Simple Harmonic Motion

Pull a mass on a spring to one side and let go, and it does not just drift back and stop. It overshoots equilibrium, swings to the other side, and returns, repeating with a steady rhythm. That clean, repeating swing is simple harmonic motion (SHM): the motion you get whenever the restoring force grows in direct proportion to how far the object is displaced.

The formulas and what each symbol means

For a mass on an ideal spring, the restoring force is

where $x$ is the displacement from equilibrium and $k$ is the spring constant. The minus sign means the force points opposite the displacement. Feeding this into Newton's second law $F = ma$ gives the defining equation of SHM:

The motion that solves it has angular frequency, period, and frequency

and displacement written as

where $A$ is the amplitude and $\phi$ is the phase constant set by the starting conditions.

Why SHM repeats: the force-acceleration loop

The equation $\frac{d^2x}{dt^2} = -\frac{k}{m}x$ already contains the whole story. When the mass is far from equilibrium, the restoring force is large, so it accelerates hard back toward the center. As it moves inward the force shrinks, but the mass keeps speeding up because it has already been accelerated. It blows through equilibrium at maximum speed, the force then reverses and decelerates it, and it stops at the far turning point before the cycle restarts. Because acceleration is always proportional to and opposite the displacement, the timing never changes, which is why the period stays constant no matter the amplitude.

This also tells you what counts as genuine SHM. All three of these must hold: the motion is about an equilibrium position, the restoring force points toward equilibrium, and the restoring force is proportional to displacement over the range you model. If one fails, the motion may still oscillate, but it is not SHM in the strict sense.

Worked example: period of a spring-mass system

A mass of $0.50\ \mathrm{kg}$ is attached to an ideal spring with $k = 200\ \mathrm{N/m}$. Find the angular frequency and the period.

So this system completes one full oscillation every $0.314$ seconds. A stiffer spring makes the oscillation faster, while a larger mass makes it slower.

Practice it yourself

Keep the same $200\ \mathrm{N/m}$ spring but double the mass to $1.0\ \mathrm{kg}$ and solve for $T$ again. You should get $\omega = \sqrt{200} \approx 14.1\ \mathrm{rad/s}$ and $T \approx 0.444\ \mathrm{s}$. Notice the period did not double: it grew by a factor of $\sqrt{2}$, because $T \propto \sqrt{m}$. That single change makes the mass dependence of the period concrete. For a deeper look at how a force law fixes a motion, compare this with Newton's second law.

Calculation traps to avoid

• Calling any oscillation SHM. Oscillation alone is not enough; the restoring force must be proportional to displacement.
• Dropping the minus sign in $F = -kx$. Without it the force would point away from equilibrium.
• Mixing up amplitude and period. Amplitude is how far the object moves; period is how long one cycle takes. Note that the period is independent of amplitude.
• Assuming a pendulum is always SHM. A simple pendulum is only approximately SHM for small angular displacements.

Where simple harmonic motion is used

SHM is the standard starting model for springs, vibrating molecules, electrical oscillators, and small oscillations near any stable equilibrium. It is also a useful approximation whenever a more complicated system behaves linearly near its equilibrium point. Real systems often add damping, driving forces, or nonlinear effects, so the motion stops being ideal SHM once those grow important.