# Homework Help: Generalization of the bohr rule for harmonic oscillators

1. Nov 28, 2012

### uppiemurphy

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 bolded letters represent vectors.

Using the generalized, show that the spectrum for the one-dimensional harmonic oscillator, for which E = p2/2m + mw2x2/2 is E = nh(bar)w.

2. Relevant equations

2∏x = nλ, px = nh(bar)

3. The attempt at a solution

Basically I know how to get E = nh(bar)w for a harmonic oscillator without using integrals, but I'm confused as to how to express E as an integral which is what I assume they're asking for.

I know that the total energy of the system when the spring is fully stretched is Etot = mw2x2/2. Do I somehow have to write this in terms of momentum and then integrate? I'm probably missing something fairly obvious here, but how would I write that in terms of P?

2. Nov 28, 2012

### TSny

Hello, uppiemurphy.

It's not that you need to express E as an integral. Rather, you need to express the result of the integraton ∫pdx in terms of E. The quantization of E will then come from the condition ∫pdx = nh.

To perform the integration ∫pdx you'll need to express p as a function of x for a specific energy E. Note E = KE + PE. See if you can express KE in terms of p and PE in terms of x. [Edit: I see that the expression for E in terms of p and x is already given in the problem. Use it to get p as a function of x so you can do the integration.]

Last edited: Nov 28, 2012
3. Nov 29, 2012

### uppiemurphy

I'm just confused as to what to do with E mainly... when I isolate my expression for p I'm left with p = sqrt(2mE-m^2w^2x^2) how do I integrate this when I have E in my expression for p?

4. Nov 29, 2012

### TSny

E is just a constant of the motion. So, you have $p = \sqrt{a-bx^2}$ where a and b are constants.

5. Nov 29, 2012

### uppiemurphy

I don't see how that integral would yield the correct answer at all though... You end up with a pretty complicated expression that is difficult to simplify

6. Nov 29, 2012

### uppiemurphy

nevermind I got it! thank you for your help!