHM: any periodic motion that does not satisfy this relationship is not SHM. The relationship is written mathematically as:

a = -\omega^2 x

📐 Worked Example

A motion has acceleration given by $a = - 0.25 x$ . State whether it is SHM, and find the angular frequency.

1
Check against the SHM definition: acceleration is proportional to displacement, with the correct negative sign, so this satisfies the SHM condition.
2
Compare the given equation to the standard SHM form $a = - ω^{2} x$ :
3
$\omega^2 = 0.25$
4
Solve for angular frequency:
5
$\omega = \sqrt{0.25} = 0.5 \text{ rad s}^{-1}$

2. Kinematic Equations of SHM★★★☆☆⏱ 5 min

Solving the SHM differential equation gives two common forms for displacement as a function of time, depending on the starting position of the object at $t = 0$ :

If motion starts at maximum displacement $x = A$ at $t = 0$ : $x (t) = A cos (ω t)$
If motion starts at equilibrium $x = 0$ with positive velocity at $t = 0$ : $x (t) = A sin (ω t)$

Velocity and acceleration are found by differentiating displacement. Maximum velocity is $v_{ma x} = ω A$ and maximum acceleration is $a_{ma x} = ω^{2} A$ .

📐 Worked Example

A mass in SHM has amplitude $A = 0.2 m$ and period $T = 2.0 s$ . It starts at maximum positive displacement at $t = 0$ . Calculate displacement and velocity at $t = 0.5 s$ .

1
First calculate angular frequency:
2
$\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi \text{ rad s}^{-1}$
3
Use the displacement equation for starting at maximum displacement:
4
$x(0.5) = 0.2 \cos(\pi \times 0.5) = 0.2 \cos\left(\frac{\pi}{2}\right) = 0 \text{ m}$
5
Differentiate displacement to get velocity:
6
$v(t) = -A\omega \sin(\omega t)$
7
Substitute values to find velocity:
8
$v(0.5) = -0.2 \times \pi \times \sin\left(\frac{\pi}{2}\right) \approx -0.63 \text{ m s}^{-1}$

3. Energy in Undamped SHM★★☆☆☆⏱ 3 min

In undamped SHM, there are no resistive forces, so total mechanical energy is conserved. Energy is continuously converted between kinetic energy (KE) and potential energy (PE) over one full oscillation.

📘 Definition

Undamped SHM

SHM with no energy loss to resistive forces, so amplitude remains constant over time.

For any undamped SHM system, total energy $E_{t o t a l} = K E + P E = constant$ . Kinetic energy is maximum at equilibrium ( $x = 0$ ) and potential energy is maximum at maximum displacement ( $x = \pm A$ ).

📐 Worked Example

A 0.5 kg mass on a spring has SHM with amplitude 0.1 m and angular frequency 4 rad s⁻¹. Find the maximum kinetic energy of the mass.

1
For a mass-spring system, $ω^{2} = k / m$ , so rearrange to find spring constant $k$ :
2
$k = m \omega^2 = 0.5 \times (4)^2 = 8 \text{ N m}^{-1}$
3
Maximum kinetic energy equals total energy, which equals maximum potential energy at maximum displacement:
4
$KE_{max} = \frac{1}{2} k A^2 = 0.5 \times 8 \times (0.1)^2 = 0.04 \text{ J}$

4. Graphical Representation of SHM★★★☆☆⏱ 3 min

A common exam question asks to interpret or draw graphs of SHM quantities (displacement, velocity, acceleration, energy) against time or displacement. The table below summarises key phase relationships:

Quantity	Phase relative to $x (t)$	Maximum value
Displacement $x$	0 rad	$A$
Velocity $v$	$+ π /2$ rad ahead	$ω A$
Acceleration $a$	$π$ rad out of phase	$ω^{2} A$

📐 Worked Example

Describe the shape of the kinetic energy vs displacement graph for undamped SHM.

1
Use conservation of energy to write KE in terms of displacement:
2
$KE(x) = E_{total} - PE(x) = \frac{1}{2}kA^2 - \frac{1}{2}kx^2$
3
This is a quadratic function in $x$ that opens downwards. It has a maximum value of $\frac{1}{2} k A^{2}$ at $x = 0$ , and equals zero at $x = \pm A$ .
4
The graph is therefore an inverted parabola, symmetric about the equilibrium position $x = 0$ .

5. Common Pitfalls

Wrong move:

Forgetting the negative sign in the SHM definition $a = - ω^{2} x$

Why:

The negative sign defines the restoring direction of acceleration; without it acceleration would push the object further from equilibrium.

Correct move:

Always include the negative sign when writing the SHM defining condition.

Wrong move:

Using the sine form of displacement when motion starts at maximum displacement

Why:

Starting conditions determine the equation form; mixing them up gives wrong values for all time.

Correct move:

Check position at $t = 0$ : use cosine for $x = A$ , sine for $x = 0$ (positive initial velocity).

Wrong move:

Claiming total energy changes with displacement in undamped SHM

Why:

Undamped SHM has no energy loss, only conversion between kinetic and potential energy.

Correct move:

Remember total energy is constant in undamped SHM.

Wrong move:

Confusing angular frequency $ω$ for SHM with angular velocity of circular motion

Why:

While they share the same symbol, $ω$ for SHM describes the rate of oscillation, not rotation.

Correct move:

For any SHM, always use $ω = 2 π / T$ regardless of the system type.

6. Quick Reference Cheatsheet

Concept	Formula	Key Note
SHM Definition	$a = - ω^{2} x$	Acceleration proportional to -displacement
Angular Frequency	$ω = 2 π f = 2 π / T$	Units: rad s⁻¹
Displacement (t=0 at x=A)	$x (t) = A cos (ω t)$	Maximum displacement start
Displacement (t=0 at x=0)	$x (t) = A sin (ω t)$	Equilibrium start
Maximum Velocity	$v_{ma x} = ω A$	Occurs at $x = 0$
Maximum Acceleration	$a_{ma x} = ω^{2} A$	Occurs at $x = \pm A$
Total Energy (undamped)	$E_{t o t a l} = \frac{1}{2} k A^{2} = constant$	Energy conserved

7. Frequently Asked

Is all periodic motion SHM?

No. Only periodic motion that satisfies $a = - ω^{2} x$ is SHM. For example, uniform circular motion is periodic but not SHM.

Why do starting conditions matter for SHM equations?

The form of the displacement equation (sine vs cosine) depends on where the object is at $t = 0$ . Choosing the wrong form will give incorrect results for all time.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
SHM acceleration vs displacement
2024 · Paper 2
Energy calculation for SHM mass-spring
2023 · Paper 1
Phase difference of SHM quantities

Going deeper

SyllabusIB Physics 2025 C.1 SyllabusOfficial topic outline

What's Next

Simple harmonic motion is the foundation for all wave behaviour covered in the rest of Theme C. Mastery of SHM concepts is required to analyse more complex oscillatory systems, including damped and forced oscillations, as well as all travelling wave phenomena that IB Physics HL exams assess heavily. Many exam questions combine SHM with energy conservation and kinematics, so it is critical to be comfortable with the core definitions and problem-solving approaches before moving on to more advanced topics.