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What is the potential energy of the spring?

  1. Jan 7, 2010 #1
    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. jcsd
  3. Jan 7, 2010 #2
    Re: Kinetic

    x=Acos(wt)

    This solution only satisfies the differential equation [tex]m\ddot{x}=-kx[/tex] if [tex]\omega^2 = k/m[/tex]
     
  4. Jan 7, 2010 #3

    Char. Limit

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    Gold Member

    Re: Kinetic

    What is [tex]\omega^2[/tex]?
     
  5. Jan 7, 2010 #4
    Re: Kinetic

    angular frequency squared
     
  6. Jan 8, 2010 #5

    Char. Limit

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    Gold Member

    Re: Kinetic

    OK.

    If I may ask, what is the independent variable in x=Acos(wt)? I'd like to solve it...
     
  7. Jan 8, 2010 #6
    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.
     
  8. Jan 8, 2010 #7

    Char. Limit

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    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!
     
  9. Jan 8, 2010 #8

    ideasrule

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    Homework Helper

    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.
     
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