Area of a Harmonic Oscillator in phase space

Question

This might be a dumb question, but its art of my assignment and im stuck at the last part. So here it goes.

I have a mass-spring system, i am supposed to workout its equation using conservation of energy, and then graph the final result and prove that the area enclosed by the curve obtained after graphing, is proportional to the total energy of the oscillator.

so here is what i have so far..

Total energy of the simple harmonic system (considering a mass spring system in 1D)

E = Kinetic energy + Potential Energy = $0.5 v^2 + 0.5 \omega^2x^2 $ (where $ \omega^2= k/m$)

now i know that this is the equation of an ellipse, now i am not sure how to proceed and show the proportionality between, the area of the curve and the total energy of the system?

Answer 1

Phase space is really momentum p and location x, but you can use v and x because p=mv so it's just a proportionality anyway. So think of your equation for E (which is really E/m, but again, proportionality is fine) as a constraint on a set of (x,v) that make an ellipse. All you need is a formula for the area of an ellipse (https://www.math.hmc.edu/funfacts/ffiles/10006.3.shtml) as a function of its major and minor axis lengths. Now express the major and minor axis lengths in terms of E, by looking at special points on that ellipse where x=0 or v=0, and you are done.

Answer 2

First of all the total energy, $$E=\frac{1}{2}mv^2+\frac{1}{2}kq^2$$ Or, $$E=\frac{p^2}{2m}+\frac{kq^2}{2}$$ Or, $$1=\frac{p^2}{2mE}+\frac{q^2}{2E/k}$$ This is an equation of ellipse with semi-major axis of $\sqrt{2E/k}$ and semi-minor axis $\sqrt{2mE}$.
So the area of the ellipse $= \pi\cdot\sqrt{2E/k}\cdot\sqrt{2mE} =2\pi E\sqrt{m/k}$.
Hence the area is proportional to $E$, total energy.