Where can I learn Simple harmonic motion for IB Physics?

This free lesson on MarkScheme covers syllabus point C.1 (Simple harmonic motion) for IB Diploma Physics (Standard Level). It includes visual revision, key points, and links to criterion-based practice marking at markscheme.app/ib/courses/physics-sl/c-1-simple-harmonic-motion.

Simple harmonic motion — IB Physics SL

Simple harmonic motion

From the swing of a pendulum to the vibration of a guitar string and the jiggle of atoms in a solid, oscillatory motion is everywhere. Simple harmonic motion is the fundamental model for it — the cleanest possible oscillation, and the building block for describing more complicated vibrations and waves. This lesson defines SHM precisely, links its timing to angular frequency, applies it to the two systems you must know, and follows the energy as it flows between kinetic and potential. As in every calculation topic, the worked examples are laid out exactly the way our engine marks your own practice: the method is on the page, mark by mark.

BY THE END, YOU CAN…

exam checklist — tick these off

1Define simple harmonic motion in terms of a restoring force and acceleration proportional to and opposite the displacement, $a = -\omega^2 x$ .
2Relate angular frequency, period and frequency, and apply the period formulae for the simple pendulum and the mass–spring system.
3Describe how displacement, velocity and acceleration vary sinusoidally with time and how they are related in phase.
4Explain the interchange of kinetic and potential energy in SHM and that, without damping, the total energy is constant.
5Present working so that method marks and error-carried-forward (ECF) protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Rhythm of Repetition

Simple harmonic motion is a special, predictable back-and-forth motion, like a pendulum's swing or a mass on a spring. The key is that the force pulling the object back to the middle grows in step with how far it has strayed — furthest away means strongest pull back.

THINK OF IT LIKE…

Think of pushing a child on a swing. The further you pull them back from the lowest point (equilibrium), the stronger the urge to return to the bottom. As they sweep through the lowest point they are moving fastest; at the very top of the swing they stop for an instant before reversing. This constant trade between position and speed, driven by a restoring force that grows with displacement, is the essence of SHM.

1
Find the equilibrium position — the point of zero net force. Every displacement is measured from here.
2
Check the restoring force. For SHM it must point back towards equilibrium and be directly proportional to the displacement, giving $a = -\omega^2 x$ .
3
Read off the timing. Angular frequency $\omega$ sets the period $T = 2\pi/\omega$ and frequency $f$ ; for a pendulum or spring these depend on the system, not the amplitude.
4
Track the energy. Without damping the total stays constant, swapping between kinetic energy (maximum at the centre) and potential energy (maximum at the extremes).

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure PhET simulation1 / 4

Loading simulation…

Fetching from PhET…

Masses and Springs

Time the oscillations with the stopwatch, then hang a heavier mass: the period grows with √m but does not change with amplitude.

If the embed does not load, use Open on PhET.

Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

Step 1

Find the equilibrium position — the point of zero net force. Every displacement is measured from here.

34 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

34 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

3JCN PhysicsStart here · 19702 17.1 · 9702 17.2 · IB C.1
Simple Harmonic Motion (SHM)
Set mass, spring constant and amplitude; watch x, v, a graphs and the energy exchange of a block on a spring
Why this one: Watch x, v and a on one time axis: v leads x by a quarter cycle and a is always opposite x.
Try this
1. Set mass, spring constant and amplitude and read the period from the x graph.
2. Double the amplitude and compare the period and the peak velocity.
3. Compare the x, v and a graphs: note where each crosses zero and peaks.
4. Watch the KE and PE exchange across one cycle.
Look for a is proportional to minus x, v peaks at x = 0, and the period is independent of amplitude.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 29702 17.1 · 9702 17.2 · 9702 12.1
Circular Motion & SHM Relationship
Run uniform circular motion beside its projection to see SHM as a shadow of circular motion
Why this one: The shadow of uniform circular motion is SHM, so omega means the same thing in both.
Try this
1. Run the circular motion and watch its projection move alongside.
2. Compare the projection's displacement with the angle turned.
Look for The projection of uniform circular motion is SHM with x = r cos ωt, so ω of the SHM equals the angular speed.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 39702 17.1 · 9702 17.2 · IB C.1
Simple Pendulum
Set length and amplitude of a simple pendulum; read the period and compare with the formula
Why this one: Change the length, read the period and check T = 2-pi-root(l/g); amplitude and mass do nothing.
Try this
1. Set a length and amplitude and read the period.
2. Quadruple the length and compare the period with the formula.
3. Change the amplitude only and compare.
Look for T = 2π√(L/g): the period doubles when the length quadruples and is independent of small amplitudes.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
PhETStart here · 49702 17.1 · IB C.1
Pendulum Lab
Swing a pendulum; vary length, mass, gravity and amplitude, with a period timer and energy graph.
Why this one: Open the energy graph and watch kinetic and potential trade places while the total stays flat.
Try this
1. Set L = 1 m, small angle — time 10 swings; then try 2 m.
2. Change the mass — does T change at all?
3. Set the angle to 60° — notice T is no longer quite what the formula gives.
Look for T = 2π√(L/g): independent of mass and (for small angles) amplitude.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
3JCN PhysicsStart here · 59702 17.1 · 9702 17.2 · 9702 1.3
SHM for Lab
Timed mass-spring oscillator for lab: change mass and k and measure the period
Why this one: Time ten oscillations for several masses and test T-squared against m to find k.
Try this
1. Set a mass and k, time ten oscillations and divide for the period.
2. Double the mass and measure the period again.
3. Double k instead and measure the period again.
Look for T = 2π√(m/k): quadrupling the mass doubles the period.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

