Simple Harmonic Motion of a Mass-Spring System

For a mass–spring system, the motion is governed by

$m x'' + kx = 0.$

Here, $m=2\,\text{kg}$ and $k=200\,\text{N/m}$, so

$x'' + \frac{200}{2}x = 0 \quad \Rightarrow \quad x'' + 100x = 0.$

This is simple harmonic motion with angular frequency

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = 10\,\text{rad/s}.$

So the period is

$T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \approx 0.628\,\text{s}.$

If the mass is displaced by $0.3\,\text{m}$ and released from rest, then the displacement is

$x(t) = 0.3\cos(10t).$

The mass oscillates back and forth between $+0.3\,\text{m}$ and $-0.3\,\text{m}$ from equilibrium, with frequency

$f = \frac{1}{T} = \frac{10}{2\pi} \approx 1.59\,\text{Hz}.$

The visualization shows the spring, the mass, and its oscillating motion about equilibrium.