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Simple Harmonic Motion

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Simple Harmonic Motion

The big idea: SHM is a special "back-and-forth" wiggle where the push pulling something back to its resting spot gets stronger the further away it gets — like a spring always yanking harder the more you stretch it.

Quick Overview

Before diving deep, here's the whole chapter in one scan. Come back to this after you've studied everything — if every line makes sense, you're ready.

What counts as SHM

Acceleration ∝ displacement, and always points back toward the centre.

The core equations

a = −ω²x, x = A cos(ωt) or A sin(ωt), v = ±ω√(A² − x²)

Time periods

Pendulum: T = 2π√(l/g). Mass-spring: T = 2π√(m/k)

The graphs

Displacement-time and velocity-time graphs are 90° out of phase.

1. What Actually Makes Something "SHM"?

Loads of things in the world wobble, swing, or bounce back and forth. But not all of them are . SHM is a very specific, mathematically "clean" type of oscillation, and there are exactly two conditions something must satisfy to earn that label:

The acceleration is proportional to the displacement. The further you pull something from its resting position, the harder it accelerates back.
The acceleration is always in the opposite direction to the displacement. It never accelerates away from home — it's always being yanked back.

Think of it like...

A dog on an elastic lead. The further it wanders from you, the harder the lead tugs it back — and it always tugs you, never away. That tug getting stronger with distance, and always pointing home, is the essence of SHM.

1aReal Examples of SHM

Some classic systems that genuinely obey both rules above:

The pendulum of a clock
A mass bouncing on a spring
A vibrating guitar string
Electrons in an alternating current flowing through a wire

1bThe Restoring Force

The thing that actually this acceleration is called the restoring force, F. It's the force always trying to drag the object back to its equilibrium (resting) position, and its size depends directly on how far away the object currently is.

Restoring Force Equation

F = −kx

"The restoring force equals minus a constant, times how far you are from home."

F = restoring force (N)

x = displacement from equilibrium position (m)

k = a constant that depends on the specific system (e.g. spring constant)

the minus sign = tells you the force (and hence acceleration) always points back toward the centre, opposite to the displacement

Notice this looks exactly like Hooke's Law — that's not a coincidence, and it's why a mass on a spring is the classic textbook example of SHM.

Worked Example — Restoring Force

A 200 g toy robot is attached to a pole by a spring with spring constant 90 N m⁻¹, oscillating horizontally.

(a) Force at amplitude position of 5 cm:

Convert: 5 cm = 0.05 m

F = −kx = −(90)(0.05) = −4.5 N

A force of 4.5 N acts on the robot, pulling it back toward equilibrium.

(b) Acceleration at this position:

Convert mass: 200 g = 0.2 kg

Using F = ma → a = F/m = −4.5 / 0.2 = −22.5 m s⁻²

Classic Trap: The Trampoline

A person jumping on a trampoline is NOT SHM! Why? While they're in the air (not touching the trampoline), the restoring force on them is just their constant weight — it doesn't get bigger the higher they jump. Since the force isn't proportional to displacement the whole time, it fails condition 1.

Practice Question 1.1

A mass on a spring has spring constant k = 45 N m⁻¹. When displaced 8 cm from equilibrium, what restoring force acts on it, and in which direction?

Practice Question 1.2

Explain, using the two conditions for SHM, why a ball bouncing on a hard floor (bouncing elastically, over and over) is NOT simple harmonic motion.

2. The Equations of SHM

This is the heart of the chapter — three linked equations that describe acceleration, position, and speed at any moment during the motion. They can look intimidating, but each one is just describing a different "snapshot" of the same wiggle.

2aAcceleration and Displacement

Defining Equation of SHM

a = −ω²x

"Acceleration is minus (a constant squared) times displacement."

a = acceleration (m s⁻²)

ω = angular frequency (rad s⁻¹)

x = displacement (m)

This is basically the maths version of the two conditions from Topic 1 — proportional (ω² is just a constant) and opposite direction (the minus sign). If you graph acceleration against displacement, you get a straight line through the origin sloping , with gradient −ω². The line crosses the x-axis at the amplitude values, −A and +A.

2bDisplacement and Time

Rearranging that acceleration equation (using calculus, which you don't need to reproduce) gives you a formula for exactly where the object is at any given time t. There are two versions, and which one you use depends entirely on at t = 0:

Starts at Maximum Displacement (pulled out and released)

x = A cos(ωt)

Use this when the object begins at its amplitude position (x = A or x = −A) at t = 0.

Starts at Equilibrium (moving through the centre)

x = A sin(ωt)

Use this when the object begins at the equilibrium position (x = 0) at t = 0.

Why cos vs sin?

