# Linear Oscillator without Friction

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?

• Imagine swinging a frictionless pendulum and plot the position vs. the velocity (that is a Lissajous figure). I would guess that the point of the question is to show that the motion continues indefinitely. – copper.hat Feb 16 '14 at 23:42

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.
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 $\sqrt{V(0)}$.