# Using Generalization of Bohr Rule for 1D Harmonic Oscillator

1. Oct 15, 2013

### Shallac

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