Cosine starts at its maximum value (cos 0 = 1), which matches an object starting fully displaced. Sine starts at zero (sin 0 = 0), matching an object starting at the centre, mid-swing. Same motion — just a different "starting photo."

Worked Example — Displacement Equation

A 55 g mass on a spring is pulled down 4.3 cm and released at t = 0. It performs SHM with period 0.8 s. Find its displacement at t = 0.3 s.

Step 1: Released from maximum displacement → use x = A cos(ωt)

Step 2: ω = 2π/T = 2π/0.8 = 7.85 rad s⁻¹

Step 3: x = 4.3 cos(7.85 × 0.3) = −3.0 cm (2 s.f.)

The negative sign tells us the mass is now 3.0 cm on the side of equilibrium from where it started.

Calculator Check!

Your calculator MUST be in radians mode for these equations, because ω is calculated in rad s⁻¹, not degrees. This is one of the single most common ways students lose easy marks — always double check before you calculate.

2cSpeed and Displacement

How fast is the object moving at any given displacement x? This equation connects them directly, without needing to know the time:

Speed Equation

v = ±ω√(A² − x²)

"Speed depends on how far you are from the edges (amplitude) squared, minus how far you already are from the centre, squared."

v = speed (m s⁻¹)

± = the value can be positive or negative depending on direction of travel

A = amplitude (m)

x = displacement (m)

Notice: when x = 0 (at equilibrium), v is at its absolute maximum — this makes total sense, since that's where the object has had the most "room to speed up." When x = A (at the amplitude), v = 0 — the object has momentarily stopped to turn around.

Worked Example — Speed Equation

A pendulum oscillates in SHM with amplitude 15 cm and frequency 6.7 Hz. Find its speed at a displacement of 12 cm from equilibrium.

Step 1: A = 0.15 m, x = 0.12 m, f = 6.7 Hz

Step 2: ω = 2πf = 2π × 6.7 = 42.1 rad s⁻¹

Step 3: v = ω√(A² − x²) = 42.1 × √(0.15² − 0.12²)

v = 3.8 m s⁻¹ (2 s.f.)

Practice Question 2.1

A mass oscillates in SHM with amplitude 6 cm and angular frequency 12 rad s⁻¹. It starts at the equilibrium position at t = 0, moving in the positive direction. Write down its displacement equation and find x at t = 0.1 s.

Practice Question 2.2

Using a = −ω²x, calculate the acceleration of an oscillator with ω = 5 rad s⁻¹ when its displacement is −3 cm. Comment on the direction of the acceleration.

3. Period of Common SHM Systems

Two systems come up again and again in exams: the simple pendulum and the mass-spring system. Each has its own formula for the time period T (time for one complete oscillation) — and crucially, these two formulas look similar but use completely different variables, so don't mix them up.

3aThe Simple Pendulum

A simple pendulum is just an object swinging side to side on a string fixed at a point above it.

Period of a Simple Pendulum

T = 2π √(l / g)

l = length of the pendulum (m)

g = gravitational field strength (m s⁻², = 9.81 on Earth)

Notice mass doesn't appear anywhere — a heavy pendulum bob and a light one swing with exactly the same period, as long as the length is the same!

Worked Example — Pendulum Period

A child sits on a swing that is 200 cm long. Find the period of oscillation.

Convert: 200 cm = 2 m

T = 2π√(l/g) = 2π√(2/9.81) = 2.84 s

3bThe Mass-Spring System

An object attached to a spring, oscillating up-down or side-to-side.

Period of a Mass-Spring System

T = 2π √(m / k)

m = mass on the end of the spring (kg)

k = spring constant of the spring (N m⁻¹)

How Would You Actually Observe This?

A classic practical: tie a pencil to a mass hanging on a spring, set it oscillating, and pull a strip of paper sideways underneath it at a steady rate. The pencil traces a curved, periodic wave — a real sine/cosine curve drawn by the physics itself! As energy is lost, you'll notice the amplitude shrinking over time (this is damping, which is covered in a related chapter).

Practice Question 3.1

A spring-mass system has a mass of 0.4 kg and spring constant 25 N m⁻¹. Calculate the period of oscillation.

Practice Question 3.2

A pendulum has a period of 3.2 s on Earth (g = 9.81 m s⁻²). What length is it? What would its period be if taken to the Moon, where g = 1.62 m s⁻²?

4. Reading the Graphs

4aDisplacement-Time Graph

Since undamped SHM is described by sine and cosine functions, the displacement-time graph is always a smooth, repeating wave — a "periodic function." Two things you can read straight off it:

Amplitude A — the maximum value of x reached (the peak height)
Period T — the time taken for one full repeating cycle

The graph's exact shape (whether it looks like sine or cosine) just depends on where the object was at t = 0 — remember from Topic 2b.

