Let the vertical be the \(x\) - direction. When the spring is neither stretched nor compressed, the mass is at rest and at the position called the equilibrium position (at \(x=0\) )
When the mass is displaced (set into vertical oscillations) from the equilibrium position and released, it is now a particle under a net force ( i.e a restoring force \(f_s\) that is proportional to the displacement \(\mathrm{F}_{\mathrm{s}}=-\mathrm{Kx}\) ) and consequently undergoes an acceleration.
Thus:
\begin{array}{l}
\sum \mathrm{F}_x=\mathrm{ma}_x \\
\text { or simply } \\
\sum \mathrm{F}=\mathrm{ma} \\
\mathrm{F}=-\mathrm{k} x \\
\Rightarrow-\mathrm{k} x=\mathrm{ma} \\
\mathrm{a}=\frac{-\mathrm{k} x}{\mathrm{~m}} \Rightarrow \mathrm{a} \alpha x \text { (a varies with } x \text { ) ____________ (1) }
\end{array}
which means that the acceleration of the mass is proportion to its position and the direction of the acceleration is opposite the direction of the displacement of the mass from its equilibrium position.
Thus, the acceleration will have varying magnitude \((a \alpha x)\) but a constant direction (acceleration always directed to the equilibrium position).
We recall from linear motion \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}^2 x}{\mathrm{dt}^2}\), so we can express equation (1) as \(\mathrm{a}=\frac{\mathrm{d}^2 x}{\mathrm{~d} \mathrm{t}^2} \frac{-\mathrm{kx}}{\mathrm{m}}\),
If we let \(\omega^2=\mathrm{k} / \mathrm{m}\), then equation (2) can be written in the form \(a=\frac{\mathrm{d}^2 x}{d \mathrm{~d}^2}=-\omega^2 x\)
equation (3) is a second order differential equation whose solution (though beyond the present scope) is
where \(\mathrm{A}, \omega\), and \(\emptyset\) are constants
and the quantity \((\omega t+\emptyset)\) is called the phase of the motion.