Simple Spring Help

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

3. 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?

 

It'd go from -A, through its equilibrium position, to A, back through equilibrium, then to –A and repeat.

Did I at least solve for “k” correctly?

Also, why care about amplitude? The period has been derived and proved to be independent of amplitude.

The time required for the mass to go from state to the exact same state again (same position and velocity) is one period.

So instead of doing T/2 it’s just T right?

 

Each position change takes T/4, and there are 4 of them...

The time required for it to go from some initial velocity and position back to the exact same velocity and position is one period.

How is that not what the problem is asking for?

 

I'm assuming when the problem states "new" equilibrium position that they mean the .290+0.197 position, which would be the starting and ending position, and it would take a period to go from start to finish.

 

Ohhhh. So it's 3/4T?

 

Also, so you think when they say "new" equilibrium position they mean the .29?

 

Please tell me it's 1/4T then...