4bVelocity-Time Graph

Since velocity is the rate of change of displacement, the velocity-time graph is the of the displacement-time graph at every point. This produces a really important relationship:

Key Insight

The velocity graph is 90° out of phase with the displacement graph

v = Δx / Δt (velocity = gradient of the displacement-time graph)

Practically, this means: wherever the displacement graph crosses zero (equilibrium), the velocity graph is at a peak or trough (maximum speed). And wherever the displacement graph is at a peak or trough (maximum displacement), the velocity graph crosses zero (momentarily stationary, turning around).

Quick Way to Remember

Think of a swing: it moves fastest at the bottom of its arc (equilibrium, x = 0) and is momentarily still at the very top of each side (maximum displacement, x = ±A). That's the whole relationship in one image.

Worked Example — Reading a Graph

A swing is pulled 5 cm and released. Its displacement-time graph shows x starting at 5 cm (t=0), crossing zero around t = 0.2 s, reaching −5 cm around t = 0.4 s, and back to +5 cm at t = 0.8 s. At what time is the velocity of the swing first at its maximum?

Step 1: Velocity is at its maximum when displacement x = 0

Step 2: Reading the graph, x = 0 first occurs at t = 0.2 s

Practice Question 4.1

A displacement-time graph for an oscillator shows a full cycle taking 0.6 s, with a peak displacement of 4 cm. At what times within the first cycle (starting at t=0 at the equilibrium moving positive) is the speed at its maximum, and what is true about the displacement at those moments?

What to Memorise

These are the facts and formulas that need to be instantly recallable — no hesitation, no looking them up mid-exam.

Concept	Formula / Fact
Two conditions for SHM	a ∝ x, and a is always opposite in direction to x
Restoring force	F = −kx
Defining SHM equation	a = −ω²x
Displacement (starts at amplitude)	x = A cos(ωt)
Displacement (starts at equilibrium)	x = A sin(ωt)
Speed at displacement x	v = ±ω√(A² − x²)
Angular frequency	ω = 2π/T = 2πf
Period — simple pendulum	T = 2π√(l/g)
Period — mass-spring system	T = 2π√(m/k)
Max speed occurs at	equilibrium position, x = 0
Max acceleration occurs at	amplitude position, x = ±A (v = 0 there)
Velocity vs displacement graphs	90° out of phase with each other

Concepts Checklist

Tick each one off honestly — only once you could explain it out loud to someone else, no notes.

I can state the two conditions required for motion to be SHM

I can explain why a bouncing ball or trampoline jumper is NOT SHM

I can use F = −kx to calculate restoring force

I can use a = −ω²x and explain what each symbol means

I know when to use x = A cos(ωt) vs x = A sin(ωt)

I can use v = ±ω√(A² − x²) to find speed at any displacement

I can calculate the period of a simple pendulum using T = 2π√(l/g)

I can calculate the period of a mass-spring system using T = 2π√(m/k)

I can read amplitude and period off a displacement-time graph

I understand why velocity and displacement graphs are 90° out of phase

I always set my calculator to radians mode before using ωt in cos/sin

I convert all units to SI (cm→m, g→kg) before substituting into formulas

Exam Tips & Common Mistakes

Radians, not degrees

ω is always in rad s⁻¹. If your calculator is in degrees mode when you compute cos(ωt) or sin(ωt), your answer will be completely wrong. Check this before every SHM calculation involving time.

Always convert to SI units first

Convert cm to m and g to kg before substituting into any equation. Examiners frequently set numbers in cm or g specifically to catch students who forget.

Don't drop the negative sign

Displacement is a vector. If a calculation gives a negative value for x, keep it — it tells you which side of equilibrium the object is on. Dropping it can cost you a mark even if your magnitude is correct.

Cos vs sin — check the starting condition

Always check what the question says happens at t = 0. "Released from rest at a displaced position" → cosine. "Passing through equilibrium" or "starts at the centre" → sine.

Don't confuse the two period formulas

T = 2π√(l/g) is for pendulums (needs length and gravity). T = 2π√(m/k) is for springs (needs mass and spring constant). Mixing them up is a very common slip under exam pressure.

"Explain why X is not SHM" questions

These always want you to reference the restoring force and show it is NOT proportional to displacement throughout the whole motion (like the trampoline example) — a vague answer without mentioning force-displacement proportionality won't get full marks.

Graph-reading questions

When asked to find "the first time velocity is maximum" from a displacement-time graph, look for where the curve crosses the x-axis (x=0) — not where it peaks. Students often mix this up under time pressure.

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