Finding distribution for a pendulum

Question

We are considering the equation of pendulum $$ x''+x=0, $$ with the initial conditions $x(0)=A \sin \varphi$, $x'(0)=A \cos \varphi$ ($A$ is an amplitude, $\varphi$-intitial phase). How to determine distribution of probability for coordinate $x(t)\in[-A,A]$ for $t>0$ knowing that initial phase $\varphi$ has uniform distribution on interval $(-\pi,\pi)$? For which $x \in [-A,A]$ this density (for obtained distribution) is little and for which big?

Any advices will be strongly desirable, because I'm weak in physics.

Thanks.

Answer 1

It can be shown quite easily that at any $t>0$ we have $X=A\sin(\varphi+t)$. Now suppose $\Theta \sim \mathcal{U}(0,2\pi)$ and fix $x\in[-A,A]$. It follows that $$F_{X}(x)=\mathbb{P}(A\sin(\varphi+t)\leq x)=\mathbb{P}\Big(\sin(\Theta) \leq \frac{x}{A}\Big)=\frac{1}{2}+\frac{1}{\pi}\arcsin\Big(\frac{x}{A}\Big)$$ Therefore for $x\in(-A,A)$ $$f_{X}(x)=\frac{d}{dx}\Big[F_X(x)\Big]=\frac{1}{\pi\sqrt{A^2-x^2}}$$ This is the formulated density on $(-A,A)$. The density vanishes elsewhere.

Answer 2

The density of $x$ is inversely proportional to the velocity there because when it is moving fast it gets away from a given point quickly. The velocity is proportional to the square root of the kinetic energy at a point.