IB Physics Topic 3.1: Simple harmonic motion | Notes & Flashcards | Aimnova

IB Physics — Simple harmonic motion

Topic 3.1 of IB Physics covers Simple harmonic motion, which is part of Unit 3: Wave behaviour. Students explore key concepts including Conditions for simple harmonic motion, Period and frequency of SHM oscillators, SHM graphs, phase and timing, Energy in simple harmonic motion. A strong understanding of simple harmonic motion is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Simple harmonic motion

Key Idea: Topic 3.1 is about oscillations — anything that swings back and forth through a middle position: a mass on a spring, a pendulum, a floating cork. It ties together four ideas: what makes motion 'simple harmonic' (the rule a = −ω²x), how long one swing takes (the period formulas), what the displacement, velocity and acceleration graphs look like (and their phase relationships), and how the energy keeps swapping between kinetic and potential. It is examined on Paper 1A (quick MCQs — identify SHM from a graph or its condition, ratio problems on the period, where the speed/acceleration peaks) and on Paper 2 (substitute into T = 2π√(m/k) or T = 2π√(l/g), use quarter-cycle timing, and energy steps like setting the maximum KE equal to the total energy).

📐 Key formulas

Four of these are given in the data booklet (C.1) — you choose the right one rather than memorising it. The energy formula is the only one you must remember.

a = -\omega^{2}x

The defining condition for SHM (given). The minus sign means a always points back toward equilibrium; the slope of an a-against-x line is −ω².

$a$: acceleration of the object (m s⁻²)
$\omega$: angular frequency — how fast it oscillates (rad s⁻¹)
$x$: displacement from equilibrium (m)

T = 2\pi\sqrt{\frac{m}{k}}

Mass-spring period (given). Depends only on the mass m and stiffness k — NOT on gravity.

$T$: period — time for one full oscillation (s)
$m$: mass on the spring (kg)
$k$: spring constant — stiffness (N m⁻¹)

T = 2\pi\sqrt{\frac{l}{g}}

Simple-pendulum period (given). Depends only on the length l and gravity g — NOT on the mass of the bob.

$T$: period — time for one full oscillation (s)
$l$: length of the pendulum (m)
$g$: gravitational field strength (m s⁻²)

T = \frac{1}{f} = \frac{2\pi}{\omega}

Links period, frequency and angular frequency (given). Rearranged: f = 1 ÷ T and ω = 2πf.

$T$: period — time for one full oscillation (s)
$f$: frequency — oscillations per second (Hz)
$\omega$: angular frequency (rad s⁻¹)

E_{total} = \tfrac{1}{2}kA^{2}

Total energy of a mass-spring oscillation. NOT in the data booklet — remember it. It also equals the maximum kinetic energy, ½mv_{max}².

$E_{total}$: total energy of the oscillation (J)
$k$: spring constant — stiffness (N m⁻¹)
$A$: amplitude — the largest displacement (m)

🧭 Which equation, and when?

The most-tested decision in this topic: what is the question actually asking for — to test for SHM, find a period, or find an energy/speed?

🌀 The two oscillators — what each period depends on

📊 The three SHM graphs — phase and where each peaks

⚡ Energy through one swing

✍️ IB-style worked examples

IB-style question — confirm SHM and find ω (a = −ω²x)

The acceleration a of an oscillating mass is measured at several displacements x from equilibrium. The data lie on a straight line through the origin, a = −81x (a in m s⁻², x in m). State whether the mass moves with SHM, and find its angular frequency ω.

Solution:

Compare the data with the given defining condition:
$a = -\omega^{2}x$
The data a = −81x has the same form (a ∝ x with a minus sign), so it is SHM. Match the constants:
$\omega^{2} = 81$
Take the square root of the slope — keep the unit:
$\omega = \sqrt{81} = 9.0\ \text{rad s}^{-1}$

Final answer:

Yes — it is SHM (a proportional to −x), and ω = 9.0 rad s⁻¹. The slope of an a-against-x line is −ω², so square-root it.

IB-style question — period and frequency of a mass-spring (T = 2π√(m/k))

A 0.80 kg mass hangs from a spring of spring constant k = 320 N m⁻¹ and is set oscillating. Calculate the period and the frequency of the oscillation.

