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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.

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    $\begingroup$ 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. $\endgroup$ – probably_someone Sep 18 at 20:28
  • $\begingroup$ The curves in 3-d space are just the projection of the Lissajous figures on the sphere . Or did I misunderstand your question.? $\endgroup$ – trula Sep 18 at 20:40
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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.

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