More simulations29 more on this topic — core ones first

oPhysicsCore9702 5.1 · 9702 5.2 · 9702 17.2
Conservation of Mechanical Energy: Mass on a Vertical Spring
Mass oscillating on a vertical spring with live KE / GPE / EPE bar graphs; adjust mass and spring constant
Try this
1. Watch the three bars through one full oscillation.
2. Increase the mass and compare the bar heights.
3. Increase the spring constant and compare the period.
Look for KE, GPE and EPE trade against each other while their total stays constant.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · 9702 7.1 · IB C.1
Simple Harmonic Motion, Circular Motion, and Transverse Waves
Link SHM, uniform circular motion and a transverse wave side by side with sliders and checkboxes
Try this
1. Use the checkboxes to show SHM and circular motion together.
2. Change the frequency slider and watch all three panels.
3. Show the transverse wave and compare one particle with the SHM.
Look for The SHM is the projection of circular motion, and each wave particle performs the same SHM with a phase lag.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · 9702 17.2 · IB C.1
Simple Harmonic Motion: Mass on a Spring
Mass on a horizontal spring: set initial displacement, mass and k; step through motion with x, v, a graphs
Try this
1. Set an initial displacement and step through one period.
2. Double the mass and compare the period.
3. Double k instead and compare again.
Look for Acceleration is always opposite to displacement, and the period grows with mass and shrinks with spring constant.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Oscillation Graphs Quiz
Quiz: match or interpret displacement/velocity/acceleration graphs of an oscillator
Try this
1. Work through one question and check the answer.
2. Identify which graph is the gradient of which.
3. Try another question.
Look for Velocity is zero where displacement is largest, and the acceleration graph is the displacement graph inverted.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Simple Harmonic Motion Tutorial
Step-by-step tutorial on SHM with embedded interactive panels (Next/Back navigation)
Try this
1. Step through the tutorial with Next.
2. Use the interactive panel on each page before moving on.
3. Go Back to recheck a definition.
Look for Acceleration in SHM is proportional to displacement and directed towards equilibrium.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Oscillations
SHM related to uniform circular motion with x-t, v-t and a-t graphs toggled by checkbox
Try this
1. Toggle the x-t graph and watch it against the circular motion.
2. Add the v-t graph and compare its phase.
3. Add the a-t graph.
Look for Velocity leads displacement by a quarter cycle and acceleration is in antiphase with displacement.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)

Key formulas

Tap any symbol to reveal exactly what it means and its units.

KEY RELATION

$a = -\omega^2 x$

KEY RELATION

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

KEY RELATION

$\text{Simple pendulum:} \quad T = 2\pi\sqrt{\dfrac{l}{g}}$

KEY RELATION

$x = x_0\cos(\omega t), \quad v = -\omega x_0\sin(\omega t), \quad a = -\omega^2 x_0\cos(\omega t) = -\omega^2 x$

KEY RELATION

$E_{total} = E_k + E_p = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \text{constant} = \tfrac{1}{2}kx_0^2 = \tfrac{1}{2}m\omega^2 x_0^2$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Defining simple harmonic motion

For an object to undergo SHM, two things must be true. First, it oscillates about a fixed equilibrium position — the point where it would rest undisturbed, because the net force there is zero. Second, whenever it is displaced, it feels a restoring force that is directly proportional to the displacement and always directed back towards equilibrium. The further you pull it away, the harder it is pulled back. Together these give the single defining condition of SHM: the acceleration is proportional to the displacement and points in the opposite direction.

$a = -\omega^2 x$

This is why a constant force (like a steady push) or a force that does not point back to a centre cannot produce SHM. It is the proportional, restoring nature of the force — captured entirely by $a = -\omega^2 x$ — that makes the motion repeat with a fixed rhythm.

The minus sign is the whole point: the acceleration $a$ is always opposite to the displacement $x$ , so the force is restoring — it drives the object back to the centre.
$\omega^2$ is the constant of proportionality. $\omega$ is the angular frequency, a fixed property of the particular oscillating system.
Force follows the same rule. Since $F = ma$ , the restoring force is $F = -m\omega^2 x$ , i.e. $F \propto -x$ — the test for whether a motion is simple harmonic at all.

