Displacement x of simple harmonic oscillation

[deezy]

Homework Statement

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.

Homework Equations

[tex]KE = \frac{1}{2} k A^2[/tex]
[tex]U = \frac{1}{2} k x^2[/tex]

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:

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

 

[vela]

Your formula for kinetic energy isn't correct.

 

[Smapple]

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

 

[deezy]

This may be it, I think:

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

 

[vela]

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.

 

[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?

 

[vela]

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.