# Displacement x of simple harmonic oscillation

1. Nov 20, 2011

### deezy

1. The problem statement, all variables and given/known data
The amplitude of simple harmonic oscillation is A = 10 cm. Find a displacement x when K = 1/6 U. Here K is a kinetic energy and U is a potential energy.

2. Relevant equations
$$KE = \frac{1}{2} k A^2$$
$$U = \frac{1}{2} k x^2$$

3. The attempt at a solution

I'm not sure the correct method of doing this problem but here is what I have attempted based on an example from my notes:

$$KE = \frac{1}{6} U$$
$$\frac{1}{2} k A^2 = \frac{1}{6} \frac{1}{2} k x^2$$
$$A^2 = \frac{1}{6} x^2$$
$$(0.1)^2 = \frac{1}{6} x^2$$
$$x \approx 0.24 m$$

2. Nov 20, 2011

### vela

Staff Emeritus
Your formula for kinetic energy isn't correct.

3. Nov 20, 2011

### Smapple

You also need another equation (or law) regarding energy.

4. Nov 21, 2011

### deezy

This may be it, I think:

$$K = \frac{1}{6} U$$
$$U = \frac{1}{2} kx^2$$
$$KE = U + K = U + \frac{1}{6} U = \frac{7}{6} U = \frac {7}{6} \frac {1}{2} kx^2$$
$$KE = \frac {1}{2} kA^2$$
$$\frac{1}{2} kA^2 = \frac {7}{6} \frac {1}{2} kx^2$$
$$A^2 = \frac{7}{6}x^2$$
$$(0.1)^2 = \frac {7}{6}x^2$$
$$x \approx 0.0926 m$$

5. Nov 21, 2011

### vela

Staff Emeritus
Looks good.

By the way, what does KE stand for? In my first post, I mistakenly thought you were referring to the kinetic energy as KE is a common abbreviation for it.

6. Nov 21, 2011

### deezy

I was using KE as the kinetic energy of the spring... wasn't sure if energy in a spring should be referred to as kinetic energy or just energy.

Should it just be E for energy?

7. Nov 22, 2011

### vela

Staff Emeritus
You're using the approximation that the spring is massless, so it has no kinetic energy. It only has potential energy. Your KE is the total energy of the system.