The equation of simple harmonic motion

Consider a mass $m$ attached to a spring of stiffness $K$, on a smooth horizontal plane. Let $x$ denote the displacement of the mass from its rest position. The elastic force is restoring, $F = -Kx$: it always points towards equilibrium, with a magnitude proportional to how far we have moved away from it. Newton’s second law then gives

$m\,\frac{\dd^2 x}{\dd t^2} = -K\,x$

This is a second-order linear differential equation with constant coefficients, and it defines simple harmonic motion. Its structure — “the acceleration is proportional to the position, with the opposite sign” — is what makes the motion oscillatory: as soon as the body moves away, a force arises that pushes it back, carries it through equilibrium, slows it down on the other side, and so on.

Principle — Equation of simple harmonic motion (SHM)

A system obeys simple harmonic motion if its acceleration is proportional to the coordinate, with the opposite sign: $\ev{\frac{\dd^2 x}{\dd t^2} = -\omega^2\,x}$ where $\omega$ is the angular frequency (in rad/s).

For the mass-spring system, comparing $m\ddot x = -Kx$ with $\ddot x = -\omega^2 x$, we immediately read off the angular frequency:

$\ev{\omega = \sqrt{\frac{K}{m}}}$

Intuition

The stiffer the spring ($K\uparrow$) or the lighter the mass ($m\downarrow$), the faster the oscillation. Stiffness supplies the “urge to go back”, the mass’s inertia opposes it: their contest sets the rhythm.

Collegamenti

Argomenti: Oscillations and harmonic motion Concetti: Simple harmonic motion · Elastic force and Hooke’s law Competenze: Symbolic setup Oggetti: Spring

Esercizi collegati: Problem — Mass on two springs in parallel · Problem — Mass between two springs, equivalent K · Problem — Mass hanging from a spring suspension