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Forced oscillation/resonance

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Derive a formula for ##x_p## if the equation is ##mx''+cx'+kx=F_0\cos(\omega t)+F_1\color{red}{\cos}(\color{red}{3}\omega t)##. Assume ##c>0##.

    2. Relevant equations

    3. The attempt at a solution

    I've started off using a guess and the undetermined coefficients method, but that doesn't seem to get me anywhere.

    Try ##x_p=A\cos(\omega t)+B\sin(\omega t)+C\cos(3\omega t)+D\sin(3\omega t)##

    ##x_p'=-A\omega\sin(\omega t)+B\omega\cos(\omega t)-3C\omega\sin(3\omega t)+3D\omega\cos(3\omega t)##

    ##x_p''=-A\omega^2\cos(\omega t)-B\omega^2\sin(\omega t)-9C\omega^2\cos(3\omega t)-9D\omega^2\sin(3\omega t)##

    And this gives me the system
    ##\begin{cases}-Am\omega^2+Bc\omega+Ak=F_0\\ -Bm\omega^2-Ac\omega+Bk=0\\ -9Cm\omega^2+3Dc\omega+Ck=F_1\\ -9Dm\omega^2-3Cc\omega+Dk=0\end{cases}##

    I'm not sure where to go from here, or if I'm on the right track in the first place.
     
    Last edited: Mar 1, 2013
  2. jcsd
  3. Mar 1, 2013 #2

    rude man

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    I am wondering why you assumed sin(3wt) and cos(3wt) terms in x_p?
     
  4. Mar 1, 2013 #3
    I was under the impression that that's the usual way to go about a nonhomogenous equation like this one.
     
  5. Mar 1, 2013 #4

    rude man

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    It's not, although maybe if you carried it out correctly the 3w terms would not have had any impact.

    Fact: if it's a linear ODE and your forcing functions are sin(wt) and/or cos(wt) then you cannot get anything but sin(wt) and/or cos(wt) terms in the solution. I mean, no harmonic terms like sin(3wt) or cos(3wt) etc. (You can still get exp(kt) and/or constant terms however. But not in this case where there were no initial conditions given.)
     
  6. Mar 1, 2013 #5
    So, my guess for x_p should have been ##x_p=A\cos(\omega t)+B\cos(3\omega t)##?
     
  7. Mar 1, 2013 #6

    rude man

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    No, here you go again with the 3w terms! :smile:
    Try one more time ...
     
  8. Mar 1, 2013 #7
    Just the ##A\cos(\omega t)+B\sin(\omega t)## then? Or without the sine term? The examples in my book are confusing and not very well-organized, so it's hard to tell what the author's doing. At one point, he omits the sine term and uses only the cosine term, but doesn't explain it.
     
  9. Mar 1, 2013 #8

    rude man

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    Got it!

    Even if there had been only a F1sin(wt) or only a F2cos(wt) forcing function you should sstill have assumed x_p = F1 sin + F2 cos. Initial conditions could generate both a sin and a cos term in the solution even if the forcing function were only sin OR cos.

    Gotta go for a couple hrs.
     
  10. Mar 1, 2013 #9
    Ah, I think I get it now. Thanks for the help!
     
  11. Mar 1, 2013 #10

    rude man

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    Good! Shoot me over your answer if you want to compare.
     
  12. Mar 1, 2013 #11
    I've altered my original post, but I don't see a way to solve for A and B here. I tried eliminating the first term of each equation and I get
    ##(A^2+B^2)c\omega=F_0 B##, but I don't know what I can do with that.

    (I multiplied the first equation by B and the second by -A, then added them together.)
     
  13. Mar 1, 2013 #12

    rude man

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    Why would anyone give you that forcing function? F_0 + F_1 = F why not?

    I believe you need to look at the posed question a third time ...
     
  14. Mar 1, 2013 #13
    Actually, there should be a 3 in front of the second omega... Darn typos. I'll fix it right away.
     
  15. Mar 1, 2013 #14

    rude man

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    It's just algebra. Since it's just that and your equations are all correct Ill give you what I got:
    A = F_0(k - mw2)/Δ
    B = wcF_0/Δ
    where Δ = (k - mw2)2 + w2c2.

    (BTW I used software to solve those equations. Do you know the determinant way to solve simultaneous algebraic linear equations? I did it tyat way also.)
     
  16. Mar 1, 2013 #15

    rude man

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    That totally changes things. Makes it messier.

    Now you need x = Asin(wt) + Bcos(wt) + Csin(3wt) + Dcos(3wt).

    Note again that, this being a linear equation, you know ahead of time that you will only get sin/cos(wt) and sin/cos(3wt) terms. No other frequencies than w and 3w. I presume you know that w = 2pi f so the only two frequencies present in x(t) and all its derivatives will be w/2pi and 3w/2pi Hz.
     
  17. Mar 1, 2013 #16
    So, everything I had in my first post is still right? And I'm back to somehow finding expressions for A through D?
     
  18. Mar 1, 2013 #17

    rude man

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    That is sadly correct.
     
  19. Mar 1, 2013 #18
    Well, at least it's nice to know I wasn't doing anything wrong in the first place...:frown:

    I'll see what I can do. Thanks for the input
     
  20. Mar 2, 2013 #19
    I haven't tried it out yet, but is it possible variation of parameters could be a better approach than undetermined coefficients?
     
  21. Mar 2, 2013 #20
    Alright, so here's what I've got using variation of parameters.
    The characteristic polynomial gives me the solutions

    ##x_1=\exp\left(\frac{-c+\sqrt{c^2-4mk}}{2m}t\right)##
    ##x_2=\exp\left(\frac{-c-\sqrt{c^2-4mk}}{2m}t\right)##

    I let ##a = \frac{-c+\sqrt{c^2-4mk}}{2m}## and ##b=\frac{-c-\sqrt{c^2-4mk}}{2m}##, and determine the Wronskian:

    ##W(x_1,x_2)=\left|\begin{matrix}\exp(at)&\exp(bt)\\ a\exp(at)& b\exp(bt)\end{matrix}\right|=(b-a)\exp((a+b)t)\not=0##, since ##a\not=b##.

    ##W(x_1,x_2)=-\frac{\sqrt{c^2-4mk}}{m}\exp\left(-\frac{c}{m}t\right)##

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
    ##x_p=x_1u_1+x_2u_2##, where

    ##\displaystyle u_1=-\int\frac{x_2(F_0\cos(\omega t)+F_1\cos(3\omega t))}{W(x_1,x_2)}dt##

    and

    ##\displaystyle u_2=\int\frac{x_1(F_0\cos(\omega t)+F_1\cos(3\omega t))}{W(x_1,x_2)}dt##

    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    ##\displaystyle u_1=\frac{1}{a-b}\int e^{-at}(F_0\cos(\omega t)+F_1\cos(3\omega t)) \;dt##

    ##\displaystyle u_2=\frac{1}{b-a}\int e^{-bt}(F_0\cos(\omega t)+F_1\cos(3\omega t)) \;dt##

    Then integration by parts... I'll be back when I finish that.
     
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