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Simple Harmonic Motion problem

  1. Feb 12, 2007 #1
    I'm supposed to derive x as a function of time for a simple
    harmonic oscillator (ie, a spring). According to my textbook
    this is done by using newton's second law and hooke's law
    as this: ma=-kx and one gets a differential equation in
    the second order. I can follow the calculations until this
    happens: (see attached picture)

    (where omega is the frequency)

    I do get the equation (1) when I solve the differential
    equation myself but I don't understand how equation (1) translates
    to (2)????
    I assume this must be done by using some trigonometric law?
    if so then which one and how??

    Attached Files:

  2. jcsd
  3. Feb 12, 2007 #2
    Spring mass systems often use omega to represent sqrt(k/m). It isn't a trigonometric law, though if your text book eventually (I can't see the picture so I don't know) represents the motion as x(t) = Acos(wt + ø) then you will need to use trig.
  4. Feb 12, 2007 #3
    That's right.. but exactly which trig law do I use and how do I use it to go from equation (1) above in the attached picture to x(t) = Asin(wt + ø).

    Or in other words HOW do I go FROM x(t)=x0cos(wt)+(v0/w)sin(wt) ---- (w being = sqrt(k/m) TO x(t) = Asin(wt + ø)
    how? HOW? HOW?!?!?!?!??!!??!? HOW?!?!?!?
  5. Feb 12, 2007 #4
    That is for you to find out. :p

    Try working backwards, it might be a little easier.
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