Harmonic Oscillator

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

Andrew Mason

I don't think those answers are right.

The potential energy for a harmonic oscillator is:

[tex]PE = \frac{1}{2}kx^2[/tex]

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

[tex]E = \frac{1}{2}kA^2[/tex]

So [tex]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)[/tex]

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 [itex]PE = E/1.5 = 1/3kA^2[/itex]. You can work out the displacement from that.

AM

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

Andrew Mason

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

The answer to b), if [itex]kA^2/3 = kx^2/2[/itex] then

[tex]x = \sqrt{\frac{2}{3}}A = .8165A[/tex]

What does your book say?

AM

lydster

Because my online quiz thing automatically says if I'm right or wrong, and those answers were off. It said that I was within 10%-100% of the actualy answer. My book doesn't really say anything on that.