Graphs of Simple Harmonic Motion showed that a mass on a spring traces sine and cosine curves. With Derivatives of Trig Functions and Higher Derivatives we can see why: Newton's second law for a spring is a differential equation, and cosine is the function that solves it.
Why it works
For a mass $m$ on a spring of stiffness $k$, with $x$ measured from equilibrium, $F = -kx$, so $$m\ddot{x} = -kx \quad\Longrightarrow\quad \ddot{x} = -\dfrac{k}{m}x.$$ We need a function whose second derivative is a negative multiple of itself. Try $x = A\cos(\omega t + \phi)$: then $\dot{x} = -A\omega\sin(\omega t + \phi)$ and $\ddot{x} = -A\omega^2\cos(\omega t + \phi) = -\omega^2x$. It works exactly when $\omega^2 = k/m$, for any amplitude $A$ and phase $\phi$.
Key idea
Simple harmonic motion obeys $\ddot{x} = -\omega^2x$, solved by $x = A\cos(\omega t + \phi)$ with $$\omega = \sqrt{\dfrac{k}{m}}, \qquad f = \dfrac{\omega}{2\pi}, \qquad T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}}.$$ The greatest speed is $A\omega$ (at equilibrium) and the greatest acceleration $A\omega^2$ (at the turning points). The period doesn't depend on the amplitude.
Worked example: Frequency and period
A $0.5\ \text{kg}$ block on a spring with $k = 50\ \text{N/m}$ oscillates on a frictionless surface. Find $\omega$, $f$ and $T$.
- $\omega = \sqrt{50/0.5} = 10\ \text{rad/s}$.
- $f = \dfrac{10}{2\pi} = 1.59\ \text{Hz}$ (cycles per second).
- $T = \dfrac{1}{f} = 0.628\ \text{s}$.
Answer: $\omega = 10\ \text{rad/s}$, $f = 1.59\ \text{Hz}$, $T = 0.628\ \text{s}$.
Worked example: Reading an equation
A quantity obeys $\ddot{x} = -49x$ (seconds). Find its period, and its greatest speed if the amplitude is $0.2$ units.
- Match $\ddot{x} = -\omega^2x$: $\omega^2 = 49$, so $\omega = 7\ \text{rad/s}$.
- $T = 2\pi/7 = 0.898\ \text{s}$.
- $v_{\max} = A\omega = 0.2(7) = 1.4$ units per second.
Answer: $T = 0.898\ \text{s}$; $v_{\max} = 1.4$ units/s.
Common mistake
Mixing up the angular frequency $\omega$ (radians per second) with the frequency $f$ (cycles per second, Hz). The block above has $\omega = 10\ \text{rad/s}$ but completes only $1.59$ cycles each second, because each cycle is $2\pi$ radians of phase. Check: if your "frequency" times the period isn't 1, you have $\omega$; $\omega T$ should be $2\pi$.
Energy gives a quick check. With $x = A\cos(\omega t + \phi)$, $\tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2 = \tfrac{1}{2}kA^2$ at every instant (use $\sin^2 + \cos^2 = 1$ and $m\omega^2 = k$). So the speed at position $x$ is $v = \omega\sqrt{A^2 - x^2}$: largest at $x = 0$, zero at $x = \pm A$.
The same equation, $\ddot{q} = -\omega^2q$, describes a pendulum's small swings, an LC circuit's charge and a vibrating molecule. Near any stable equilibrium, as in Potential Energy Diagrams, small motions are simple harmonic.