posted by david on .
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.
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.