Given that the position of a one-dimensional harmonic oscillator is given by
$$x(t) = Acos(\omega t + \phi)$$
where $A,\phi,\omega \geq 0$ are constants of real numbers I'm trying in some sense
find the $p(x)\mathrm dx$ probability of finding the oscillator between position $x$ and $x+\mathrm d x$.
There is a hint that this is the same as calculate $\mathrm dT/T$ where $T$ is the period of oscillation and $\mathrm dT $ is an interval of time within the period.
My attempt: So, I'm trying to calculate using $\mathrm dT/T$ but I do not know how to do it. If we have that $x(t+T) = x(t)$ then we use $\mathrm dx / \mathrm dT = (\mathrm d x/\mathrm dt)(\mathrm dt/\mathrm dT )$ but this do not go much further. I also tryed to look at other derivatives as a trick of considering the constants as variables but for all the constants I got stuck in calculations that didn't go anywhere; for exemple for $\omega$ I just found $\mathrm d T/ T = -\mathrm d \omega/\omega$. Do not give a full answer, give just a hint so that I can conclude the question.
Physical concept involved: This is a statistical mechanics half-problem; the harmonic oscillator is one of the few examples of problems that it is possible to chek the validity of two important elements of the theory: Ergodic Hypothesis and equal-a-priori postulate.
Answer
Start with energy conservation $$ \frac{1}{2} m \left(\frac{dx}{dt}\right)^2 + \frac{1}{2} k x^2 = E $$ Note that the amplitude $A$ can be found from setting $dx/dt = 0$ in the above and solving for $x$.
Now solve the above for $dt$ and consider the motion from $x=-A$ to $x=+A$.
You should find that the probability density is $$ p\left(x\right)dx = \frac{dt}{T/2} = \frac{1}{\pi A}\frac{dx}{\sqrt{1-\left(\frac{x}{A}\right)^2}} $$
where $T$ is the period.
You should get the same result if you start with your expression for $x\left(t\right)$, note that $dt = -dx / \left[A \omega \sin\left(\omega t + \phi\right)\right]$, and use $\cos^2\theta + \sin^2\theta = 1$.
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