Linear Oscillator without Friction

Question

I already figured out the majority of the solution to this problem but I just need help on the last part. The question is:

Consider the linear oscillator without friction: $$m\frac{d^2x}{dt^2}=-kx$$ (a) Sketch the solution in the phase plane
(b) Interpret the solution

So I already kow that the sketch on the phase plane are going to be concentric circles kind of like this

However, I don't understand what that means. Does that mean the mass will oscillate on the spring forever if it is on one of those lines?

Answer 1

Every circle centered at the origin is a possible route of the solution of your harmonic oscillator, i.e., of $$ \big(x(t),x'(t)\big), \quad t\in\mathbb R. $$ The fact that the routes are circles (in general, closed curves) means that the equation describes a periodic phenomenon.

Answer 2

Note that if you take the quantity $V(t) = {1 \over 2} (x(t)^2 + {m \over k}\dot{x}(t)^2)$, then $\dot{V}(t) = 0$, hence the points $(x(t), \sqrt{m \over k} \dot{x}(t))$ lie on a circle of radius $\sqrt{V(0)}$.