Solution:

It is a mass on a spring, so use the given spring formula (gravity does not appear):
$T = 2\pi\sqrt{\frac{m}{k}}$
Substitute m = 0.80, k = 320:
$T = 2\pi\sqrt{\frac{0.80}{320}}$
Work out the inside, then the root:
$T = 2\pi\sqrt{0.0025} = 2\pi \times 0.050$
So the period is:
$T = 0.31\ \text{s}$
Frequency is 1 ÷ period (the given link):
$f = \frac{1}{T} = \frac{1}{0.31} = 3.2\ \text{Hz}$

Final answer:

T ≈ 0.31 s and f ≈ 3.2 Hz. The square root means quartering m/k only halves T — changes scale by the root, not the whole factor.

IB-style question — quarter-cycle timing (T = 1/f = 2π/ω)

A pendulum oscillates with SHM at a frequency of 0.50 Hz. Find the period, then the time it takes to travel from one extreme of its swing to the equilibrium (centre) position.

Solution:

Get the period from the frequency using the given link:
$T = \frac{1}{f} = \frac{1}{0.50} = 2.0\ \text{s}$
One extreme → the centre is exactly one quarter of a full cycle:
$t = \frac{T}{4}$
Substitute the period:
$t = \frac{2.0}{4}$
Work it out — keep the unit:
$t = 0.50\ \text{s}$

Final answer:

T = 2.0 s, and end → centre takes T/4 = 0.50 s. A full cycle splits into four equal quarters, each T/4.

IB-style question — energy and maximum speed (E_{total} = ½kA²)

A 0.50 kg mass on a spring of spring constant k = 200 N m⁻¹ oscillates with an amplitude of 0.10 m. Calculate the total energy of the oscillation and the maximum speed of the mass.

Solution:

Total energy = the energy stored at the amplitude (remember this — it is not in the booklet):
$E_{total} = \tfrac{1}{2}kA^{2}$
Substitute k = 200, A = 0.10:
$E_{total} = \tfrac{1}{2}\times200\times0.10^{2} = 1.0\ \text{J}$
At the centre the energy is all kinetic, so the maximum KE equals the total energy:
$\tfrac{1}{2}mv_{max}^{2} = E_{total}$
Substitute m = 0.50, Eₜₒₜₐₗ = 1.0, and rearrange for vₘₐₓ²:
$v_{max}^{2} = \frac{2\times1.0}{0.50} = 4.0$
Square-root — keep the unit:
$v_{max} = 2.0\ \text{m s}^{-1}$

Final answer:

Eₜₒₜₐₗ = 1.0 J and vₘₐₓ = 2.0 m s⁻¹. The key move: at the centre all the energy is kinetic, so max KE = total energy.

✅ Quick self-check

Tap each card to reveal the answer.

🎯 Highest-yield exam reminders

Exam Tips

SHM needs BOTH: acceleration proportional to displacement AND directed back to equilibrium — the minus sign in a = −ω²x carries that direction. The slope of an a-against-x line is −ω², so square-root it to get ω.
Pick the period formula by the system: spring → T = 2π√(m/k) (no gravity); pendulum → T = 2π√(l/g) (no bob mass). Both have a square root, so multiplying a quantity by 4 only changes T by √4 = 2.
Switch freely between T, f and ω with T = 1/f = 2π/ω: f = 1 ÷ T (in Hz) and ω = 2πf (in rad s⁻¹).
A full cycle splits into four equal quarters of T/4: centre → end → centre → other end → centre. End-to-centre (or centre-to-end) is one quarter, end-to-end is half a period.
Velocity leads displacement by 90° (a quarter-cycle); acceleration is antiphase (180°). Speed peaks at the centre, acceleration at the ends.
Energy keeps swapping: KE is maximum at the centre, PE at the ends, but the total stays constant at Eₜₒₜₐₗ = ½kA². Bigger amplitude → more total energy (E ∝ A²).
Eₜₒₜₐₗ = ½kA² is NOT in the data booklet — remember it. To get the maximum speed, set the maximum KE equal to the total energy (½mvₘₐₓ² = ½kA²).

IB Physics — Simple harmonic motion

Key concepts in Simple harmonic motion

Key Idea: Topic 3.1 is about oscillations — anything that swings back and forth through a middle position: a mass on a spring, a pendulum, a floating cork. It ties together four ideas: what makes motion 'simple harmonic' (the rule a = −ω²x), how long one swing takes (the period formulas), what the displacement, velocity and acceleration graphs look like (and their phase relationships), and how the energy keeps swapping between kinetic and potential. It is examined on Paper 1A (quick MCQs — identify SHM from a graph or its condition, ratio problems on the period, where the speed/acceleration peaks) and on Paper 2 (substitute into T = 2π√(m/k) or T = 2π√(l/g), use quarter-cycle timing, and energy steps like setting the maximum KE equal to the total energy).

