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Simple Harmonic Motion of a Block on a Spring

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    There's a block attached to a spring on a frictionless surface that oscillates back and forth. (Assume no damping).
    At t=0, the potential energy in the spring is 25% of the maximum potential energy.
    Kinetic energy decreases with time at t=0, and at t=2, the kinetic energy becomes 0 for the first time.

    Determine the frequency of this motion.

    2. Relevant equations
    [tex]\frac{1}{2}[/tex]mv2 + [tex]\frac{1}{2}[/tex]kx2 = Enet
    x(t) = Acos([tex]\omega[/tex]*t+[tex]\phi[/tex]o)

    3. The attempt at a solution
    Ok. In order to get Frequency, we need the Period. To get the period, we solve for [tex]\omega[/tex] by using the position function.
    To use the position function, we must first find [tex]\phi[/tex].

    So, according to the data, at t=0, the PE is 25% of the maximum potential energy. We know that the maximum potential energy is actually equal to [tex]\frac{1}{2}[/tex]kA2. Thus, 25% of the maximum potential energy is equal to [tex]\frac{1}{4}[/tex]*[tex]\frac{1}{2}[/tex]*kA2.
    Therefore, at t=0, [tex]\frac{1}{4}[/tex]*[tex]\frac{1}{2}[/tex]*kA2=[tex]\frac{1}{2}[/tex]kx2.

    Solving for x, we get that x=[tex]\frac{1}{2}[/tex]A.

    Putting that into the position function,
    @t=0
    [tex]\frac{1}{2}[/tex]A=Acos([tex]\omega[/tex]*0+[tex]\phi[/tex])
    [tex]\frac{1}{2}[/tex]=cos([tex]\phi[/tex])
    acos([tex]\frac{1}{2}[/tex]=[tex]\phi[/tex])

    [tex]\phi[/tex]=[tex]\pi[/tex]/3

    Now, solving for [tex]\omega[/tex] and the period.

    And this is where I have a problem...

    x(t)=A*cos([tex]\omega[/tex]*t+[tex]\phi[/tex])
    At t=2, the potential energy is max, meaning the kinetic energy is 0. Thus, the position x is equal to the amplitude A.

    A=A*cos([tex]\omega[/tex]*2+[tex]\pi[/tex]/3)
    Divide by A, and [tex]\omega[/tex] = 2[tex]\pi[/tex]/T
    1=cos(4[tex]\pi[/tex]/T+[tex]\pi[/tex]/3)
    Acos(1) = 0
    0 = 4[tex]\pi[/tex]/T+[tex]\pi[/tex]/3
    Subtracting by [tex]\pi[/tex]/3
    -[tex]\pi[/tex]/3 = 4[tex]\pi[/tex]/T
    -1/3=4/T
    This gives us:
    T=-12

    Which, I'm 90% sure, we cannot have.
    A negative period gives a negative frequency. Where did I mess up in this problem? :/
     
  2. jcsd
  3. Feb 7, 2010 #2
    I had a look at your solution .. maybe I cant be helpful .. but you know that cos(60)=cos(-60)=0.5 .. if you use the -60 instead of 60 (pi/3) you will end up with the right sign , since it is impossible (100%) to have a negative frequency or period they alwaye positive ..
     
  4. Feb 7, 2010 #3
    I think, expanding from this idea, it's possible to use, instead of
    0=4[tex]\pi[/tex]/3+[tex]\pi[/tex]/3
    you can do
    2[tex]\pi[/tex]=4[tex]\pi[/tex]/3+[tex]\pi[/tex]/3

    This gives f=5/12

    With a -[tex]\pi[/tex]/3, it gives a f=1/12

    Which is correct then? Hmm... Maybe someone can comment on the theories behind each of these approaches?
     
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