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Underdamped harmonic motion

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a particular solution to the differential equation using undetermined coefficients.

    x[itex]^{''}[/itex]+5x[itex]^{'}[/itex]+4x=2sin2t
    x(0)=1
    x'(0)=0


    I know that the equation is underdamped because c<W[itex]_{0}[/itex], and that W[itex]_{0}[/itex]=2.

    I know that the particular solution is x(t)=acos(2t)+bsin(2t)=Asin(2t)/(W[itex]_{0}[/itex][itex]^{2}[/itex]-w[itex]^{2}[/itex])

    Plugging the initial conditions in to x(t) and x[itex]^{'}[/itex](t) gives me a=1 and b=0.

    However, my professors answer is:

    x(t)=[itex]\frac{8}{5}[/itex]e[itex]^{-t}[/itex]-e(2/5)[itex]^{-4t}[/itex]

    How did he get this final answer? My book seems to set the solutions up differently, my professor hasn't been returning my e-mail, and my exam is tomorrow morning! Any help would be appreciated!
     
    Last edited: Jun 4, 2012
  2. jcsd
  3. Jun 4, 2012 #2
    Do you remember how to solve differential equations? First you should solve the homogeneous equation,
    [tex] x'' + 5x' + 4x = 0 [/tex]
    Easiest way in this case is to try to find solutions of the form [itex] x = e^{rt} [/itex] where r is a constant.
     
  4. Jun 4, 2012 #3

    LCKurtz

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    Something is wrong here. The two sides of your professor's answer aren't equal to each other, and neither one could possibly give a sine function when you plug it into the equation. You should find the general solution of the homogeneous equation first, then look for a particular solution of the NH equation of the form ##y_p=A\cos(2t)+B\sin(2t)##.
     
  5. Jun 4, 2012 #4

    vela

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    What do you mean by c? The system is actually overdamped. This is clear because the roots to the characteristic equation are real.
     
  6. Jun 4, 2012 #5
    Sorry in advanced for poor formatting. My internet is down and I have to use my phone. I agree that the equation is overdamped, I had under written in my notes. I found -4/3((e^-t)+(e^-4t)) for the homogeneous equation. I have no idea how my professor got his answer. Also, there was a 0 sin 2t at the end of his answer if that helps. I had to edit the original answer of his that I posted. There was an equal sign where there should have been a minus.
     
  7. Jun 4, 2012 #6
    I just got -2/5(cos2t+sin2t) as my particular solution
     
  8. Jun 4, 2012 #7

    LCKurtz

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    That is incorrect. You should get ##x_p=-\frac 1 5 \cos(2t)##. If you add that to the general solution of the homogeneous equation ##x_h = Ae^{-4t} + Be^{-t}## you will have the general solution to the DE. When you apply the initial conditions to figure out ##A## and ##B## you should find the unique solution to the initial value problem is$$
    x=-(\frac 2 5)e^{-4t}+(\frac 8 5)e^{-t}-\frac 2 5 \cos(2t)$$Your professor just left off the last term.
     
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