Generalization of the bohr rule for harmonic oscillators

[uppiemurphy]

Homework Statement

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.

Homework Equations

2∏x = nλ, px = nh(bar)

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?

 

[TSny]

Hello, uppiemurphy.

It's not that you need to express E as an integral. Rather, you need to express the result of the integration ∫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.]

 

[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?

 

[TSny]

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

 

[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

 

[uppiemurphy]

nevermind I got it! thank you for your help!