A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. What is the speed of the 1.0 kg mass at the equilibrium position?

Answer 1

Maximum kinetic energy (which occurs at the equilbrium position) equals maximum stored spring potential energy.

Therefore

(1/2) M Vmax^2 = (1/2) k X^2

Vmax = X*sqrt(k/M)

X is the amplitude, 0.100 m.