# Harmonic Oscillator kinetic and potential energies

1. Apr 12, 2006

### lydster

A simple harmonic oscillator has a total energy of E.

(a) Determine the kinetic and potential energies when the displacement is three-fourths the amplitude. (Give your answer in terms of total energy E of the oscillator.)

Kinetic energy ______________ x E <----(times E)

Potential energy _____________ x E <-----(times E)

(b) For what value of the displacement does the kinetic energy equal one half the potential energy? (Give your answer in terms of the amplitude A of the oscillator.)

_________________ A

I followed an example from the book, which was the same question, except for A and B it was one-half, and their answers are for (a) Kinetic is 3/4 E and potential is 1/4 E. and I have no clue on B

Thanks

Last edited: Apr 12, 2006
2. Apr 13, 2006

### Andrew Mason

I don't think those answers are right.

The potential energy for a harmonic oscillator is:

$$PE = \frac{1}{2}kx^2$$

The total energy is the PE when KE=0 which occurs at maximum amplitude. ie total E is:

$$E = \frac{1}{2}kA^2$$

So $$KE = E - PE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kA^2(1 - (\frac{x}{A})^2) = E(1 - (\frac{x}{A})^2)$$

where x is the displacement and A is the maximum amplitude.

So for a), if displacement is 3/4 of A, then KE = 7/16 of E and PE is 9/16 of E

For b) if KE = .5PE, then $PE = E/1.5 = 1/3kA^2$. You can work out the displacement from that.

AM

Last edited: Apr 13, 2006
3. Apr 13, 2006

### lydster

Yeah I got a friend to try those numbers, and they didn't work out. He got the same answers as you, and they are wrong. Hmmmm...I dunno. Everything that you said makes sense

4. Apr 13, 2006

### Andrew Mason

What makes you think the 7/16 , 9/16 answer is wrong?

The answer to b), if $kA^2/3 = kx^2/2$ then

$$x = \sqrt{\frac{2}{3}}A = .8165A$$