A 1.0 kg object is suspended from a vertical spring whose spring constant is 105 N/m. The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.

Answer 1

The original position is the equilibrium position, and the velocity is a maximum there. Let that velocity be Vmax

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

Vmax = X*sqrt(k/M)

k is the spring constant
X = 0.25 m
m = 1.0 kg

I have not considered gravitational potential energy in this derivation. As I recall, its effect cancels out when you define X as being measured from the equilibrium position, with stretching due to the weight taken into account. I am too lazy to verify this.