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Undamped Forced Oscillation

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex] \ddot x +\omega_0 x = \frac{F_0}{m}sin{\omega_0 t} [/tex]

    2. Relevant equations
    [tex] y_h=C_1 cos{\omega_0 t} + C_2 sin{\omega_0 t} [/tex]

    3. The attempt at a solution

    So, I complexified this problem, hoping to make it easier. I saw that I couldn't let [tex] X_p = Ae^{i \omega_0 t - \frac{\pi}{2}} [/tex] because it was contained in the homogeneous solution. So I tried [tex] X_p = Ate^{i \omega_0 t - \frac{\pi}{2}} [/tex]

    I went through the entire song and dance and ended up with [tex] \frac{Ft}{2m \omega_0} cos{\omega_0 t} [/tex]
    The answer that works it looks like though is the negative of this solution. I've checked my algebra over and over. I must have maybe made a mistake with my initial pick? Is there something else I should have tried?

    Thanks.
     
  2. jcsd
  3. Oct 23, 2015 #2

    BvU

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    Yes, the solution to the homogeneous equation can stil be added -- to deal with the intial conditions.
    The minus sign is a phase difference that you somehow left out. [edit] sorry, did not leave out: there's a ##\pi/2## in there !
     
  4. Oct 23, 2015 #3
    Do you think I made a mistake in the algebra to come up with the wrong answer? Or is it to do with my "guess"?
     
  5. Oct 23, 2015 #4

    BvU

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    Will check, but have to run now...
    (Initial: I asssume we can assume :smile: C1=C2 = 0 )
     
  6. Oct 23, 2015 #5
    Thank you for helping! there are initials x(0)=0 and x'(0) =0.

    I'm really just after the general form now. With my "luck" I was able to solve the later parts of the problem I'm just interested in how I messed up the answer at the end of this part.
     
  7. Oct 23, 2015 #6

    ehild

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    You have a typo in the first equation. ω0 should be squared.
    You can not assume the particular solution as A t exp(iω0t-π/2).
    Why do you subtract pi/2 from the exponent?
    Keep the real notations and assume real particular solution.
     
  8. Oct 23, 2015 #7

    ehild

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    You have a typo in the first equation. ω0 should be squared.
    You can not assume the particular solution as A t exp(iω0t-π/2).
    Why do you subtract pi/2 from the exponent?
    Keep the real notations and assume real particular solution.
     
  9. Oct 23, 2015 #8
    You are correct. The omega should be squared. I was having a lot of trouble early on keeping the particular real. I guess I could try again since I've learned a good deal about this problem over the last several hours, (haha).
     
  10. Oct 23, 2015 #9

    ehild

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    If you want to work with complex variables, you need to rewrite the original equation also in complex form. And the trial function should include both exp(iw0t) and exp(-iw0t).
     
  11. Oct 23, 2015 #10
    Actually, I'm curious as to why I can't assume an answer of that form? I've seen it a lot. Unless I'm missing a subtle difference here.
    I got the idea to try it after failing at trying real answers because they just seemed too messy. I kind of verified that I could try answers of that form from this site http://isites.harvard.edu/fs/docs/icb.topic251677.files/notes22.pdf

    They actually have a very similar problem. I knew I couldn't use their answer because they had a cos originally. I was ok with that though as I was already using a sine in my problem. I just wanted to make sure it was plausible first before spending a lot of time on it.
     
  12. Oct 23, 2015 #11
    Oh, I think I see... I shouldn't try to make the exponential only sine before hand right? I don't know why I was trying that... Ok let me try this and see if it works out better.
     
  13. Oct 23, 2015 #12

    ehild

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    If you do it with real variables, you should include both sin and cosine into the particular solution, but one of them will cancel.
     
  14. Oct 23, 2015 #13
    I know scans/pictures aren't really liked here, but, it would take me an age to type this out in latex.

    I wrote it out neat. Hoping maybe you can point out my mistake. I'm getting confused at some steps here.
    [PLAIN]http://[ATTACH=full]199950[/ATTACH] [PLAIN][PLAIN]http://[ATTACH=full]202428[/ATTACH] [url=http://postimg.org/image/ebhk70ndl/][ATTACH=full]199952[/ATTACH]
     

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    Last edited by a moderator: May 7, 2017
  15. Oct 23, 2015 #14

    ehild

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    It depends how you write the original equation in complex form. If you take the right-hand side as the imaginary part of F0/m sin(w0t), You need exp(iw0t) only. But don't forget to take the imaginary part at the end.
     
  16. Oct 23, 2015 #15
    What do you mean, "take the imaginary part at the end" Can I just say I want the imaginary part and take that as my real answer???
     
  17. Oct 23, 2015 #16

    ehild

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    Last edited by a moderator: May 7, 2017
  18. Oct 23, 2015 #17

    ehild

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    Yes. But you have to write the RHS in complex form.
     
  19. Oct 23, 2015 #18
    That is the part I'm struggling with. You're saying I should just write that expression without the phase shift?
     
  20. Oct 23, 2015 #19

    ehild

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    sin(wt) is the imaginary part of what?
     
  21. Oct 23, 2015 #20
    Mind blowing. Ok... then I just write the expression without the phase shift, I keep the unreal cosine part, the answer is negative and it works. Why can I do that????
     
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