A harmonic oscillator with a vertical mass on a string has a hanging mass of 2m and a spring constant of K. It oscillates with an amplitude of Z. When its position is at a distance Z/2 of the equilibrium point, its potential energy is Ui. What is the maximum kinetic energy Km?
E=1/2mV2 + 1/2mA2
The Attempt at a Solution
I realized that the max kinetic energy must be equal to the to the max potential energy because total energy is constant. I just don't know where to put the equations to make it fit.
Can someone show me the set up? The answer give was Km = 4Ui
Thanks in advance!