# A simple harmonic oscillator has total energy E= ½ K A^2

1. May 3, 2014

### Myr73

A simple harmonic oscillator has total energy
E= ½ K A^2
Where A is the amplitude of oscillation.
 E= KE+PE
a) Determine the kinetic and potential energies when the displacement is one half the amplitude.
b) For what value of the displacement does the kinetic energy equal the potential energy?

For a) I got ,
KE=?= 0.5mv^2
PE=?= mgy= 0.5 Kx^2
X= 0.5 A
0.5KA^2= 0.5mv^2 + 0.5Kx^2
0.5KA^2= 0.5mv^2 + 0.5K (0.5A)^2
KE= 0.5KA^2- 0.5K (0.5A)^2 = 0.5k(A^2- 0.5A^2)
PE= 0.5KA^2-0.5mv^2

And for b) all I have so far is
,Kinetic Energy= 1/2mv^2
Potential Energy = 1/2mx^2
So when KE=PE, then 1/2mv^2=1/2kx^2.
If you times by 2 then mv^2= kx^2 , therefore→ SquareRoot(mv^2/k)= x

But I don't know if that is correct or if it missing information

2. May 3, 2014

### dauto

Your answers should be in terms of variables given by the problem. The speed v is not given so you must correct any of your answers that depend explicitly on v. Make sure your answers is in terms of A and k. That's all that's given.

3. May 3, 2014

### dauto

Also, make sure you simplify your final answer as much as possible.

4. May 3, 2014

### Myr73

sorry.. is that for b or a as well?

5. May 3, 2014

### dauto

You did not finish simplifying the expression for the kinetic energy K

6. May 3, 2014

### Myr73

How do I do that?

7. May 4, 2014

8. May 9, 2014

### Myr73

Oh I think I got it.. a)KE= E-PE= 0.5KA^2- 0.5K (0.5A)^2=0.5KA^2- 0.25(0.5KA^2) (Note; 0.5KA^2=E) E-1/4E= 3/4E
PE= E-KE= E-3/4= 1/4E
b)Kinetic Energy= ½ mv^2
Potential Energy = ½ kx^2
So when KE=PE, PE is half of the energy,
therefore ½ Kx^2= ½ { ½ KA^2} x= SquareRoot{½ A^2} = A/ (SquareRoot 2)

9. May 10, 2014

### dauto

Looks right now