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Forced Damped Oscillator

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the frequency that gives the maximum amplitude response for the forced damped oscillator d[tex]^{2}[/tex]x/dt[tex]^{2}[/tex] + 6dx/dt + 45x = 50cos([tex]\omega[/tex]t)

    2. Relevant equations

    I'm really confused by this problem, but I know that the amplitude can be found by taking the [tex]\sqrt{c_{1}^2+c_{2}^2}[/tex] with c[tex]_{1}[/tex] and c[tex]_{2}[/tex] being parameters of the general solution...

    3. The attempt at a solution

    I suppose I want to maximize my c[tex]_{1}[/tex] and c[tex]_{2}[/tex] values. And this can be done by modifying the value of [tex]\omega[/tex]. So, my only guess as to how I could solve this problem is through manipulation of the Method of Undetermined Coefficients, and see for what values of [tex]\omega[/tex] my c[tex]_{1}[/tex] and c[tex]_{2}[/tex] become largest...

    If anyone could offer me any suggestions involving different strategies for solving this problem, i would greatly appreciate it

    the superscripts above some of my "c" parameters should be subscripts, i'm not sure why they keep getting turned into superscripts, sorry =(
    Last edited: Jan 26, 2010
  2. jcsd
  3. Jan 27, 2010 #2


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    Hi gpax42! :smile:

    (have an omega: ω and a square-root: √ and try using the X2 and X2 tags just above the Reply box :wink:)

    What do you have as the general solution for the full equation (ie, including ω)?
  4. Jan 27, 2010 #3
    thats my first gray area... im fine with the general solution of the homogenous equation, but I can't use the method of undetermined coefficients to solve the full equation seeing that [tex]\omega[/tex] isn't a constant

    the general solution for the homogenous part is e[tex]^{-3t}[/tex](C_1*cos(6t) + C_2*sin(6t))
  5. Jan 27, 2010 #4


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    (what happened to that ω i gave you? :confused: and try using the X2 and X2 tags just above the Reply box :wink:)

    Yes, your general solutiuon is correct. :smile:

    Now look for a particular solution of the form Acosωt + Bsinωt. :wink:

    (and ω is constant … it's a constant you can choose, but once you choose it it's constant)
  6. Jan 27, 2010 #5
    taking that trial solution and its respective first and second derivatives and plugging those back into the original oscillation equation i get...

    -ω[tex]^{2}[/tex]Acos(ωt)-ω[tex]^{2}[/tex]Bsin(ωt)-ω6Asin(ωt)+ω6Bcos(ωt)+45Acos(ωt)+45Bsin(ωt) = 50cos(ωt)

    when i isolate out the common terms I'm left with the equations...

    -ω[tex]^{2}[/tex]A+6ωB+45A = 50

    -ω[tex]^{2}[/tex]B-6ωA+45B = 0


    I then tried solving or A and B in terms of ω but got ridiculous solutions.
    Is it simply supposed to be A = 50/45 = 10/9 and B = 0

    which would give me a particular solution of

    [tex]\frac{10}{9}[/tex]cos(ωt) ?
  7. Jan 27, 2010 #6


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    Hi gpax42! :smile:
    Yes, the solution for A and B is pretty horrible …

    but if you look at http://en.wikipedia.org/wiki/Damped_harmonic_oscillator#Sinusoidal_driving_force",

    you'll find that Zm2 = ((45 - ω2)2 + 36ω2)/ω2,

    which is the denominator of A and B …

    so I think it is correct. :smile:
    Last edited by a moderator: Apr 24, 2017
  8. Jan 27, 2010 #7
    ahhh, i understand the problem completely now :smile:... thank you very much for all your help tiny tim!
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