U = (1/2)kx^2
T = 2 * π * (m/k)^ (1/2)
The Attempt at a Solution
I said that the angular velocity is
w = (2 * π) / t
And the equation for velocity would be:
v(t) = -Aw * cos (w * t)
However, I don't know how to relate this to the speed at the bottom. I was thinking of using conservation of energy, but I don't have the height.