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Homework Help: DE - mass-spring system

  1. Feb 23, 2008 #1
    (I had made a thread with a problem similar to this one, but it turned a bit messy after finding out the professor made some mistakes and the wording of his problem was awkward ... however, this is it)


    Differential equation governing a forced, mass-spring system:
    [tex]X\text{''}+4*X=0.04*\cos (\omega *t)[/tex]

    Spring constant = 4
    Mass = 1Kg
    Mass starts from rest, at equilibrium

    Find the range of [tex]\omega[/tex] in which the system doesn't break given that the spring breaks if stretched more than 0.06 m from the equilibrium position.


    Equilibrium position = 2.45 m

    [tex]X(t) = \left(2.45-\frac{0.04}{4-\omega ^2}\right) \text{cos}(2*t)+\frac{0.04 *\text{cos}(t *\omega )}{4-\omega ^2}[/tex]

    So, now, how would I find all [tex]\omega[/tex] such that [tex]X(t) > 2.51 = 2.45 + 0.06[/tex] for all [tex]t[/tex]?
    Don't know of any exact way of calculating this. Used MATLAB to approximate the range of \omega's and it is ~ [tex]1.997379 < \omega < 2.303155[/tex]

    Last edited: Feb 23, 2008
  2. jcsd
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