Not clear?

Angular frequency, period and frequency

Three quantities describe the timing of an oscillation, and they are all connected. The period $T$ is the time for one complete oscillation (in seconds). The frequency $f$ is the number of oscillations per second (in hertz, Hz), so $f = 1/T$ . The angular frequency $\omega$ measures the rate of oscillation in radians per second and ties directly to the defining equation. All three are linked by a single relationship.

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

Because $\omega$ appears in $a = -\omega^2 x$ , knowing the period of an oscillator immediately tells you its acceleration at any displacement, and vice versa. Everything about the timing of the motion is carried by this one constant.

Not clear?

The two systems: pendulum and mass–spring

Two systems dominate SHM problems, and each has a period formula you must be able to use. A simple pendulum — a small bob on a light string — swings with SHM as long as the angle stays small. A mass on a spring oscillates with SHM because the spring supplies a restoring force proportional to extension (Hooke's law). Notice what each period depends on, and — just as importantly — what it does not.

$\text{Simple pendulum:} \quad T = 2\pi\sqrt{\dfrac{l}{g}}$ $\text{Mass-spring:} \quad T = 2\pi\sqrt{\dfrac{m}{k}}$

Pendulum: the period depends only on the length $l$ and the gravitational field strength $g$ — not on the mass of the bob and not (for small swings) on the amplitude.
Mass–spring: the period grows with mass $m$ and shrinks with spring stiffness $k$ ; a heavier mass or a softer spring oscillates more slowly.
Neither depends on amplitude. This isochronism — same period whatever the swing size — is exactly what makes oscillators useful as clocks.
Link to $\omega$ : comparing with $\omega = 2\pi/T$ gives $\omega = \sqrt{g/l}$ for the pendulum and $\omega = \sqrt{k/m}$ for the spring.

Not clear?

How displacement, velocity and acceleration vary

In SHM the displacement traces out a sinusoidal curve in time — a sine or cosine, depending only on where you start the clock. If timing begins when the object is at maximum displacement (the amplitude $x_0$ ), the displacement follows a cosine, the velocity follows a negative sine, and the acceleration follows a negative cosine. You do not need to derive these at SL, but you must know how the three quantities relate in size and in phase.

$x = x_0\cos(\omega t), \quad v = -\omega x_0\sin(\omega t), \quad a = -\omega^2 x_0\cos(\omega t) = -\omega^2 x$

At the extremes ( $x = \pm x_0$ ): the speed is zero and the acceleration is maximum, $a_{max} = \omega^2 x_0$ . The object is momentarily at rest as it turns around.
At the centre ( $x = 0$ ): the acceleration is zero and the speed is maximum, $v_{max} = \omega x_0$ . This is where the object moves fastest.
Phase relationships: displacement and acceleration are exactly out of step (opposite signs, peaking together), while velocity peaks a quarter-cycle away from both, as the object passes through the centre.
Velocity from displacement: at any point, $v = \pm\omega\sqrt{x_0^2 - x^2}$ , so speed falls smoothly from $v_{max}$ at the centre to zero at the extremes.

Not clear?

Energy in simple harmonic motion

In an ideal, undamped oscillator the total mechanical energy is constant; it only shuttles between kinetic energy $E_k$ and potential energy $E_p$ . At the extremes the object is momentarily at rest, so all the energy is potential. As it accelerates towards the centre that potential energy converts into kinetic energy, reaching all-kinetic at the equilibrium position, where the speed is greatest. The cycle then reverses. Because the total is fixed by the amplitude, $E_{total} \propto x_0^2$ : doubling the amplitude quadruples the energy.

$E_{total} = E_k + E_p = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \text{constant} = \tfrac{1}{2}kx_0^2 = \tfrac{1}{2}m\omega^2 x_0^2$

Kinetic energy: maximum at the centre ( $x = 0$ ); zero at the extremes ( $x = \pm x_0$ ).
Potential energy: zero at the centre; maximum at the extremes.
Total energy: constant throughout the motion (no damping), and proportional to the square of the amplitude, $x_0^2$ .
Damping (friction, air resistance) removes energy, so a real oscillator's amplitude and total energy decay over time.

Not clear?

