Displacement x of simple harmonic oscillation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
6 replies · 3K views
deezy
Messages
18
Reaction score
0

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]
 
Physics news on Phys.org
Your formula for kinetic energy isn't correct.
 
You also need another equation (or law) regarding energy.
 
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]
 
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.
 
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?
 
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.