Springs and Simple Harmonic Motion

Here $x$ is the displacement from the spring's natural (equilibrium) length, in metres, and $k$ is the spring constant in newtons per metre: a stiff spring has a large $k$. The negative sign says the spring force always opposes the displacement: stretch it and it pulls back, compress it and it pushes out. That is why it is called a restoring force. The force you must apply to hold the spring at $x$ is equal and opposite, $F=kx$. Hooke's law holds only up to the elastic limit; stretch a spring beyond that and it is permanently deformed.

Why a half? The force you apply is not constant: it grows from $0$ to $kx$ as you stretch. Work is the area under a force-displacement graph, and here that area is a triangle with base $x$ and height $kx$, so $W=\frac{1}{2}(x)(kx)=\frac{1}{2}k{x}^2$. That work is stored as elastic potential energy. Because $x$ is squared, doubling the stretch stores four times the energy, and a compression stores exactly as much as an equal stretch.

Attach a mass $m$ to a spring on a frictionless horizontal surface, pull it out to a distance $A$ (the amplitude) and release it. At the ends it stops for an instant, so all the energy is elastic. As it passes equilibrium ($x=0$) the spring is relaxed, so all the energy is kinetic. In between it is shared. Setting the total at the end equal to the total anywhere else gives $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$, and at $x=0$ this gives the maximum speed, $v_{max}=A\sqrt{k/m}$.

Imagine a point moving at constant speed around a circle of radius $A$, and look at its shadow on a line through the centre. The shadow moves back and forth between $-A$ and $+A$, fastest in the middle and momentarily stopped at the ends, exactly like the mass on a spring. This comparison gives the period without calculus. The point covers the circumference in one period, so its speed is $2\pi A/T$. At the centre the shadow moves at that full speed, which must equal $v_{max}=A\sqrt{k/m}$. Setting $2\pi A/T=A\sqrt{k/m}$ and solving for $T$ gives the result below. A simple pendulum swinging through a small angle is also close to SHM, with $T=2\pi \sqrt{L/g}$.