Common mistakes examiners penalise

Forgetting the defining condition — SHM requires a restoring force proportional to displacement and directed towards equilibrium ( $F \propto -x$ ). A constant force, or a force that does not point back to the centre, is not SHM, however 'wobbly' the motion looks.
Dropping or misreading the minus sign in $a = -\omega^2 x$ — the acceleration is always opposite to the displacement. The sign shows direction (restoring), not merely 'slowing down'.
Swapping where speed and acceleration peak — speed is maximum at the centre and zero at the extremes; acceleration is maximum at the extremes and zero at the centre. Mixing these up costs marks in both graph and calculation questions.
Thinking the period depends on amplitude — for a spring, and a pendulum at small angles, the period is independent of amplitude. A bigger swing changes the energy and the top speed, not the timing.
Assuming total energy changes during the motion — without damping the total is constant; only the split between kinetic and potential energy changes. Don't say energy is 'lost' at the extremes — it is all potential there.
Confusing pendulum and spring variables — a pendulum period depends on length and $g$ (not mass); a spring period depends on mass and $k$ (not $g$ ). Check which formula the system needs before substituting.
Forgetting to convert centimetres to metres, or leaving the calculator in degrees — convert every length to metres before substituting, carry extra figures through the working, and always attach the correct unit to the final answer.

Not clear?

Model answer — marked the way our engine marks it

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on does not have to cost you the marks that follow. But that protection only exists if your method is written down. Study how each mark below is earned by a specific line in a two-part SHM calculation.

Not clear?

Where this leads

SHM is the seed of wave physics. The sinusoidal displacement of a single oscillator, repeated across many connected particles, is exactly what a travelling wave is — so the definitions of amplitude, period, frequency and phase carry straight into the wave topics that follow in unit C. The energy picture reappears too: standing waves, resonance and damping all build on the idea of energy stored and exchanged in an oscillator. Master the habit here — check the defining condition, read the timing from $\omega$ , keep speed and acceleration at the right places, and follow the energy — and wave behaviour becomes variations on a motion you already own.

Not clear?

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

reveal slowly — mark each step

EXAMPLE 1

A simple pendulum has a length of 0.65 m. Taking $g = 9.81$ m s⁻², calculate (a) its period of oscillation and (b) its frequency. [3]

1
List what you have. $l = 0.65$ m, $g = 9.81$ m s⁻².

EXAMPLE 2

A trolley oscillates with SHM of amplitude 12 cm and period 0.80 s. Calculate (a) the angular frequency, (b) the maximum speed of the trolley and (c) the magnitude of its acceleration when the displacement is 6.0 cm. [5]

1
List what you have. $x_0 = 12$ cm $= 0.12$ m, $T = 0.80$ s.

EXAMPLE 3

A 0.30 kg mass oscillates on a spring of spring constant $k = 120$ N m⁻¹ with an amplitude of 5.0 cm. Calculate (a) the total energy of the oscillator and (b) the speed of the mass when its displacement is 3.0 cm from equilibrium. [4]

1
List what you have. $m = 0.30$ kg, $k = 120$ N m⁻¹, $x_0 = 5.0$ cm $= 0.050$ m.

EXAMPLE 4

A mass of 0.20 kg on a spring of spring constant 50 N m⁻¹ oscillates with SHM. Calculate the period and the maximum speed if the amplitude is 4.0 cm. [4]

1
Model answer — full working.

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEASimple harmonic motion

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Key terms for this topic — skim now; the Check step will test them.

Definition of simple harmonic motion: Oscillation in which the acceleration is directly proportional to the displacement from a fixed equilibrium position and always directed towards it. Equivalently, the restoring force obeys $F \propto -x$ .
The defining equation of SHM: $a = -\omega^2 x$ . The acceleration is proportional to displacement $x$ and the minus sign makes it point opposite to $x$ — back towards equilibrium. $\omega^2$ is the constant of proportionality.
Angular frequency $\omega$: The rate of oscillation in rad s⁻¹. Linked to period and frequency by $\omega = \dfrac{2\pi}{T} = 2\pi f$ . It is a scalar characteristic of the oscillating system.
Period vs frequency: Period $T$ is the time for one complete oscillation (seconds); frequency $f$ is the number of oscillations per second (Hz). They are reciprocals: $f = 1/T$ .
Period of a simple pendulum: $T = 2\pi\sqrt{\dfrac{l}{g}}$ , where $l$ is the length and $g$ is the gravitational field strength. The period depends on length and $g$ only — not on the mass of the bob nor (for small swings) on the amplitude.
Period of a mass–spring system: $T = 2\pi\sqrt{\dfrac{m}{k}}$ , where $m$ is the mass and $k$ is the spring constant. A heavier mass or a softer spring gives a longer period.
Phase of displacement, velocity and acceleration: Displacement and acceleration are exactly out of step ( $a$ is largest and opposite when $x$ is largest); velocity is a quarter-cycle out of phase with both, peaking as the object passes through the centre.
Energy interchange in SHM: Kinetic energy is maximum at the centre and zero at the extremes; potential energy is zero at the centre and maximum at the extremes. Without damping the total mechanical energy is constant.
Total energy of an SHM oscillator: $E_{total} = \tfrac{1}{2}m\omega^2 x_0^2$ ; for a spring this equals $\tfrac{1}{2}kx_0^2$ . The total energy is proportional to the square of the amplitude $x_0$ .