📐 Key formulas

Four of these are given in the data booklet (C.1) — you choose the right one rather than memorising it. The energy formula is the only one you must remember.

a = -\omega^{2}x

The defining condition for SHM (given). The minus sign means a always points back toward equilibrium; the slope of an a-against-x line is −ω².

$a$: acceleration of the object (m s⁻²)
$\omega$: angular frequency — how fast it oscillates (rad s⁻¹)
$x$: displacement from equilibrium (m)

T = 2\pi\sqrt{\frac{m}{k}}

Mass-spring period (given). Depends only on the mass m and stiffness k — NOT on gravity.

$T$: period — time for one full oscillation (s)
$m$: mass on the spring (kg)
$k$: spring constant — stiffness (N m⁻¹)

T = 2\pi\sqrt{\frac{l}{g}}

Simple-pendulum period (given). Depends only on the length l and gravity g — NOT on the mass of the bob.

$T$: period — time for one full oscillation (s)
$l$: length of the pendulum (m)
$g$: gravitational field strength (m s⁻²)

T = \frac{1}{f} = \frac{2\pi}{\omega}

Links period, frequency and angular frequency (given). Rearranged: f = 1 ÷ T and ω = 2πf.

$T$: period — time for one full oscillation (s)
$f$: frequency — oscillations per second (Hz)
$\omega$: angular frequency (rad s⁻¹)

E_{total} = \tfrac{1}{2}kA^{2}

Total energy of a mass-spring oscillation. NOT in the data booklet — remember it. It also equals the maximum kinetic energy, ½mv_{max}².

$E_{total}$: total energy of the oscillation (J)
$k$: spring constant — stiffness (N m⁻¹)
$A$: amplitude — the largest displacement (m)

🧭 Which equation, and when?

The most-tested decision in this topic: what is the question actually asking for — to test for SHM, find a period, or find an energy/speed?

🌀 The two oscillators — what each period depends on

📊 The three SHM graphs — phase and where each peaks

⚡ Energy through one swing

✍️ IB-style worked examples

IB-style question — confirm SHM and find ω (a = −ω²x)

Solution:

Compare the data with the given defining condition:
$a = -\omega^{2}x$
The data a = −81x has the same form (a ∝ x with a minus sign), so it is SHM. Match the constants:
$\omega^{2} = 81$
Take the square root of the slope — keep the unit:
$\omega = \sqrt{81} = 9.0\ \text{rad s}^{-1}$

Final answer:

Yes — it is SHM (a proportional to −x), and ω = 9.0 rad s⁻¹. The slope of an a-against-x line is −ω², so square-root it.

IB-style question — period and frequency of a mass-spring (T = 2π√(m/k))

A 0.80 kg mass hangs from a spring of spring constant k = 320 N m⁻¹ and is set oscillating. Calculate the period and the frequency of the oscillation.

Solution:

It is a mass on a spring, so use the given spring formula (gravity does not appear):
$T = 2\pi\sqrt{\frac{m}{k}}$
Substitute m = 0.80, k = 320:
$T = 2\pi\sqrt{\frac{0.80}{320}}$
Work out the inside, then the root:
$T = 2\pi\sqrt{0.0025} = 2\pi \times 0.050$
So the period is:
$T = 0.31\ \text{s}$
Frequency is 1 ÷ period (the given link):
$f = \frac{1}{T} = \frac{1}{0.31} = 3.2\ \text{Hz}$

Final answer:

T ≈ 0.31 s and f ≈ 3.2 Hz. The square root means quartering m/k only halves T — changes scale by the root, not the whole factor.

IB-style question — quarter-cycle timing (T = 1/f = 2π/ω)

A pendulum oscillates with SHM at a frequency of 0.50 Hz. Find the period, then the time it takes to travel from one extreme of its swing to the equilibrium (centre) position.

Solution:

Get the period from the frequency using the given link:
$T = \frac{1}{f} = \frac{1}{0.50} = 2.0\ \text{s}$
One extreme → the centre is exactly one quarter of a full cycle:
$t = \frac{T}{4}$
Substitute the period:
$t = \frac{2.0}{4}$
Work it out — keep the unit:
$t = 0.50\ \text{s}$

Final answer:

T = 2.0 s, and end → centre takes T/4 = 0.50 s. A full cycle splits into four equal quarters, each T/4.

