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Suspension system, linear differential equation

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    In the limit as t→∞, the solution approaches [tex]x(t) =K \sin[ω(t − t_0)][/tex]
    where K and t0 depend on ω. A>0 and ω≥0. Show that

    [tex]K(ω) = \frac{A}{\sqrt{ω^4 + 2ω^2 + 1}}[/tex]
    .

    2. Relevant equations
    Here is the differential equation

    [tex]\frac{\textrm{d}^{2}x}{\textrm{d}t^{2}}+2\frac{\textrm{d}x}{\textrm{d}t}+x=A\sin(\omega t)[/tex]

    3. The attempt at a solution
    Here is the general solution

    [tex]x=\left(C+Dt\right)e^{-t}-\frac{A\left(2\omega\cos\left(\omega t\right)+\omega^{2}\sin\left(\omega t\right)-\sin\left(\omega t\right)\right)}{\omega^{4}+2\omega^{2}+1}[/tex]

    As t get large the exponential term vanishes but cannot see how the solution approaches [tex]x(t) =K \sin[ω(t − t_0)][/tex]

    Any pointers?
     
  2. jcsd
  3. Oct 27, 2014 #2

    pasmith

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    Homework Helper

    Use the identity [tex]K\sin(\theta - \phi) = K\sin \theta \cos\phi - K \cos \theta \sin \phi[/tex] and set [tex]
    K \sin (\omega t - \omega t_0) = -\frac{A(2 \omega \cos(\omega t) + (\omega^2 - 1)\sin (\omega t))}{\omega^4 + 2\omega^2 + 1}.[/tex] Compare the coefficients of [itex]\sin(\omega t)[/itex] and [itex]\cos(\omega t)[/itex] of each side. This gives you two simultaneous equations of the form [tex]
    C = K \cos (\omega t_0) \\
    D = K \sin (\omega t_0)[/tex] to be solved for [itex]K \geq 0[/itex] and [itex]- \frac \pi 2 \leq \omega t_0 \leq \frac \pi 2[/itex].
     
  4. Oct 31, 2014 #3
    Thanks pasmith, need to try and remember those trig identities!
     
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