09Remember

Name it

Read the meaning, then pick which of this lesson’s terms it describes. Miss one and you see what your choice really means.

1 of 6 · set 1 of 2

Which of these is being described?

Oscillation in which the acceleration is directly proportional to the displacement from a fixed equilibrium position and always directed towards it. Equivalently, the restoring force obeys $F \propto -x$ .

Choose the term this describes.

10Apply

Quick check

Write your answer first, then compare it with the model one — the gap is what you would have lost.

0 of 5 answered

11Understand

Teach it back

If you can explain it simply, you own it — gaps here are marks you’d lose.

Teach it back

Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.

Your explanation — no notes

12Remember

Revision flashcards

Guess first, then flip — retrieval beats re-reading.

Your guess — write it before you look

1 / 14 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

close the tab only after these stick

The minus sign is the whole point: the acceleration $a$ is always opposite to the displacement $x$ , so the force is restoring — it drives the object back to the centre.
$\omega^2$ is the constant of proportionality. $\omega$ is the angular frequency, a fixed property of the particular oscillating system.
Force follows the same rule. Since $F = ma$ , the restoring force is $F = -m\omega^2 x$ , i.e. $F \propto -x$ — the test for whether a motion is simple harmonic at all.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 2 calculation marked: solve an SHM problem with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

IB Physics Data Booklet↗

Frequently asked

What actually makes a motion 'simple harmonic'?

Why is there a negative sign in

a = -\omega^2 x

Does the period of a pendulum or spring depend on how far it swings?

Where is the object fastest, and where does it accelerate most?

Does the total energy of an SHM system stay constant?

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 calculation marked: solve an SHM problem with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUSB.5 Current and circuits NEXT →C.2 Wave model

⚛

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Simple harmonic motion

BY THE END, YOU CAN…

exam checklist — tick these off

1Define simple harmonic motion in terms of a restoring force and acceleration proportional to and opposite the displacement, $a = -\omega^2 x$ .
2Relate angular frequency, period and frequency, and apply the period formulae for the simple pendulum and the mass–spring system.
3Describe how displacement, velocity and acceleration vary sinusoidally with time and how they are related in phase.
4Explain the interchange of kinetic and potential energy in SHM and that, without damping, the total energy is constant.
5Present working so that method marks and error-carried-forward (ECF) protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Rhythm of Repetition

THINK OF IT LIKE…

1
Find the equilibrium position — the point of zero net force. Every displacement is measured from here.
2
Check the restoring force. For SHM it must point back towards equilibrium and be directly proportional to the displacement, giving $a = -\omega^2 x$ .
3
Read off the timing. Angular frequency $\omega$ sets the period $T = 2\pi/\omega$ and frequency $f$ ; for a pendulum or spring these depend on the system, not the amplitude.
4
Track the energy. Without damping the total stays constant, swapping between kinetic energy (maximum at the centre) and potential energy (maximum at the extremes).

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure PhET simulation1 / 4

Loading simulation…

Fetching from PhET…

Masses and Springs

Time the oscillations with the stopwatch, then hang a heavier mass: the period grows with √m but does not change with amplitude.

If the embed does not load, use Open on PhET.

Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

Step 1

Find the equilibrium position — the point of zero net force. Every displacement is measured from here.

34 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

34 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

3JCN PhysicsStart here · 19702 17.1 · 9702 17.2 · IB C.1
Simple Harmonic Motion (SHM)
Set mass, spring constant and amplitude; watch x, v, a graphs and the energy exchange of a block on a spring
Why this one: Watch x, v and a on one time axis: v leads x by a quarter cycle and a is always opposite x.
Try this
1. Set mass, spring constant and amplitude and read the period from the x graph.
2. Double the amplitude and compare the period and the peak velocity.
3. Compare the x, v and a graphs: note where each crosses zero and peaks.
4. Watch the KE and PE exchange across one cycle.
Look for a is proportional to minus x, v peaks at x = 0, and the period is independent of amplitude.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 29702 17.1 · 9702 17.2 · 9702 12.1
Circular Motion & SHM Relationship
Run uniform circular motion beside its projection to see SHM as a shadow of circular motion
Why this one: The shadow of uniform circular motion is SHM, so omega means the same thing in both.
Try this
1. Run the circular motion and watch its projection move alongside.
2. Compare the projection's displacement with the angle turned.
Look for The projection of uniform circular motion is SHM with x = r cos ωt, so ω of the SHM equals the angular speed.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
3JCN PhysicsStart here · 39702 17.1 · 9702 17.2 · IB C.1
Simple Pendulum
Set length and amplitude of a simple pendulum; read the period and compare with the formula
Why this one: Change the length, read the period and check T = 2-pi-root(l/g); amplitude and mass do nothing.
Try this
1. Set a length and amplitude and read the period.
2. Quadruple the length and compare the period with the formula.
3. Change the amplitude only and compare.
Look for T = 2π√(L/g): the period doubles when the length quadruples and is independent of small amplitudes.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
PhETStart here · 49702 17.1 · IB C.1
Pendulum Lab
Swing a pendulum; vary length, mass, gravity and amplitude, with a period timer and energy graph.
Why this one: Open the energy graph and watch kinetic and potential trade places while the total stays flat.
Try this
1. Set L = 1 m, small angle — time 10 swings; then try 2 m.
2. Change the mass — does T change at all?
3. Set the angle to 60° — notice T is no longer quite what the formula gives.
Look for T = 2π√(L/g): independent of mass and (for small angles) amplitude.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
3JCN PhysicsStart here · 59702 17.1 · 9702 17.2 · 9702 1.3
SHM for Lab
Timed mass-spring oscillator for lab: change mass and k and measure the period
Why this one: Time ten oscillations for several masses and test T-squared against m to find k.
Try this
1. Set a mass and k, time ten oscillations and divide for the period.
2. Double the mass and measure the period again.
3. Double k instead and measure the period again.
Look for T = 2π√(m/k): quadrupling the mass doubles the period.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0