IB-style question — energy and maximum speed (E_{total} = ½kA²)

A 0.50 kg mass on a spring of spring constant k = 200 N m⁻¹ oscillates with an amplitude of 0.10 m. Calculate the total energy of the oscillation and the maximum speed of the mass.

Solution:

Total energy = the energy stored at the amplitude (remember this — it is not in the booklet):
$E_{total} = \tfrac{1}{2}kA^{2}$
Substitute k = 200, A = 0.10:
$E_{total} = \tfrac{1}{2}\times200\times0.10^{2} = 1.0\ \text{J}$
At the centre the energy is all kinetic, so the maximum KE equals the total energy:
$\tfrac{1}{2}mv_{max}^{2} = E_{total}$
Substitute m = 0.50, Eₜₒₜₐₗ = 1.0, and rearrange for vₘₐₓ²:
$v_{max}^{2} = \frac{2\times1.0}{0.50} = 4.0$
Square-root — keep the unit:
$v_{max} = 2.0\ \text{m s}^{-1}$

Final answer:

Eₜₒₜₐₗ = 1.0 J and vₘₐₓ = 2.0 m s⁻¹. The key move: at the centre all the energy is kinetic, so max KE = total energy.

✅ Quick self-check

Tap each card to reveal the answer.

🎯 Highest-yield exam reminders

Exam Tips

SHM needs BOTH: acceleration proportional to displacement AND directed back to equilibrium — the minus sign in a = −ω²x carries that direction. The slope of an a-against-x line is −ω², so square-root it to get ω.
Pick the period formula by the system: spring → T = 2π√(m/k) (no gravity); pendulum → T = 2π√(l/g) (no bob mass). Both have a square root, so multiplying a quantity by 4 only changes T by √4 = 2.
Switch freely between T, f and ω with T = 1/f = 2π/ω: f = 1 ÷ T (in Hz) and ω = 2πf (in rad s⁻¹).
A full cycle splits into four equal quarters of T/4: centre → end → centre → other end → centre. End-to-centre (or centre-to-end) is one quarter, end-to-end is half a period.
Velocity leads displacement by 90° (a quarter-cycle); acceleration is antiphase (180°). Speed peaks at the centre, acceleration at the ends.
Energy keeps swapping: KE is maximum at the centre, PE at the ends, but the total stays constant at Eₜₒₜₐₗ = ½kA². Bigger amplitude → more total energy (E ∝ A²).
Eₜₒₜₐₗ = ½kA² is NOT in the data booklet — remember it. To get the maximum speed, set the maximum KE equal to the total energy (½mvₘₐₓ² = ½kA²).

Key concepts in Simple harmonic motion

📐 Key formulas

🧭 Which equation, and when?

🌀 The two oscillators — what each period depends on

📊 The three SHM graphs — phase and where each peaks

⚡ Energy through one swing

✍️ IB-style worked examples

IB-style question — confirm SHM and find ω (a = −ω²x)

IB-style question — period and frequency of a mass-spring (T = 2π√(m/k))

IB-style question — quarter-cycle timing (T = 1/f = 2π/ω)

IB-style question — energy and maximum speed (E_{total} = ½kA²)

✅ Quick self-check

🎯 Highest-yield exam reminders

Exam Tips

What you'll learn in Topic 3.1

Study resources — 3.1 Simple harmonic motion

Conditions for simple harmonic motion

Period and frequency of SHM oscillators

SHM graphs, phase and timing

Energy in simple harmonic motion

Ready to study Simple harmonic motion?

Ready to practice?

Key concepts in Simple harmonic motion

📐 Key formulas

🧭 Which equation, and when?

🌀 The two oscillators — what each period depends on

📊 The three SHM graphs — phase and where each peaks

⚡ Energy through one swing

✍️ IB-style worked examples

IB-style question — confirm SHM and find ω (a = −ω²x)

IB-style question — period and frequency of a mass-spring (T = 2π√(m/k))

IB-style question — quarter-cycle timing (T = 1/f = 2π/ω)

IB-style question — energy and maximum speed (E_{total} = ½kA²)

✅ Quick self-check

🎯 Highest-yield exam reminders

Exam Tips

What you'll learn in Topic 3.1

Study resources — 3.1 Simple harmonic motion

Conditions for simple harmonic motion

Period and frequency of SHM oscillators

SHM graphs, phase and timing

Energy in simple harmonic motion

Ready to study Simple harmonic motion?

Ready to practice?