# P(x)dx of a one-dimensional harmonic oscillator

1. Sep 15, 2008

### Buce

The displacement of a harmonic oscillator is given by

x = A Cos(wt + Phi)

The phase angle phi is equally likely to have any value in the range 0 < Phi < 2Pi, so the probability W(Phi) that Phi lies between Phi and Phi + delta-Phi is delta-Phi/(2Pi). For a fixed time t, find the probability P(x)dx that x lies between x and x+dx by summing W(Phi)delta-Phi over all angles for which x lies in this range. Express P(x) in terms of A and x.

I know that for a harmonic oscillator, the total energy is

E = p^2/(2m) + kx^2/2

where the momentum p is

p = mx' = -mAwSin(wt + Phi)

I can solve the problem using px phase space, where the area between E and E+dE forms a elliptical band of area s, and P(x)dx = (dp/dE)(dE/ds)dx, but I'm only supposed to do it that way for the second part of the problem (which I won't reproduce here). If I did that part correctly, the answer is dx/(Pi Sqrt(A^2 - x^2)), but I'm having a hard time setting up the equations needed to evaluate an integral with respect to the phase angle.

2. Oct 18, 2008

### jimmymo82

i've got the exact reverse problem. i can get your result by summing W(phi) over all angles for which x lies, but i'm off by a factor of 1/(mw^2A) when i try the ratio of areas in phase-space. let's help each other out