More simulations29 more on this topic — core ones first

oPhysicsCore9702 5.1 · 9702 5.2 · 9702 17.2
Conservation of Mechanical Energy: Mass on a Vertical Spring
Mass oscillating on a vertical spring with live KE / GPE / EPE bar graphs; adjust mass and spring constant
Try this
1. Watch the three bars through one full oscillation.
2. Increase the mass and compare the bar heights.
3. Increase the spring constant and compare the period.
Look for KE, GPE and EPE trade against each other while their total stays constant.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · 9702 7.1 · IB C.1
Simple Harmonic Motion, Circular Motion, and Transverse Waves
Link SHM, uniform circular motion and a transverse wave side by side with sliders and checkboxes
Try this
1. Use the checkboxes to show SHM and circular motion together.
2. Change the frequency slider and watch all three panels.
3. Show the transverse wave and compare one particle with the SHM.
Look for The SHM is the projection of circular motion, and each wave particle performs the same SHM with a phase lag.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · 9702 17.2 · IB C.1
Simple Harmonic Motion: Mass on a Spring
Mass on a horizontal spring: set initial displacement, mass and k; step through motion with x, v, a graphs
Try this
1. Set an initial displacement and step through one period.
2. Double the mass and compare the period.
3. Double k instead and compare again.
Look for Acceleration is always opposite to displacement, and the period grows with mass and shrinks with spring constant.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Oscillation Graphs Quiz
Quiz: match or interpret displacement/velocity/acceleration graphs of an oscillator
Try this
1. Work through one question and check the answer.
2. Identify which graph is the gradient of which.
3. Try another question.
Look for Velocity is zero where displacement is largest, and the acceleration graph is the displacement graph inverted.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Simple Harmonic Motion Tutorial
Step-by-step tutorial on SHM with embedded interactive panels (Next/Back navigation)
Try this
1. Step through the tutorial with Next.
2. Use the interactive panel on each page before moving on.
3. Go Back to recheck a definition.
Look for Acceleration in SHM is proportional to displacement and directed towards equilibrium.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
oPhysicsCore9702 17.1 · IB C.1
Oscillations
SHM related to uniform circular motion with x-t, v-t and a-t graphs toggled by checkbox
Try this
1. Toggle the x-t graph and watch it against the circular motion.
2. Add the v-t graph and compare its phase.
3. Add the a-t graph.
Look for Velocity leads displacement by a quarter cycle and acceleration is in antiphase with displacement.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)

Key formulas

Tap any symbol to reveal exactly what it means and its units.

KEY RELATION

$a = -\omega^2 x$

KEY RELATION

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

KEY RELATION

$\text{Simple pendulum:} \quad T = 2\pi\sqrt{\dfrac{l}{g}}$

KEY RELATION

$x = x_0\cos(\omega t), \quad v = -\omega x_0\sin(\omega t), \quad a = -\omega^2 x_0\cos(\omega t) = -\omega^2 x$

KEY RELATION

$E_{total} = E_k + E_p = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \text{constant} = \tfrac{1}{2}kx_0^2 = \tfrac{1}{2}m\omega^2 x_0^2$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Defining simple harmonic motion

$a = -\omega^2 x$

The minus sign is the whole point: the acceleration $a$ is always opposite to the displacement $x$ , so the force is restoring — it drives the object back to the centre.
$\omega^2$ is the constant of proportionality. $\omega$ is the angular frequency, a fixed property of the particular oscillating system.
Force follows the same rule. Since $F = ma$ , the restoring force is $F = -m\omega^2 x$ , i.e. $F \propto -x$ — the test for whether a motion is simple harmonic at all.

