Question

A 0.300 kg mass is attached to a 26.6 N/m spring. It is pulled 0.120 m and released. How much potential energy does it have when it is 0.0600 m from equilibrium? (Unit = J)

Answer 1

The sum of kinetic and potential energy is conserved in a simple harmonic oscillator.
We are told the mass is starting at an maximum elongation x=0.12 m with 0 speed.
So total energy at that point is entirely held in the potential energy kx² / 2, because the kinetic energy is 0.
E = kx² / 2 + 0

E = (26.6 N/m * 0.12²)/2 = 0.1915 J

When the mass is getting pulled towards equilibrium point, this stored potential energy is being turned into kinetic energy of the mass.
At x=0.06 m potential energy is only
Ep = kx²/ 2

(26.6 N/m * 0.06²)/2

Ep = 0.04788 J