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## Homework Statement

A mass of 3.15 kg stretches a vertical spring 0.290 m. If the spring is stretched an additional 0.197 m and released, how

long does it take to reach the (new) equilibrium position again?

## Homework Equations

F=-kx

T=2(pi)(m/k)^.5

## The Attempt at a Solution

First I solved for k using F=kx where F is the weight of the mass

mg=kx

Then I solved for the period, T...

T=2(pi)(m/k)^.5

and the time required to reach equilibrium would be half a period so T/2 right?

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