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

  1. Sep 15, 2008 #1
    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. jcsd
  3. Oct 18, 2008 #2
    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
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