# Damped simple harmonic oscillator problem

I'm supposed to calculate and draw the phase space trajectory for this:

for the two different cases when and .

I've never done this sort of question before, how are they done? I've tried googling about hamiltonians etc but can't find a clear answer

Hamiltonian formalism won't help so much because the problem is dissipative. You can solve the homogeneous linear diff. equation with constant coefficients by supposing $x=C e^{\lambda t}$. You will get complex solutions, but you should be able to add them to get a real solution. Now you should have $x(t)$ with two arbitrary constants determined by initial conditions. The general evolution will be an exponentially damped sinusoid.

Now to the phase portrait. A phase portrait just means to plot $(x(t),p(t))$ where $p = m \dot{x}$ or sometimes you can just take $\dot{x}$. For every $t$, you get one $x$ and one $p$ - plot them and when you do this for a lot of $t$s you get the phase portrait of one trajectory (a lot of plotting tools can do this with the "parametric plot" where $t$ would be the parameter). For the case of a damped oscillator this will look like a spiral converging to $x=0$, $p=0$ for large $t$.

You're simply meant to plot $x$ against $\dot{x}$. You can rewrite the DE as $p\,\mathrm{d}_q p + \gamma\, p + \omega^2 q=0$ where $p = \dot{x}$ and $q=x$, but I can't see a way of solving this one directly: without the dissipative term you'd get $p\,\mathrm{d}_q p + \omega^2 q = \mathrm{d}_q (p^2 + \omega^2 q^2/2) = 0$, which simply defines an ellipse. A lossless system cycles in phase space, which is what this ellipse is telling you. Mathematica tells me that the solution to my dissipative DE is:

$$\frac{1}{2} \log \left(\frac{\gamma\, p(q)}{q}+\frac{p(q)^2}{q^2}+\omega^2\right)-\frac{\gamma}{\sqrt{4\, \omega^2-\gamma ^2}} \arctan\left(\frac{\gamma\,q +2 p(q)}{q\,\sqrt{4 \omega^2-\gamma ^2}}\right)=c_1-\log (q)$$

where you adjust $c_1$ for your initial conditions. This is your path. I can't see off the top of my head how to get to this result: I shall update my answer if I do. But this will from memory give you a spiral: it must approach the origin as $t\to\infty$ as the system loses its energy.

The differential equation is given by,

$$\ddot{x} + \gamma \dot{x} + \omega^2 x = 0$$

If we propose the ansatz $x(t) = e^{rt}$ for some $r \in \mathbb{C}$, then we obtain a quadratic equation for $r$:

$$r^2 + \gamma r + \omega^2 = 0$$

The solutions are given by,

$$r = \frac{1}{2} \left( -\gamma \pm \sqrt{\gamma^2 -4\omega^2}\right)$$

Let us choose one of the cases, $\gamma = 2\omega$, in which case we have a unique solution $r=-\gamma$, and

$$x(t) = c_1e^{-2 \omega t} + c_2te^{-2\omega t}$$

The other case $\gamma = \omega$ gives rise to two complex solutions,

$$r = \frac{\omega}{2}\left( \pm i\sqrt{3} -1\right)$$

and with some manipulation using Euler's formula and the superposition principle, we can show,

$$x(t)=e^{-\omega t/2} \left( c_1 \sin \omega \frac{\sqrt{3}}{2}t + c_2 \cos \frac{\sqrt{3}}{2}t\right)$$

Now that you know $x(t)$, I leave it to you to draw the phase space plot, which is simply $(x,p)$. Every point in the space corresponds to a particular state parametrized by the position and momentum.

• Your solution for $\gamma=2\omega$ does not gives you enough degrees of freedom, such that all initial conditions yield a solution. In that case $c_1$ should be $c_1+c_2t$. – fibonatic Oct 13 '14 at 2:30
• @fibonatic: Yes, well spotted, I've corrected the solution. – JamalS Oct 13 '14 at 8:47