# Oscillator with equal amounts of kinetic and potential energy

1. Jan 20, 2009

### Dwrigh08

1. The problem statement, all variables and given/known data

If you have a simple harmonic oscillator based on a spring of force constant 5.79 N/m, with an attached mass of 0.828 kg, and the oscillator is initially displaced 3.5 cm from equilibrium, at what distance from the equilibrium point (in cm) will the oscillator have equal amounts of kinetic and potential energy?

2. Relevant equations

I am using Kinetic energy=1/2 mass x omega^2 and Potential energy= 1/2 spring constant x displacement^2 and omega= root of (spring constant/ mass)

3. The attempt at a solution

My first thought was to set the kinetic energy and pitential equal to each other and solve for the displacement. When I try to do this I end up with 1. My next thought was to take into account the initial displacement and subtract that from 1. Neither work for me. I feel like i am missing an obvious step. Any help would be appreciated.

2. Jan 21, 2009

### rl.bhat

Can you wright down the expressions for kinetic energy and potential energy for a SHM in terms of k, amplitude and displacement?

3. Jan 21, 2009

### Dwrigh08

I know that total energy= 1/2 k A^2. I tried to solve for total energy and then using E= K+U and rewriting E= 1/2 m w^2 + 1/2 k x^2. I come out with an energy from the first equation and then try to solve for x using the second equation. I dont think that is the right approach.

4. Jan 21, 2009

### rl.bhat

In SHM, at any instant, the kinetic energy = 1/2*m*w^2(A^2-x^2) = 1/2*k*(A^2-x^2)and potential energy = 1/2*k*x^2
Equate KE = PE and solve for x.
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