# Spring & Oscillation

## Homework Statement

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure

http://session.masteringphysics.com/problemAsset/1073872/3/knight_Figure_14_36.jpg
What is her mass if the spring constant is 240N/m

Hookes Law
F = -kx
U = (1/2)*k*x2

## The Attempt at a Solution

So the spring was pulled back 1.4 metres and then released where it went to the point 0.6
so 0.6 is the point at equilibrium, so -0.8 is the displacement and that is the x

F = (240 * 0.8)
and then im not sure what to do, im not even sure if im doing this right and help would be appreciated

## Answers and Replies

im pretty sure my approach is wrong...

F = -260 * -0.8 = 208N

mg = weight

208/9.8 = m

m = 21.224489795918367346938775510204kg

im sure thats wrong is there a better way to approach this problem and why is the way im approaching it wrong?