# Trying to Calculate k, using Hooke's Law

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1. Nov 25, 2016

### masterexploder

1. The problem statement, all variables and given/known data
The position of a 49 g oscillating mass is given by x(t)=(1.8cm)cos12t, where t is in seconds.

2. Relevant equations
k=mg/x

3. The attempt at a solution
I've tried working this problem multiple different ways and it is just not working for me.
I used k= (.049*9.8)/.018
Is this correct with the information I've been given? I used up all my attempts and the solution is apparently
7.1 N/m...but I keep coming up with roughly 26.7 N/m

2. Nov 25, 2016

### BvU

Hello ex master,
It is not. One can stretch a spring wrt its equilibrium state, and then let go. The amount of stretch becomes the amplitude of the ensuing oscillation so it has little to do with the spring constant.
You have been given another bit of info that does have a relationship with the spring constant. Can you guess which bit ?

3. Nov 25, 2016

### Cutter Ketch

kx = mg answers the question "how much does the spring stretch when I add this mass". It is an equilibrium answer. The problem didn't tell you how long the spring was before you added mass, or for that matter how long the spring was after the mass was added. What you have is how the spring oscillates about the equilibrium point. Do you have any other equations or ideas that might apply?

4. Nov 25, 2016

### haruspex

Yes, but even then that is only in a vertical context.
@masterexploder , there is nothing in the question about the spring being vertical. This could be happening on a smooth horizontal surface, so you have no basis for involving g.

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