# What is the potential energy of the spring?

1. Jan 7, 2010

### zachmgilbert

1. The problem statement, all variables and given/known data
What is the potential energy of the spring?
mass=.52 kg
velocity=-6.54sin(20.1t)
time=40.84 seconds
amplitude=.33 meters
angular frequency=20.1

2. Relevant equations
U =1/2kx2

3. The attempt at a solution
I used x=Acos(wt) and got .2 meters for x. I know this is right.I don't know how to find k.

2. Jan 7, 2010

### JaWiB

Re: Kinetic

x=Acos(wt)

This solution only satisfies the differential equation $$m\ddot{x}=-kx$$ if $$\omega^2 = k/m$$

3. Jan 7, 2010

### Char. Limit

Re: Kinetic

What is $$\omega^2$$?

4. Jan 7, 2010

### zachmgilbert

Re: Kinetic

angular frequency squared

5. Jan 8, 2010

### Char. Limit

Re: Kinetic

OK.

If I may ask, what is the independent variable in x=Acos(wt)? I'd like to solve it...

6. Jan 8, 2010

### Winzer

Re: Kinetic

Hello Char!

The independent variable is time(t) right? The position(x) depends on it.

You now have x,t & k, you should be able to solve for U.

7. Jan 8, 2010

### Char. Limit

Re: Kinetic

Ah, I have the an-word (answer comes from an-swear, meaning "an-word")

I actually wanted to prove the diff e.q. That was mentioned, but I got carried away. The answer is..

Aaaaaaaauuuggh......
Hello Winzer!

8. Jan 8, 2010

### ideasrule

Re: Kinetic

Then how come the amplitude is 0.33 m?

Anyhow, if you're absolutely sure it's 0.2 m and not 0.33 m, w^2=k/m.