Question

A .0500Kg mass is attached to a spring (with a spring constant of 23.3 N/m) and undergoing simple harmonic motion. The total mechanical energy of the system is 7.60 J. When the mass has a velocity of 4.15 m/s, how far is it from the equilibrium position?

Answer 1

$$E=K+U=0.5mv^2+0.5kx^2\\2(7.6)=(0.05)(4.15)^2+(23.3)x^2\\x=0.784\ m$$