Simple Harmonic Motion

IB Physics SL· Unit 3: Wave Behaviour, Topic 1· 45 min read

1. Core Definition and SHM Condition★★☆☆☆⏱ 10 min

📘 Definition

Simple Harmonic Motion

SHM

Periodic oscillatory motion that satisfies the core relationship $a = - ω^{2} x$ , where $a$ is acceleration, $x$ is displacement from equilibrium, and $ω$ is constant angular frequency.

This defining condition separates SHM from other periodic motions (like bouncing balls). A key property of SHM is isochronism: period is independent of amplitude, so changing how far you pull a mass does not change how fast it oscillates.

📐 Worked Example

Displacement of an oscillator is $x = 0.5 sin (3 t)$ . Show that the motion is SHM.

1
To confirm SHM, prove the motion satisfies $a = - ω^{2} x$ :
2
Calculate velocity (first derivative of displacement):
3
$v = \frac{dx}{dt} = 1.5 \cos(3t)$
4
Calculate acceleration (second derivative of displacement):
5
$a = \frac{d^2x}{dt^2} = -4.5 \sin(3t)$
6
Substitute $x = 0.5 sin (3 t) ⟹ sin (3 t) = 2 x$ into acceleration:
7
$a = -4.5 (2x) = -9x$
8
This matches the SHM condition with $ω^{2} = 9$ , so motion is SHM.

2. Equations of Motion for SHM★★★☆☆⏱ 15 min

Starting from the core condition $a = - ω^{2} x$ , we get general displacement equations that depend on initial conditions. Angular frequency is always defined as:

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

If $x = A$ (max displacement) at $t = 0$ : $x = A cos (ω t)$
If $x = 0$ (equilibrium) at $t = 0$ : $x = A sin (ω t)$
Velocity for any displacement $x$ : $v = \pm ω A^{2} - x^{2}$

📐 Worked Example

A SHM oscillator has amplitude 2 cm, period $4 π$ s. At $t = 0$ , $x = 2$ cm. Find velocity at $x = 1$ cm.

1
First calculate angular frequency:
2
$\omega = \frac{2\pi}{T} = \frac{2\pi}{4\pi} = 0.5 \text{ rad s}^{-1}$
3
Use the velocity-displacement relation:
4
$v = \pm \omega \sqrt{A^2 - x^2}$
5
Substitute values $ω = 0.5$ , $A = 2$ , $x = 1$ :
6
$v = \pm 0.5 \sqrt{2^2 - 1^2} = \pm 0.5\sqrt{3} \approx \pm 0.87 \text{ cm s}^{-1}$
7
Unless direction is specified, both positive and negative values are acceptable.

✓ Quick check

Test your understanding:

What is acceleration when displacement $x = A$ (amplitude)?
- $a = 0$
- $a = - ω^{2} A$
- $a = ω A^{2}$
- Cannot be determined
Reveal answer
1 —
When $x = A$ , substitute directly into $a = - ω^{2} x$ to get maximum acceleration opposite displacement.

3. Energy in Undamped SHM★★☆☆☆⏱ 10 min

In undamped SHM (no energy loss to friction), total mechanical energy is conserved. Energy continuously swaps between kinetic energy ( $E_{k}$ ) and potential energy ( $E_{p}$ ):

At $x = \pm A$ (max displacement): $E_{k} = 0$ , $E_{p} = E_{total}$
At $x = 0$ (equilibrium): $E_{p} = 0$ , $E_{k} = E_{total}$

📘 Definition

Total Energy of Undamped SHM

Total energy is constant and equal to $E_{total} = \frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} k A^{2}$

📐 Worked Example

A 0.2 kg mass undergoes SHM with amplitude 0.1 m and angular frequency 4 rad s⁻¹. Calculate total energy.

