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Spring Damping

  1. Nov 17, 2003 #1
    so, f = ma, and f = -kx -bv.
    rearrange to the form of a differential equation.
    i am stuck when the next line simply states the general solution.
    i have done this for first differential only, not second derivatives.
    please help!
     
  2. jcsd
  3. Nov 18, 2003 #2

    jamesrc

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    For the homogeneous solution to ma = -kx -bv, it is standard practice to find the characteristic equation:

    First, rewrite into a standard form:

    [itex] \ddot{x} + \frac{b}{m}\dot{x} + \frac{k}{m}x [/itex]

    Set
    [itex] \frac{k}{m} = \omega_n^2 [/itex]

    [itex] \frac{b}{m} = 2\zeta\omega_n [/itex]

    (the reason why should be clear by the end of the problem; natural frequency and damping ration are useful, meaningful quantities in the study of oscillations)

    characteristic equation:

    [itex] s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 [/itex]

    find the roots of the characteristic equation (it's just a quadratic in s), s 1,2 , so that the solution to the differential equation is written:

    [itex] x(t) = C_1 e^{s_1 t} + C_2 e^{s_2 t} [/itex]

    using the Euler identity and some algebra, you end up with the solution:

    [itex] x(t) = A e^{-\zeta\omega_n t}\cos\left(\omega_d t + \phi \right) [/itex]

    where the damped frequency [itex] \omega_d = \omega_n \sqrt{1-\zeta^2} [/itex] and the constants A and φ (magnitude and phase) are determined by the initial conditions. (You can solve it with a sine or cosine, you'll just end up with a different phase.) That's just the basics off the top of my head, but I hope that helps.
     
  4. Dec 3, 2003 #3
    I dont understand how you get the charateristic equation.
     
  5. Dec 3, 2003 #4

    HallsofIvy

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    That's usually one of the first things you learn in an introductory differential equation course.

    The linear homogeneous differential equation a y"+ by'+ cy= 0 has
    "characteristic equation" ar2+ br+ c= 0.


    More generally, you replace the nth derivative with rn.
     
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