Solving Spring Oscillations Homework: Stretched Distance d

[sheepy]

Homework Statement

When a mass of 0.35 kg is attached to a vertical spring and lowered slowly, the spring stretches a distance of d. The mass is now displaced from its equilibrium position and undergoes 100 oscillations in 48.9 seconds. WHat is the stretched distance d?

Homework Equations

Number of oscillations/Time = Period (T)
T = 2pi ($$\sqrt{m/k}$$)
F=kx

The Attempt at a Solution

I found by doing 100/48.9, the period is around 2 oscillations per second.
Using the second equation I figured out that k=3.45 by plugging it all the other variables. Then i know that the F=mg. So mg is 3.43.
3.43/3.45 = x (distance)
The distance comes out to around 1. But my teacher says that 5.94 cm...what did i do wrong?

 

[Leong]

the period is supposed to be 48.9/100 s