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How to find the equilibrium position and potential energy?

  1. Dec 23, 2008 #1


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    1. The problem statement, all variables and given/known data
    Given a setup with spring, the massless spring constant is k, unstretched length of the spring is [tex]L_0[/tex]. A mass m is attaching to the top of the spring. We only know the height of the mass (measuring from the top of the platform) is x, the height of the platform is h. Try to find the equilibrium position of the system, the potential energy and kinetic energy of the system in terms of x.


    2. The attempt at a solution
    I setup a coordinate system with the origin on CD line, call the vertical axis X (capital X). The hooke law gives

    [tex]mg = -k\Delta X, \qquad \Delta X = x + h - L_0[/tex]

    For finding the equilibrium position ([tex]x_0[/tex]), we let [tex]\Delta X = x + h - L_0 = 0[/tex] gives [tex]x_0 = L_0 - h[/tex]

    The kinetic energy of the mass is obvious

    [tex]E_k = \frac{m}{2}\left(\frac{dX}{dt}\right)^2 = \frac{m}{2}\left[\frac{d(x+h)}{dt}\right]^2 = \frac{m}{2}\left(\frac{dx}{dt}\right)^2[/tex]

    For the potential energy, we have
    [tex]E_p = \frac{k}{2}(\Delta X)^2 = \frac{k}{2}\left(x + h - L_0)^2[/tex]

    But the answer for the potential energy is
    [tex]E_p = \frac{k}{2}x^2[/tex]

    How is that possible?
    Last edited: Dec 23, 2008
  2. jcsd
  3. Dec 23, 2008 #2
    There are a couple of things which Im not too sure about. Is it given that x+h is the equilibrium position of the system (ie the system is in equilibrium in the diagram?). If it isnt, then your equation [tex]\Delta x=x+h-L_0[/tex] is not valid.

    Even if the system is in equilibrium, why would [tex]\Delta x=x+h-L_0=0[/tex] give you the equilibrium position? That would give you the initial position of the spring when its uncompressed.
  4. Dec 23, 2008 #3


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    Thanks for reply. The problem doesn't tell x+h is the equilibrium position. Well, for vertical spring, I think the equilibrium position is where the force (or acceleration) is ZERO. So I let mg=0 that gives [tex]\Delta X=0[/tex]. Am I doing anything wrong? If so, please help me. Thanks

  5. Dec 23, 2008 #4
    If you assume cd to be the origin line, then let the distance of the block from the line be y. As such, the potential energy of the block is given by mgy.

    Also, the kinetic energy of the block is [tex]\frac{1}{2}mv^2[/tex] where [tex]v=\frac{dy}{dt}[/tex].

    Now you have to express v2 in terms of y to get the expression for kinetic energy.

    To do this you need express find the force equation of the block in terms of y. This gives you [tex]F=mg-k(x+h-y)[/tex]. From this you can get the value of [tex]\frac{v^2}{2}[/tex] and hence the expression for kinetic energy.

    Now you know the expressions for KE and PE. Add them up and you get total energy. At equilibrium, the total energy must be minimum. You can find the minima of the system as the expression for the total energy will be a function of y, which will give you the equilibrium position.

    I know its a little long, maybe someone else can come up with a shorter way to do it.
  6. Dec 23, 2008 #5
    Youre right, the force must be zero, but that is the NET force acting on the block. In this case, the net force is the gravitational force MINUS the force of the spring. That value must be zero, not mg.
  7. Dec 23, 2008 #6


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    Ok, I just update the figure and add the coordinate system. With your hints, I reconsider the solution and note that the problem require to express all answers in terms of [tex]x[/tex].

    1) First calculate the kinetic energy, I know that (as you told) the the kinetic energy will be

    [tex]E_k = \frac{m}{2}\left(\frac{dy}{dt}\right)^2[/tex]

    Since [tex]y=x+h[/tex], so we also have

    [tex]E_k = \frac{m}{2}\left(\frac{dx}{dt}\right)^2[/tex]

    2) For finding the equilibrium position, we must set the total force to zero, let the unstretched length of spring be [tex]L_0[/tex]

    [tex]F = mg - k(x+h - L_0) = 0[/tex]

    gives the equilibrium position (measure from the top of the platform) be

    [tex]x_0 = \frac{mg}{k}(L_0 - h)[/tex]

    3) The elastic potential energy in terms of [tex]x[/tex] is
    [tex]E_{ep} = \frac{k}{2}(x - x_0)^2[/tex]

    Hope these results are correct :)

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