# Vertical Spring With Weight Reaching Equilibrium

## Homework Statement

A 1.25-kg mass stretches a vertical spring 0.215 m. If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again?

m_object = 1.25 kg
A (amplitude) = 0.130 m (since the spring has a new equilibrium at 0.215 m due to the object)
g (gravity) = 9.80 m/s^2
x_object displacement on spring = 0.215 m

## Homework Equations

k = F/x = (mg)/x
T (period) = 2*π*√(m/k)

## The Attempt at a Solution

k = F/x = [(1.25 kg)(9.80 m/s^2)]/0.215 m ≈ 56.98 N/m
T = 2*π*√(m/k) = 2*π*√(1.25 kg/56.98 N/m) ≈ 0.931 s (seconds)
0.931 s (seconds) / 4 ≈ 0.233 s
[I divided by 4 since a period is an oscillation from one amplitude to the other, and back. Since it is at one amplitude and needs only reach the equilibrium once again, that is 1/4 of the distance of a period, and therefore 1/4 of the time.]
Therefore, it will take approximately 0.233 seconds to reach equilibrium again.

^^^Does all this look correct?