Mini-Lesson

Simple Harmonic Motion

This mini-lesson covers Theme C.1 — Simple harmonic motion: the defining condition a ∝ −x, period and frequency, the pendulum and mass-spring systems, and the interchange of kinetic and potential energy.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Watch for the HL flag on higher-level extensions. Press Start when you're ready.

C.1 · definition

What makes motion "simple harmonic"?

An object performs simple harmonic motion (SHM) when its acceleration is (i) proportional to its displacement from equilibrium and (ii) always directed back towards equilibrium.

a ∝ −xthe minus sign means the restoring acceleration opposes the displacement

Examples: a pendulum (small swings) and a mass on a spring. At the extremes the displacement, restoring force and acceleration are greatest; at the centre they are zero but the speed is greatest.

Quick check

?For a mass on a spring undergoing SHM, where is the acceleration greatest and where is the speed greatest?

C.1 · period

Period, frequency and the two systems

Period T is the time for one full oscillation; frequency f is oscillations per second. They are reciprocals: T = 1/f.

T = 2π√(L ÷ g) · T = 2π√(m ÷ k)simple pendulum (length L) · mass–spring (mass m, stiffness k)

Worked example — a seconds pendulum

A pendulum of length L = 1.0 m, with g = 9.81 m s⁻².

T = 2π√(1.0 ÷ 9.81) = 2π × 0.319 = 2.0 s

Calculate

#Find the period of a simple pendulum of length 1.0 m (g = 9.81 m s⁻²).

s

Hint: T = 2π√(L ÷ g) = 2π√(1.0 ÷ 9.81).

Calculate

#An oscillator has a period of 0.50 s. Find its frequency.

Hz

Hint: f = 1 ÷ T = 1 ÷ 0.50.

Calculate

#A 0.20 kg mass oscillates on a spring of stiffness 80 N m⁻¹. Find the period.

s

Hint: T = 2π√(m ÷ k) = 2π√(0.20 ÷ 80).

Sort it

Effect on a pendulum's period T

Tap an item, then tap the group it belongs to.

⬆️ Increases T

⬇️ Decreases T

➖ No effect on T

C.1 · energy

Energy in SHM

During SHM, energy shuttles between kinetic and potential stores while the total stays constant (ignoring damping).

At the extremes: all energy is potential (KE = 0).
At equilibrium: all energy is kinetic (PE at minimum), so speed is maximum.

A graph of KE and PE against displacement shows two parabolas that cross at x = 0; their sum is a flat line — the constant total energy.

Quick check

?A child on a swing is a good approximation to SHM. As they pass through the lowest point, what is true of their energy?

Match it

Match term to meaning

Tap a statement on the left, then its match on the right.

Statement

Answer

C.1 · HL depth

The SHM equations

At HL the motion is described quantitatively using the angular frequency ω = 2πf:

a = −ω²x · v = ±ω√(x₀² − x²) · v_max = ωx₀x₀ is the amplitude

Total energy E_T = ½mω²x₀², and kinetic energy E_K = ½mω²(x₀² − x²). Displacement follows x = x₀ sin(ωt) or x = x₀ cos(ωt) depending on the start point.

v_max occurs at the centre (x = 0); a_max = ω²x₀ occurs at the extremes (x = ±x₀).

Calculate

#HL: a 0.50 kg mass on a 200 N m⁻¹ spring oscillates with amplitude 0.030 m. Find the maximum speed. (ω = √(k/m).)

m s⁻¹

Hint: ω = √(200 ÷ 0.50) = 20 rad s⁻¹; v_max = ωx₀ = 20 × 0.030.

Recap

The big ideas to know

SHM condition: a ∝ −x: acceleration proportional to, and opposite, displacement

Period: T = 1/f; pendulum T = 2π√(L/g); spring T = 2π√(m/k)

Energy: KE ↔ PE interchange; total constant; v max at centre, a max at extremes

Independence: pendulum period does not depend on mass or (small) amplitude

HL: a = −ω²x · v = ±ω√(x₀²−x²) · v_max = ωx₀

That completes Simple Harmonic Motion for IB Diploma Physics HL. Press Finish to see your score.

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