Not clear?

Angular frequency, period and frequency

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

Not clear?

The two systems: pendulum and mass–spring

$\text{Simple pendulum:} \quad T = 2\pi\sqrt{\dfrac{l}{g}}$ $\text{Mass-spring:} \quad T = 2\pi\sqrt{\dfrac{m}{k}}$

Pendulum: the period depends only on the length $l$ and the gravitational field strength $g$ — not on the mass of the bob and not (for small swings) on the amplitude.
Mass–spring: the period grows with mass $m$ and shrinks with spring stiffness $k$ ; a heavier mass or a softer spring oscillates more slowly.
Neither depends on amplitude. This isochronism — same period whatever the swing size — is exactly what makes oscillators useful as clocks.
Link to $\omega$ : comparing with $\omega = 2\pi/T$ gives $\omega = \sqrt{g/l}$ for the pendulum and $\omega = \sqrt{k/m}$ for the spring.

Not clear?

How displacement, velocity and acceleration vary

$x = x_0\cos(\omega t), \quad v = -\omega x_0\sin(\omega t), \quad a = -\omega^2 x_0\cos(\omega t) = -\omega^2 x$

At the extremes ( $x = \pm x_0$ ): the speed is zero and the acceleration is maximum, $a_{max} = \omega^2 x_0$ . The object is momentarily at rest as it turns around.
At the centre ( $x = 0$ ): the acceleration is zero and the speed is maximum, $v_{max} = \omega x_0$ . This is where the object moves fastest.
Phase relationships: displacement and acceleration are exactly out of step (opposite signs, peaking together), while velocity peaks a quarter-cycle away from both, as the object passes through the centre.
Velocity from displacement: at any point, $v = \pm\omega\sqrt{x_0^2 - x^2}$ , so speed falls smoothly from $v_{max}$ at the centre to zero at the extremes.

Not clear?

Energy in simple harmonic motion

$E_{total} = E_k + E_p = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \text{constant} = \tfrac{1}{2}kx_0^2 = \tfrac{1}{2}m\omega^2 x_0^2$

Kinetic energy: maximum at the centre ( $x = 0$ ); zero at the extremes ( $x = \pm x_0$ ).
Potential energy: zero at the centre; maximum at the extremes.
Total energy: constant throughout the motion (no damping), and proportional to the square of the amplitude, $x_0^2$ .
Damping (friction, air resistance) removes energy, so a real oscillator's amplitude and total energy decay over time.

Not clear?

Common mistakes examiners penalise

Forgetting the defining condition — SHM requires a restoring force proportional to displacement and directed towards equilibrium ( $F \propto -x$ ). A constant force, or a force that does not point back to the centre, is not SHM, however 'wobbly' the motion looks.
Dropping or misreading the minus sign in $a = -\omega^2 x$ — the acceleration is always opposite to the displacement. The sign shows direction (restoring), not merely 'slowing down'.
Swapping where speed and acceleration peak — speed is maximum at the centre and zero at the extremes; acceleration is maximum at the extremes and zero at the centre. Mixing these up costs marks in both graph and calculation questions.
Thinking the period depends on amplitude — for a spring, and a pendulum at small angles, the period is independent of amplitude. A bigger swing changes the energy and the top speed, not the timing.
Assuming total energy changes during the motion — without damping the total is constant; only the split between kinetic and potential energy changes. Don't say energy is 'lost' at the extremes — it is all potential there.
Confusing pendulum and spring variables — a pendulum period depends on length and $g$ (not mass); a spring period depends on mass and $k$ (not $g$ ). Check which formula the system needs before substituting.
Forgetting to convert centimetres to metres, or leaving the calculator in degrees — convert every length to metres before substituting, carry extra figures through the working, and always attach the correct unit to the final answer.

Not clear?

Model answer — marked the way our engine marks it

Not clear?

Where this leads

Not clear?

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

reveal slowly — mark each step

EXAMPLE 1

A simple pendulum has a length of 0.65 m. Taking $g = 9.81$ m s⁻², calculate (a) its period of oscillation and (b) its frequency. [3]

1
List what you have. $l = 0.65$ m, $g = 9.81$ m s⁻².

EXAMPLE 2

1
List what you have. $x_0 = 12$ cm $= 0.12$ m, $T = 0.80$ s.

EXAMPLE 3

1
List what you have. $m = 0.30$ kg, $k = 120$ N m⁻¹, $x_0 = 5.0$ cm $= 0.050$ m.

EXAMPLE 4

A mass of 0.20 kg on a spring of spring constant 50 N m⁻¹ oscillates with SHM. Calculate the period and the maximum speed if the amplitude is 4.0 cm. [4]

1
Model answer — full working.

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEASimple harmonic motion

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Key terms for this topic — skim now; the Check step will test them.

