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?
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)
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.