What is the potential energy of the spring?

AI Thread Summary
The discussion revolves around calculating the potential energy of a spring using the formula U = 1/2 kx², where the position x is derived from the equation x = Acos(wt). The participants confirm that the independent variable in this equation is time (t), which affects the position (x). There is confusion regarding the spring constant (k) and the relationship between angular frequency (ω) and k, as it is established that ω² = k/m. The conversation highlights the importance of correctly identifying parameters to solve for potential energy, emphasizing the need to clarify the values of x and k.
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Homework Statement


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


Homework Equations


U =1/2kx2



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.
 
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x=Acos(wt)

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


What is \omega^2?
 


angular frequency squared
 


OK.

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


Char. Limit said:
OK.

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

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.
 


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!
 


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.

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.
 
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