Simple harmonic motion

Oscillation in which the **acceleration is proportional to the displacement** from equilibrium and is always directed **back toward equilibrium**.

$a = -\omega^{2}x$ — **given** in the data booklet. a = acceleration, ω = angular frequency, x = displacement.

The acceleration points **opposite to the displacement** — always back toward equilibrium (the restoring direction).

A force that always acts to push or pull the object **back toward its equilibrium (resting) position**.

The central resting position where the object would sit still — where the displacement x = 0.

1. Acceleration **proportional to** displacement. 2. Acceleration directed **back toward equilibrium** (opposite to x).

A **straight line through the origin** with a **negative slope** equal to −ω².

There is a **restoring force** (and acceleration) directed **back to equilibrium** that is **proportional to the displacement** — exactly the condition a = −ω²x.

The steady loss of energy from an oscillation (to friction or drag), so each successive swing has a **smaller amplitude**.

The **amplitude slowly decreases** over many cycles while the **period stays almost the same**.

Yes — same form as a = −ω²x, so ω² = 25 → **ω = 5.0 rad s⁻¹**.

The **time for one complete oscillation** (one full cycle), measured in seconds.

The **number of oscillations per second**, measured in hertz (Hz). f = 1 ÷ T.

$f = \dfrac{1}{T}$ and $T = \dfrac{1}{f}$ — they are reciprocals. **Given** in the data booklet (T = 1/f).

$T = 2\pi\sqrt{\dfrac{m}{k}}$ — depends on the mass m and spring constant k. **Given**.

$T = 2\pi\sqrt{\dfrac{l}{g}}$ — depends on the length l and gravity g. **Given**.

**No** — mass does not appear in T = 2π√(l/g), so the period is unchanged.

**No** — g does not appear in T = 2π√(m/k); only the mass and stiffness matter.

**×√4 = ×2** — the period doubles, because T ∝ √l.

**×1/√2 ≈ 0.71** — a stiffer spring oscillates faster, so a shorter period (T ∝ 1/√k).

How fast the cycle turns (2π radians per cycle): **ω = 2πf = 2π ÷ T**. Unit: rad s⁻¹.

Both period formulas have a √, so a quantity ×4 inside the root comes out as ×√4 = ×2.

The spring's **stiffness** — a bigger k means a stiffer spring that pulls back harder and oscillates faster.

All three are **sinusoids** (smooth waves), but **shifted** in phase relative to one another.

Velocity **leads** displacement by a **quarter-cycle (90°)** — v is biggest at the centre, zero at the ends.

They are **antiphase (180° apart)** — a is the mirror image of x. This is the rule **a = -ω²x**.

At the **centre** (equilibrium, x = 0). It is **zero** at the turning points (maximum displacement).

At the **turning points** (maximum displacement). It is **zero** at the centre, because a = -ω²x.

The acceleration always points **back toward the equilibrium position** (a restoring acceleration), opposite to the displacement.

**T/4** — one quarter of the period (each quarter-cycle takes the same time).

**T/2** — half a period (two quarter-cycles, passing through the centre).

$a = -\omega^{2}x$ — acceleration proportional to displacement and directed back toward equilibrium.

$T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}$ — both given in the data booklet.

Thinking it is greatest at the ends — it is greatest at the **centre** and **zero** at the ends.

**Four** quarter-periods, each lasting **T/4**.

**Kinetic energy** (of motion) and **potential energy** (stored, e.g. in a stretched spring). They swap back and forth as it oscillates.

It **stays constant** — KE and PE just trade places, but their sum never changes.

At the **centre** (equilibrium position), where the object moves **fastest**.

At the **ends** (the amplitude), where the object is **momentarily at rest**.

The **greatest displacement** from the centre — the turning point where the object briefly stops.

$E_{total} = \tfrac{1}{2}kA^{2}$ — set by the amplitude A. **Not** in the data booklet, so remember it.

At the amplitude the object is at rest (KE = 0), so all the energy is the elastic PE stored at the biggest stretch, ½kx² with x = A.

Set the **maximum KE equal to the total energy**: ½mv_{max}² = ½kA², then solve for v_{max}.

It becomes **four times larger**, because E_{total} = ½kA² depends on A² (the amplitude squared).

It **equals the total energy** — at the centre the PE is zero, so all the energy is kinetic.

**PE** is an upward parabola (min at the centre); **KE** is a downward parabola (max at the centre); their sum is a flat line.

Thinking the speed is greatest at the ends — it is greatest at the **centre**; at the ends the object is momentarily still.