Definition of simple harmonic motion: Oscillation in which the acceleration is directly proportional to the displacement from a fixed equilibrium position and always directed towards it. Equivalently, the restoring force obeys $F \propto -x$ .
The defining equation of SHM: $a = -\omega^2 x$ . The acceleration is proportional to displacement $x$ and the minus sign makes it point opposite to $x$ — back towards equilibrium. $\omega^2$ is the constant of proportionality.
Angular frequency $\omega$: The rate of oscillation in rad s⁻¹. Linked to period and frequency by $\omega = \dfrac{2\pi}{T} = 2\pi f$ . It is a scalar characteristic of the oscillating system.
Period vs frequency: Period $T$ is the time for one complete oscillation (seconds); frequency $f$ is the number of oscillations per second (Hz). They are reciprocals: $f = 1/T$ .
Period of a simple pendulum: $T = 2\pi\sqrt{\dfrac{l}{g}}$ , where $l$ is the length and $g$ is the gravitational field strength. The period depends on length and $g$ only — not on the mass of the bob nor (for small swings) on the amplitude.
Period of a mass–spring system: $T = 2\pi\sqrt{\dfrac{m}{k}}$ , where $m$ is the mass and $k$ is the spring constant. A heavier mass or a softer spring gives a longer period.
Phase of displacement, velocity and acceleration: Displacement and acceleration are exactly out of step ( $a$ is largest and opposite when $x$ is largest); velocity is a quarter-cycle out of phase with both, peaking as the object passes through the centre.
Energy interchange in SHM: Kinetic energy is maximum at the centre and zero at the extremes; potential energy is zero at the centre and maximum at the extremes. Without damping the total mechanical energy is constant.
Total energy of an SHM oscillator: $E_{total} = \tfrac{1}{2}m\omega^2 x_0^2$ ; for a spring this equals $\tfrac{1}{2}kx_0^2$ . The total energy is proportional to the square of the amplitude $x_0$ .

09Remember

Name it

Read the meaning, then pick which of this lesson’s terms it describes. Miss one and you see what your choice really means.

1 of 6 · set 1 of 2

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Choose the term this describes.

10Apply

Quick check

Write your answer first, then compare it with the model one — the gap is what you would have lost.

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11Understand

Teach it back

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Teach it back

Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.

Your explanation — no notes

12Remember

Revision flashcards

Guess first, then flip — retrieval beats re-reading.

Your guess — write it before you look

1 / 14 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

close the tab only after these stick

The minus sign is the whole point: the acceleration $a$ is always opposite to the displacement $x$ , so the force is restoring — it drives the object back to the centre.
$\omega^2$ is the constant of proportionality. $\omega$ is the angular frequency, a fixed property of the particular oscillating system.
Force follows the same rule. Since $F = ma$ , the restoring force is $F = -m\omega^2 x$ , i.e. $F \propto -x$ — the test for whether a motion is simple harmonic at all.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 2 calculation marked: solve an SHM problem with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

IB Physics Data Booklet↗

Frequently asked

What actually makes a motion 'simple harmonic'?

Why is there a negative sign in

a = -\omega^2 x

Does the period of a pendulum or spring depend on how far it swings?

Where is the object fastest, and where does it accelerate most?

Does the total energy of an SHM system stay constant?

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 calculation marked: solve an SHM problem with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

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In simple terms

Explore the concept

Simulations

Simple Harmonic Motion (SHM)

Circular Motion & SHM Relationship

Simple Pendulum

Pendulum Lab

SHM for Lab

Conservation of Mechanical Energy: Mass on a Vertical Spring

Simple Harmonic Motion, Circular Motion, and Transverse Waves

Simple Harmonic Motion: Mass on a Spring

Oscillation Graphs Quiz

Simple Harmonic Motion Tutorial

Oscillations

Key formulas

Full topic notes

Defining simple harmonic motion

Angular frequency, period and frequency

The two systems: pendulum and mass–spring

How displacement, velocity and acceleration vary

Energy in simple harmonic motion

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Name it

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 calculation marked: solve an SHM problem with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Simple harmonic motion

Revise Simple harmonic motion for IB Physics (C.1)

In simple terms

Explore the concept

Simulations

Simple Harmonic Motion (SHM)

Circular Motion & SHM Relationship

Simple Pendulum

Pendulum Lab

SHM for Lab

Conservation of Mechanical Energy: Mass on a Vertical Spring

Simple Harmonic Motion, Circular Motion, and Transverse Waves

Simple Harmonic Motion: Mass on a Spring

Oscillation Graphs Quiz

Simple Harmonic Motion Tutorial

Oscillations

Key formulas

Full topic notes

Defining simple harmonic motion

Angular frequency, period and frequency

The two systems: pendulum and mass–spring

How displacement, velocity and acceleration vary

Energy in simple harmonic motion

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Name it

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 calculation marked: solve an SHM problem with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Simple harmonic motion