# Question concerning phase curves and Lissajous figures

I want to draw the "orbits" of a spherical pendulum under small oscillations. In this case its equations are given by $$\ddot{x}_{1}=-x_{1}$$ and $$\ddot{x}_{2}=-x_{2}$$. Of course the potential energy is given by $$U=\frac{1}{2}(x_{1}^2+x_{2}^2)$$, and the level sets of it will be concentric circles in the $$x_{1}x_{2}$$ plane. Consequently, by the law of conservation of energy: $$E=\frac{1}{2}(\dot{x}_{1}^2+\dot{x}_{2}^2)+\frac{1}{2}(x_{1}^2+x_{2}^2)={\rm const}.$$ And this represents a sphere in four space. Now, suppose I want to draw the orbits in the $$x_{1}$$-$$x_{2}$$ plane. Is all I have to do to make $$\dot{x}_{1}=\dot{x}_{2}=0$$? If so, I also get concentric spheres for each value of E.

But now using the Lissajous figures method: the solutions of these equations can be written as $$x_{1}=A_{1}\sin{(t+\phi_{1})}$$ and $$x_{2}=A_{2}\sin{(t+\phi_{2})}$$, where $$A_{i}=\sqrt{2E}$$ for $$i=1,2$$. So the orbits have to lie inside this region. By Lissajous figures, the shape of the curve on the $$x_{1}$$-$$x_{2}$$ plane depends entirely on the difference $$\phi_{2}-\phi_{1}$$ and this is a circle in the case where $$\phi_{2}-\phi_{1}=\pi/2$$.

So, what shape exactly does this orbit have? I may be misunderstanding some concepts, and if so, please let me know. Thanks in advance.

• Setting $\dot{x_1}=\dot{x_2}=0$ initially means that your pendulum oscillates along a line going through the origin. It does not represent a general solution. – probably_someone Sep 18 at 20:28
• The curves in 3-d space are just the projection of the Lissajous figures on the sphere . Or did I misunderstand your question.? – trula Sep 18 at 20:40

To expand on @probably_someone's comment, by setting $$\dot{x}_1(0)=0=\dot{x}_2(0)$$ you have already set the phase difference. Are a result, your solution is not the most general one. (One way to see this is that you only really have one free parameter left to "play" with, the phase difference $$\phi_2 - \phi_1$$. However the most general solution is much more flexible.)
You should be able to see that I can rewrite your solutions as \begin{aligned}x_1 &= A \sin(t),\\x_2 &= A \sin(t + \phi_2- \phi_1),\end{aligned}
and if I further impose the condition that $$\dot{x}_1(0) = 0 = \dot{x}_2(0)$$, then indeed this just means that $$\cos(\phi_2 - \phi_1) = 0 \quad \text{or} \quad \phi_2 - \phi_1 = \frac{\pi}{2},$$ as you'd expect, so your two "methods" do indeed give the same answers.