The Mass on a Spring: Hooke's Law and the Master Equation

Hooke's law: the restoring force

Picture a block on a frictionless table tied to a spring. At the spring's natural length nothing pulls — that is equilibrium. Stretch the block a distance x to the right and the spring pulls it back left; compress it and the spring pushes it back right. For small displacements the force is proportional to x and points opposite to it. That is Hooke's law.

F = -kx

Hooke's law: the restoring force is proportional to displacement; the minus sign means 'back toward equilibrium'.

Here k is the spring constant (also called the force constant), measured in newtons per metre — it is the spring's stiffness. The minus sign is the whole personality of an oscillator: the force always points home, back toward x = 0. This is our restoring force in mathematical form.

Newton's second law turns it into motion

Now apply Newton's second law F = ma to the block. On the frictionless table the only horizontal force is the spring's, so we can set the two expressions for the force equal.

ma = -kx \quad\Longrightarrow\quad a = -\frac{k}{m}\,x

Acceleration is proportional to displacement and points opposite to it.

This is the signature of simple harmonic motion: the acceleration is proportional to -x. It is convenient to bundle the constant k/m into a single symbol, the angular frequency \omega, defined by \omega^2 = k/m. The equation then takes its cleanest form.

a = -\omega^2 x, \qquad \omega^2 = \frac{k}{m}

The defining equation of simple harmonic motion.

Solving it: position as a sine wave

What function has a second derivative equal to -\omega^2 times itself? Sines and cosines do exactly that. The general solution of a = -\omega^2 x is therefore a shifted cosine.

x(t) = A\cos(\omega t + \varphi)

General SHM solution: amplitude A, angular frequency ω, phase constant φ.

Here A is the amplitude, fixed by how far you first pulled the block; \varphi (the phase constant) is fixed by the instant you chose as t = 0. Crucially, \omega is set by the spring and the mass alone — it contains no A. That is the promised isochronism: the period does not depend on amplitude.

\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Angular frequency, period and frequency of a mass-spring oscillator.

A worked example

A 0.50 kg block on a spring with k = 200 N/m is pulled 4.0 cm from equilibrium and released from rest. Find its angular frequency, period, frequency and maximum speed. Step 1 — angular frequency. Drop k and m straight into \omega = \sqrt{k/m}.

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = 20\ \text{rad/s}

Step 1: angular frequency from the spring constant and mass.

Step 2 — period and frequency. These follow immediately from \omega.

T = \frac{2\pi}{\omega} = \frac{2\pi}{20} \approx 0.31\ \text{s}, \qquad f = \frac{1}{T} \approx 3.2\ \text{Hz}

Step 2: period and frequency.

Step 3 — maximum speed. At release the block sits at x = A with zero speed. It moves fastest as it flies through equilibrium, where the peak speed is v_{\max} = A\omega. With A = 0.040 m:

v_{\max} = A\omega = (0.040\ \text{m})(20\ \text{rad/s}) = 0.80\ \text{m/s}

Step 3: maximum speed, reached at the equilibrium position.