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Problem with last step of SHM derivation

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A masss m ossilates on a spring, find the position as a function of time.


    2. Relevant equations
    F=ma
    F=-ky


    3. The attempt at a solution

    Well known problem, set
    m(dx/dt) = -kx
    get roots of characteristic equation and sub into general solution,
    x = Aexp(iwt) +Bexp(-iwt)
    or
    x = Asin(wt) + Bcos(wt)
    I'm ok down as far as here however I know this can be written as Ccos(wt-) but I cant see how?
     
  2. jcsd
  3. Oct 2, 2011 #2
    (1) probably just typo, not mx'=-kx but mx''=-kx

    (2) if you want to go from x = Asin(wt) + Bcos(wt) to x=Ccos(wt+phi), you would have to rewrite the first expression as

    x = sqrt(AA+BB) (Asin(wt)/sqrt(AA+BB) + Bcos(wt)/sqrt(AA+BB))

    ( sqrt(a) is squared root from a)

    as soon as (A/sqrt(AA+BB))^2+(B/sqrt(AA+BB))^2 = 1, you can say that there is some angle phi for which cos(phi)= B/sqrt(AA+BB), sin(phi)=A/sqrt(AA+BB). Then use formula for cos of difference.
     
  4. Oct 2, 2011 #3
    thanks
     
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