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Using Generalization of Bohr Rule for 1D Harmonic Oscillator

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data

    The generalization of the bohr rule to periodic motion more general than circular orbit states that:
    ∫p.dr = nh = 2∏nh(bar).

    the integral is a closed line integral and the "p" and "r" are vectors

    Using the generalized rule (the integral above), show that the spectrum for the one-dimensional harmonic oscillator, for which E = p^2/2m + (1/2)m*ω^2*x^2, is E = nh(bar)ω.

    2. Relevant equations

    px = nh(bar)
    p= m*ω*x

    3. The attempt at a solution

    I have put p in terms of E and x, to get p=sqrt(2mE-m^2*ω^2*x^2).

    I took the indefinite integral of that expression with respect to x, since the problem is one-dimensional, and got (1/2)x*sqrt(2mE-m^2*ω^2*x^2) + (E*arctan((m*ω*x)/(sqrt(2mE-m^2*ω^2*x^2))))/ω. Both WolframAlpha and my TI-89 give this same answer.

    I realize here that, replacing sqrt(2mE-m^2*ω^2*x^2) with p and p with m*ω*x, the arctangent simplifies to arctan(1)=pi/4 and the first term simplifies to (1/2)x*p= (1/2)nh(bar). However, I'm not sure where to go from here, or even positive if this is correct.

    I know--or rather, think I know--that I'm supposed to set this integral equal to 2∏nh(bar) and solve for E.

    However, the professor indicated we needed to find the endpoints where "the kinetic energy is zero." I know one of these points would be 0, but what's the other? Does it even matter here?

    I'm really not seeing how I can get a result of E=nh(bar)ω. Even if I plug in E=nh(bar)ω into my indefinite integral I'm not coming up with the right answer.


    Please, any help is very greatly appreciated!


    EDIT: I have also realized that, since the integral is a closed loop, my integral should essentially be multiplied by 2 to account for the trip back.
     
    Last edited: Oct 15, 2013
  2. jcsd
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