1
Maximum speed in SHM is $v_{max} = ω A$ :
2
$v_{\text{max}} = 4 \times 0.1 = 0.4 \text{ m s}^{-1}$
3
Total energy equals maximum kinetic energy, since potential is zero at equilibrium:
4
$E_{\text{total}} = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.2 \times 0.4^2 = 0.016 \text{ J}$

4. Common SHM Systems★★★☆☆⏱ 12 min

IB Physics SL regularly tests two standard SHM systems: mass-spring systems and small-angle simple pendulums. Their period formulas are summarized below:

System	Period Formula	SHM Conditions
Horizontal mass-spring	$T = 2 π \frac{m}{k}$	Spring obeys Hooke's law
Vertical mass-spring	$T = 2 π \frac{m}{k}$	Spring obeys Hooke's law (gravity shifts equilibrium only)
Simple pendulum	$T = 2 π \frac{l}{g}$	Displacement angle < 10° (small angle approximation)

📐 Worked Example

A simple pendulum has period 2 s on Earth ( $g = 9.8$ m s⁻²). Calculate its length.

1
Rearrange the pendulum period formula to solve for $l$ :
2
$l = \frac{g T^2}{4 \pi^2}$
3
Substitute values $T = 2$ s, $g = 9.8$ m s⁻²:
4
$l = \frac{9.8 \times (2)^2}{4 \pi^2} = \frac{9.8}{\pi^2} \approx 1.0 \text{ m}$

5. Common Pitfalls

Wrong move:

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

Why:

The negative sign is a core part of the SHM definition, showing acceleration opposes displacement.

Correct move:

Always include the negative sign when stating the SHM condition or writing acceleration equations.

Wrong move:

Claiming period of SHM depends on amplitude

Why:

A defining property of SHM is isochronism: period is independent of amplitude for valid SHM systems.

Correct move:

Use the system-specific period formula, which never includes amplitude as a variable.

Wrong move:

Thinking vertical mass-springs have different periods than horizontal ones

Why:

Gravity only shifts the equilibrium position, it does not change the restoring force or period.

Correct move:

Use $T = 2 π m / k$ for all mass-spring systems, regardless of orientation.

Wrong move:

Using the simple pendulum period formula for large displacements

Why:

The formula only works for SHM, which requires the small angle approximation (< 10°).

Correct move:

Recognize that for angles larger than 10°, the pendulum does not obey SHM and the formula is invalid.

Wrong move:

Confusing total energy with kinetic energy at non-zero displacement

Why:

Total energy is only equal to maximum kinetic energy at equilibrium, not at any other displacement.

Correct move:

Calculate potential energy for any non-zero displacement and add it to kinetic energy to get total energy.

6. Quick Reference Cheatsheet

Concept	Formula
SHM Core Condition	$a = - ω^{2} x$
Angular Frequency	$ω = 2 π f = \frac{2 π}{T}$
Displacement ( $t = 0, x = A$ )	$x = A cos (ω t)$
Displacement ( $t = 0, x = 0$ )	$x = A sin (ω t)$
Velocity	$v = \pm ω A^{2} - x^{2}$
Total SHM Energy	$E_{total} = \frac{1}{2} m ω^{2} A^{2}$
Period: Mass-Spring	$T = 2 π \frac{m}{k}$
Period: Simple Pendulum	$T = 2 π \frac{l}{g}$

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.

2023 · 1
Acceleration vs displacement relationship
2022 · 2
Energy transformations in SHM
2021 · 1
Period of mass-spring system

Going deeper

simulationPHET SHM SimulationInteractive SHM demonstration

What's Next

Simple harmonic motion is the foundational concept for all oscillation and wave topics in IB Physics SL. All wave motion can be modeled as a collection of coupled SHM oscillators, so mastering SHM makes understanding more advanced topics much easier. Next, you will build on this foundation to learn what happens when SHM systems lose energy or are driven by external forces, leading to the important phenomenon of resonance, which is a common exam question in Paper 1 and Paper 2. Explore the following topics to continue building your knowledge of wave behaviour.