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Homework Help: N-D spring-mass damper total energy

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    I am trying to derive the total energy for an n-dimensional linear (Hookeian) spring-mass damper system and show that the total energy is either decreasing or constant.

    2. Relevant equations
    1D spring-mass damper equation
    n-D spring-mass damper equation

    3. The attempt at a solution

    I started with the following force equation in 1D to describe the system:

    (1) m[tex]\ddot{x}[/tex] + b[tex]\dot{x}[/tex] + kx = 0

    m[tex]\ddot{x}[/tex] = Facc = m[itex]\frac{d^{2}x}{dt^{2}}[/itex]
    b[tex]\dot{x}[/tex] = Fdamp (viscous friction) = C[itex]\frac{dx}{dt}[/itex]
    kx = Fpotential (gradient of some potential) = [tex]\nabla[/tex]P(x)

    So the equation becomes:

    (2) -[tex]\nabla[/tex]P(x) - C[itex]\frac{dx}{dt}[/itex] = m[itex]\frac{d^{2}x}{dt^{2}}[/itex]

    I then modeled the total energy as follows:

    (3) Etot = P(x) + [itex]\frac{1}{2}[/itex]m[tex]\dot{x}[/tex]2


    (4) [itex]\frac{dE}{dt}[/itex]= [tex]\nabla[/tex]P(x)[tex]\dot{x}[/tex] + m[tex]\dot{x}[/tex][tex]\ddot{x}[/tex]

    In (4) I substituted m[tex]\ddot{x}[/tex] from (1) and got:

    [itex]\frac{dE}{dt}[/itex]= [tex]\nabla[/tex]P(x)[tex]\dot{x}[/tex] + [tex]\dot{x}[/tex](-[tex]\nabla[/tex]P(x) - C[tex]\dot{x}[/tex])

    => -C[tex]\dot{x}[/tex]2

    ....which shows that total energy is either decreasing (|[tex]\dot{x}[/tex]| > 0) or constant ([tex]\dot{x}[/tex] = 0)

    I then tried to derive the total energy for the n-dimensional case and relate it to an n-D force equation as I did above. To do this I looked at the 2D case and then generalized the formulas for kinetic and potential energy to n-D but I'm not sure if this was correct. Here are my kinetic and potential energy equations for 2D:
  2. jcsd
  3. May 2, 2010 #2
    My latex seems to bug out sometimes so if it looks weird please click on the source but let me continue...

    KEtot,2D = [itex]\frac{1}{2}[/itex](m11[tex]\dot{x}[/tex]12 + m22[tex]\dot{x}[/tex]22 + m12[tex]\dot{x}[/tex]1[tex]\dot{x}[/tex]2)

    PEtot = [itex]\frac{1}{2}[/itex](m11[tex]{x}[/tex]12 + m22[tex]{x}[/tex]22 + m12[tex]{x}[/tex]1[tex]{x}[/tex]2)

    I took the generalized forms to be:

    KEtot,nD = [itex]\sum_{i=1}^{n}[/itex] (mii)[tex]\dot{x}[/tex]i2 + [itex]\sum_{j=1}^{n-1}[/itex] [itex]\sum_{k=j+1}^n[/itex] (mjk)[tex]\dot{x}[/tex]j[tex]\dot{x}[/tex]k

    ...and the same for PE with [tex]{x}[/tex] replacing [tex]\dot{x}[/tex].

    My plan was to take the derivatives of these and replace the m[tex]\ddot{x}[/tex] terms with the force equivalences (as above). Is this right? I also can't think of how to make a generalized form of the force equation for an n-D vector.

    Any help would be greatly appreciated.


    Last edited: May 2, 2010
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