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Linear damping model

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data
    I am asked to write down the equation of motion of [tex]m[/tex] with respect to O, hence show that [tex]x(t)[/tex] satisfies the differential equation.


    See attached diagram.

    2. Relevant equations
    I have the force of the spring H, the Damping R and the weight W as


    3. The attempt at a solution
    Using the above forces and [tex]\bold{F}=m\ddot{x}[/tex], what I get is

    I take it I am asked to find the general solution? Where does the [tex]m\ddot{y}[/tex] come from? Do I have the forces correct?

    Thanks, James
    Last edited: Jun 23, 2010
  2. jcsd
  3. Jun 22, 2010 #2
    I think that y(t) is known function of time.
    So coordinate of m is
    as y(t) is known function
  4. Jun 23, 2010 #3
    Thanks, so this would give me

    [tex]-m\ddot{x}-r\dot{x}-kx=-mg-kl_0-m\ddot{y}[/tex] and multiplying both sides by -1


    Ok, so how would I find the particular integral seeing that the [tex]m\ddot{y}[/tex] term is present? Would it be the standard trial solution of [tex]P\cos(\Omega t) + Q\sin(\Omega t)[/tex]?

    Thanks, James
  5. Jun 23, 2010 #4
    [tex]\ddot{y}[/tex] or further conditions of the motion must be given. Otherwise, it is unsolvable.
  6. Jun 23, 2010 #5
    Can we obtain a general solution? Or because we dont know exactly the roots of the complementary function we cannot get the particular solution?

    Also is multiplying the equation of motion by -1 valid?

  7. Jun 23, 2010 #6
    I'm not sure if there is a general solution.
    If [tex]\ddot{y}=const[/tex] then [tex]x=Ae^{at}cos(\omega t+\phi)+C[/tex].
    If [tex]\ddot{y}=Ycos(\Omega t)[/tex] then [tex]x=Ae^{at}cos(\omega t+\phi)+Bcos(\Omega t + \Phi)+C[/tex].
    So the solution varies depending on [tex]\ddot{y}[/tex]. Anyway the problem only asks you to deduce the equation, not to solve it. And multipyling the equation by any number is always valid.
  8. Jun 23, 2010 #7

    I was pre-empting things a little. The course text has a similar example, I have adapted it slightly.

    I am given [tex]y=a\cos(\Omega t)[/tex] so [tex]m\ddot{y}=-ma\Omega^{2}\cos(\Omega t)[/tex]? So let [tex]P=ma\Omega^{2}[/tex] changing the origin to the equilibrium position we have

    [tex]m\ddot{x}+r\dot{x}+kx=-P\cos(\Omega t)[/tex]

    The text goes on to find the coefficients of the trial solution [tex]B\cos(\Omega t)+C\sin(\Omega t)[/tex] in terms of r, k, m etc starting out with [tex]m\ddot{x}+r\dot{x}+kx=P\cos(\Omega t)[/tex] Is what I have done so far ok?

    Thanks, James
  9. Jun 23, 2010 #8
    The solution given by the text only describes the motion in steady state. You may look at my previous post for the general solution :)
  10. Jun 23, 2010 #9

    Just one last thing, is what I have done with




    replacing [tex]y(t)[/tex] with [tex]\ddot{y}(t)[/tex] valid?
  11. Jun 23, 2010 #10
    No :(
    For example, if [tex]y=at^2[/tex] then [tex]\ddot{y}=2a[/tex]. They're obviously different, right? :)
  12. Jun 23, 2010 #11
    Ok, sorry if I'm being slow, thanks for your patience.

    This is the equation of motion from equilibrium.


    I am told that [tex]y=P\cos(\Omega t)[/tex]

    but as the rhs has the [tex]\ddot{y}[/tex] so should it now be

    [tex]m\ddot{x}+r\dot{x}+kx=-mP\Omega^{2}\cos(\Omega t)[/tex] which is the second derivative of [tex]y=P\cos(\Omega t)[/tex].

    Thanks again
  13. Jun 23, 2010 #12
    I'm not sure what you were asking about when substituting y with y" in post #9, but you're right at post #11 :)
  14. Jun 23, 2010 #13

    I was trying to ask the same question as post 11, obviously not very wel :-)

    Thank you very much for all of you help.

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