Simple Harmonic Motion

CIE A-Level Physics· 9702 2025 Syllabus 18.1· 15 min read

1. Defining Characteristics of SHM★★☆☆☆⏱ 4 min

📘 Definition

Simple Harmonic Motion

a = -\omega^2 x

A periodic motion that satisfies two core conditions: 1) Acceleration is directly proportional to displacement from a fixed equilibrium position, 2) Acceleration is always directed towards the equilibrium position (opposite direction to displacement).

📐 Worked Example

A ball bounces elastically between two parallel walls, with constant speed between the walls. Is this motion simple harmonic? Explain your answer.

1
Recall the two required conditions for SHM: $a \propto x$ (displacement from equilibrium) and $a$ is directed towards equilibrium.
2
For the bouncing ball, acceleration is zero when the ball is between the walls, regardless of how far it is from the equilibrium midpoint.
3
Acceleration is only non-zero when the ball hits the wall, so it is not proportional to displacement at all points.
4
Conclusion: The motion is not simple harmonic.

Exam tip:

Always check both conditions when asked to confirm if a motion is SHM

2. Kinematic Equations for SHM★★★☆☆⏱ 6 min

If an oscillator starts at maximum displacement ( $x = + A$ ) when $t = 0$ , displacement is given by $x = A cos (ω t)$ . If it starts at equilibrium ( $x = 0$ ) moving in the positive direction when $t = 0$ , displacement is given by $x = A sin (ω t)$ . Differentiating displacement gives velocity, and differentiating velocity gives acceleration, which confirms the defining equation $a = - ω^{2} x$ .

v = \frac{dx}{dt} = -A\omega \sin(\omega t) = \pm \omega\sqrt{A^2 - x^2}

📐 Worked Example

A SHM oscillator has amplitude $0.20 m$ and angular frequency $4.0 rad s^{- 1}$ . It starts at maximum displacement at $t = 0$ . Calculate displacement, velocity and acceleration at $t = 0.50 s$ .

1
Use the displacement equation for starting at maximum displacement:
2
$x = A\cos(\omega t) = 0.20 \cos(4.0 \times 0.50) = 0.20 \cos(2)$
3
Calculate $cos (2) \approx - 0.416$ , so $x \approx 0.20 \times (- 0.416) = - 0.083 m$
4
Velocity is the derivative of displacement:
5
$v = -A\omega \sin(\omega t) = -(0.20 \times 4.0) \sin(2) = -0.80 \times 0.909 = -0.73 \text{ m s}^{-1}$
6
Check acceleration using the defining equation $a = - ω^{2} x$ :
7
$a = -(4.0)^2 (-0.083) = 1.3 \text{ m s}^{-2}$

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

Graphs of displacement, velocity and acceleration against time for SHM are all sinusoidal, with consistent phase differences between them:

Acceleration is 180° ( $π$ radians) out of phase with displacement
Velocity is 90° ( $π /2$ radians) out of phase with displacement, leading displacement
The gradient of a displacement-time graph equals velocity, and the gradient of a velocity-time graph equals acceleration

📐 Worked Example

For an oscillator starting at $x = + A$ at $t = 0$ , sketch displacement-time and acceleration-time graphs, and state their phase difference.

1
Displacement is $x (t) = A cos (ω t)$ , so the graph starts at $+ A$ at $t = 0$ , oscillates between $+ A$ and $- A$ with period $T = 2 π / ω$ .
2
Acceleration is $a (t) = - ω^{2} x (t)$ , so $a (t) = - ω^{2} A cos (ω t)$ .
3
This means the acceleration graph is an inverted version of the displacement graph, starting at $- ω^{2} A$ at $t = 0$ .
4
The phase difference between displacement and acceleration is $π$ radians (180°): they are always out of phase with each other.

4. Common SHM Systems★★☆☆☆⏱ 5 min

Two common examples of SHM tested in CIE exams are horizontal mass-spring systems and small-angle simple pendulum oscillations. Their period formulas are derived from the defining SHM equation:

System	Angular Frequency	Period
Mass-spring (mass $m$ , spring constant $k$ )	$ω = \frac{k}{m}$	$T = 2 π \frac{m}{k}$
Simple pendulum (length $l$ , small angles)	$ω = \frac{g}{l}$	$T = 2 π \frac{l}{g}$

📐 Worked Example

A 0.50 kg mass is attached to a spring of spring constant 20 N m⁻¹, and set into SHM. Calculate the period of oscillation.

1
Use the period formula for a mass-spring SHM system:
2
$T = 2\pi \sqrt{\frac{m}{k}}$
3
Substitute values $m = 0.50 kg$ , $k = 20 N m^{- 1}$ :
4
$T = 2\pi \sqrt{\frac{0.50}{20}} = 2\pi \sqrt{0.025} \approx 2\pi (0.158) \approx 1.0 \text{ s}$

5. Common Pitfalls

Wrong move:

Omitting the negative sign in the defining equation $a = - ω^{2} x$

Why:

The negative sign is required to show acceleration is opposite in direction to displacement, a core condition of SHM

Correct move:

Always include the negative sign in the defining equation, and confirm direction when asked

Wrong move:

Using $sin$ for displacement when starting at maximum displacement (or vice versa)

Why:

This gives the wrong initial value for displacement, leading to incorrect calculations for all other quantities

Correct move:

Always check the initial position at $t = 0$ before selecting the displacement equation

Wrong move:

Assuming all periodic motion is SHM

Why:

Only motions that satisfy the two conditions of SHM are classed as simple harmonic; many periodic motions do not meet this

Correct move:

Always check both defining conditions when asked to confirm if a motion is SHM

Wrong move:

Using the simple pendulum period formula for large angle oscillations

Why:

The formula is derived using the small angle approximation, which does not hold for angles larger than ~10°

Correct move:

Recognize that the formula is only valid for small amplitude pendulum oscillations

Wrong move:

Confusing angular frequency for SHM with angular velocity for circular motion

Why:

While they share the same symbol and units, they describe different physical quantities

Correct move:

Remember for SHM, $ω = 2 π f = 2 π / T$ , a constant describing oscillation speed

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Defining SHM	$a = - ω^{2} x$	Two conditions must be satisfied
Displacement (x = A at t=0)	$x = A cos ω t$
Displacement (x = 0 at t=0)	$x = A sin ω t$
Maximum velocity	$v_{ma x} = ω A$	Occurs at $x = 0$ (equilibrium)
Maximum acceleration	$a_{ma x} = ω^{2} A$	Occurs at maximum displacement
Mass-spring period	$T = 2 π \frac{m}{k}$	Independent of amplitude
Simple pendulum period	$T = 2 π \frac{l}{g}$	Only for small angles <10°
Phase difference (x vs a)	$π$ rad (180°)	Always completely out of phase

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.

2022 · 1
Identify SHM from given motions
2023 · 2
Calculate period of mass-spring SHM
2024 · 1
Phase difference between x and a

What's Next

Now that you have mastered the core definition and kinematics of simple harmonic motion, you can build on this knowledge to explore energy changes in SHM, then damped and forced oscillations, the remaining topics in the CIE A-Level Oscillations unit. SHM is the foundational model for all wave phenomena, so a solid understanding of this sub-topic is critical for later topics including progressive waves, standing waves, and interference. Exam questions frequently combine SHM concepts with energy and circular motion, so practice applying the defining equations and graphical analysis to consolidate your understanding.