# Spring Oscillations

1. Dec 16, 2007

### sheepy

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

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

3. 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?

2. Dec 16, 2007

### Leong

the period is supposed to be 48.9/100 s