# Physics

.A mass of 1.61 kg stretches a vertical spring 0.313 m. If the spring is stretched an additional 0.132 m and released, how long does it take to reach the (new) equilibrium position again?

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1. Assuming this is a physics question, and not a differential equation problem, I will just apply the established formulas:
y(t)=A sin(ωt)
where
A=maximum displacement from equilibrium position, = 0.132m
ω=frequency=√(k/m)
m=mass = 1.61 kg
k=stiffness of string=m/d=1.61/0.313

From the extreme lowest point to the equilibrium position, it is one-quarter of the cycle, or (1/4)*2π=(1/2)π.

Thus, solve for t in
ωt=(1/2)π

I get ω=1.787
and consequently t=0.88 s.

Check my calculations

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