Is Probability of Oscillator Displacement x?

[catsonmars]

I am pre studying for Statistical Mechanics class in the fall and need help with this problem. I’ve already spent some time with it.

Let the displacement x of an oscillator as a function of time t be given by X=Acos(wt+ϕ). Assume that the phase angle ϕ is equally likely to assume any value in the range 0 < ϕ < 2pi. The probability w(ϕ)d ϕ that ϕ lies in the range between ϕ and ϕ +d ϕ is then simply w(ϕ) dϕ=(2pi)^-1d ϕ. For any fixed time t, find the probability P(x)dx that x lies between x and x+dx by summing w(ϕ) over all angles ϕ for which x lies in this range. Express P(x) in termas of A and x.


Relevant equations[/b]
X=Acos(wt+ϕ).
w(ϕ)d ϕ=(2pi)^-1d ϕ

The Attempt at a Solution

The only thing I can come up with is integrating
∫P(x)dx = ∫(2pi)^-1d ϕ and inegrating over x and x+dx
Or ƩP((x)dx* w(ϕ) dϕ)/p(x)

 

[TSny]

Hello catsonmars. Welcome to PF!

You have a good start with $w(\phi) = \frac{1}{2\pi}$.

Since you are looking for the probability that $x$ lies in an infinitesimal range from $x$ to $x+dx$, you will not need to integrate. The probability is just $\small P(x)dx$. This is given by the probability $w(\phi)|d\phi|$ that $\phi$ lies in the range $\phi$ to $\phi + d\phi$, where $\phi$ is the value of the phase angle that corresponds to $x$ and $\phi + d\phi$ corresponds to $x+dx$. [Caution: think about whether or not there is more than one value of $\phi$ that corresponds to the same $x$. If so, you will need to make an adjustment for that.]

So, the probability that $x$ lies between $x$ and $x+dx$ could be expressed as $\small P(x)dx$ or as $w(\phi)|d\phi|$ (if there is only one value of $\phi$ that corresponds to a value of $x$). That is, $\small P(x)dx =$ $w(\phi)|d\phi|$ [I leave it to you to think about what to do if there is more than one value of $\phi$ corresponding to the same value of $x$. Perhaps this has something to do with the word "summing" in the statement of the problem.]

Since you already know how to express $w(\phi)$, all you need to do is find an expression for $d\phi$ in terms of $x$ and $dx$. Hint: $d\phi = \frac{d\phi}{dx}dx$.

 

[catsonmars]

There should be more x values than ϕ's because the range of ϕ is much smaller than x. Second I'm still not sure how I should right the summation. I have ϕ=(Ʃw(ϕ)dϕ)/(P(x)(dx)) but that still seems wrong. I've also thought about relating ϕ+dϕ to x+dx somehow but I'm can't think of what would make them equal so I can get ϕ in terms of x. Also I've looked at the answer key and I have no idea how the amplitue "A" would fit into the equation.

 

[TSny]

Basically, you need to solve $\small P(x)|dx| = w(\phi)|d\phi|$ for $\small P(x)$. That is,

$P(x) = w(\phi)\frac{d\phi}{dx}$

Use $x = Acos(\omega t + \phi)$ to find $\frac{d\phi}{dx}$ as a function of $A$ and $x$.

There's the additional task of dealing with the fact that there might be two different values of $\phi$ corresponding to the same value of $x$.