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Solving ode of forced oscillator with dumping

  1. Dec 9, 2009 #1
    Hi i have to solve this ODE which descirbes motion of forced oscillator with dumping and constant friction :p

    I'm already solving it numerically with Runge-Kutta 4 yet I'm totaly puzzeled how to do it analytically.


    [tex]mx'' + kx' + w^2_0x + F_f = A cos(\delta t)[/tex] Ff delta k and w are constant

    moving acceleration x'' to one side we get

    [tex]x'' = \frac 1 m (- kx' - w^2_0x - F_f + A cos(\delta t))[/tex]

    i need to solve this equation twice to get velocity x' than position x. Yet i have no clue i know only how to solve x' = f(x) first order ODE :/
  2. jcsd
  3. Dec 10, 2009 #2
    first of all rearrange the equation

    [tex]mx''+kx'+w^{2}_{0}x=Acos(\delta t)-F_{f}[/tex]

    Divide the solution into two steps.

    1) Homogenous solution
    This is the behaviour of the system without the external force acting on it


    This is a constant coefficient equation, and solved with the characteristic polynomial:


    The roots are given by:

    [tex]r_{1,2}=\frac{-k\pm \sqrt{k^{2}-4mw^{2}_{0}}}{2m}[/tex]

    Now I'm gonna assume [tex]4mw^{2}_{0}>k^{2}[/tex] and define:

    [tex] \alpha = \frac{k}{2m}; \omega=\sqrt{-\frac{k^{2}}{4m}+w^{2}_{0}} [/tex]

    [tex] r_{1,2}=\alpha\pm i\omega[/tex]

    Therefore the set of homogenous solutions is
    [tex]x_{1}(t)=e^{-\alpha t}cos(\omega t); x_{2}(t)=e^{-\alpha t}sin(\omega t)[/tex]

    The general homogenous solution is given by superposition.

    2) Particular Solution
    The response of the system to the external force

    I'm gonna use "guessing", but first we will divide the response into two:

    2.1) Response to the constant force [tex]F_{f}[/tex]
    The easiest guess is that x itself will be constant. So if we choose x=c, we'll have


    2.2) Response to the harmonic force
    Now we gonna choose some [tex]x=Ccos(\delta t)+Dsin(\delta t)[/tex]

    Substituting this into the equation gives:

    [tex]-mC\delta^{2} cos(\delta t) -mD\delta^{2}sin(\delta t)-kC\delta sin(\deltat) + kD\delta cos(\delta t)+w^{2}_{0}Ccos(\delta t)+w^{2}_{0}Dsin(\delta t)=Acos(\delta t) [/tex]

    Comparing coefficient of cosines and sines gives:


    which gives

    [tex] C=\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}; D=\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}[/tex]

    So [tex]x_{p,2}=\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}cos(\delta t)+\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}sin(\delta t)[/tex]

    Now finally we compose all of the results together to get the general solution of the ODE:

    [tex]x(t)=c_{1}e^{-\alpha t}cos(\omega t)+c_{2}e^{-\alpha t}sin(\omega t)-\frac{F_{f}}{w^{2}_{0}}+\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}cos(\delta t)+\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}sin(\delta t)[/tex]

    The first & second term are superposition of the homogoneous solutions.
    The third & fourth term are sum of the responses to the external force.

    The coefficients c1 & c2 are arbitrary but given initial conditions you can find them.
  4. Dec 10, 2009 #3


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    Science Advisor

    Nicely done.
  5. Dec 10, 2009 #4
    big thx it took me some time how to get this results but u helped me a lot thx mate :)
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