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Solve eqution for oscillation

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve: ##\ddot{x}+\Omega^{2} x=D+\frac{C}{2}+Ecos\omega t+\frac{C}{2}cos2\omega t##


    2. Relevant equations



    3. The attempt at a solution
    I got a hint to use ##x=\alpha sin\omega t+\beta cos\omega t## so ##\ddot{x}=-\alpha ^{2}\omega ^{2}sin\omega t-\beta ^{2}\omega ^{2}cos\omega t## in the equation above than:
    ##(-\alpha ^{2}\omega ^{2}sin\omega t-\beta ^{2}\omega ^{2}cos\omega t)+\Omega^{2}x=\alpha sin\omega t+\beta cos\omega t=D+\frac{C}{2}+Ecos\omega t+\frac{C}{2}cos2\omega t##
    Which gives me 4 separate equations depending on ##sin\omega t##, ##cos\omega t##, ##cos2\omega t## and constant:

    first: ##-\alpha ^{2}\omega ^{2}+\Omega ^{2}\alpha=0##
    second: ##-\beta ^{2}\omega ^{2}+\Omega ^{2}\beta =E##
    third: ##\frac{C}{2}=0##
    fourth: ##D+\frac{C}{2}=0##

    Forth and third together say that ##D=0## and ##C=0##
    First says that:
    ##\alpha ^{2}\omega ^{2}=\Omega ^{2}\alpha##
    ##\alpha =(\frac{\Omega }{\omega })^{2}##
    But for second I am not sure, whether I can divide it with ##\beta## (probably not since it could be equal to 0) or how do I solve it?
    PLEASE HELP

    BTW, if everything is completely wrong and this is not how usually this kind of equations are solved, please let me know.
     
  2. jcsd
  3. May 8, 2013 #2

    haruspex

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    C and D are given, so if you deduce values for them you have gone wrong. Seems to me you need more flexibility in your proposed solution, probably terms involving sin and cos of 2ωt.
     
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