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Shifting the origine of time in isotropic oscillator

  1. Feb 26, 2006 #1
    I am really lost here :(

    The equation of motion X(t)=Ax Cos(wt-delta(x))
    Y(t)=Ay Cos(wt-delta(y))
    by shifting the origine of time ( t'=t+to where I need to figure out what is appropriate for time to ) and ( Delta=Delta(y)-Delta(x) )
    I am suposed to find X(t)=Ax Cos(wt) and Y(t)=Ay Cos(wt-Delta)

    I wish I could give some of my work but I cant even start it right now since I cannot think of the appropriate time (to) for X , I know that it can not be 0 because then I come to the same point again.
    anyhelp appreciated
    thanks
     
  2. jcsd
  3. Feb 26, 2006 #2

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    I think you mean for the argument of that last function you gave to be t'. What do you get when t'=0? Compare this to what you get with the original function at the same time as measured in t (ie, in terms of to).
     
    Last edited: Feb 26, 2006
  4. Feb 26, 2006 #3
    Yes t to be t' but in the question itself says once you find the solution with t',convert it into t .
    I thought I can give to= dx/w then my equation become for X=Ax Cos( wt'-(wto+Deltax)) and when I unpack it, X/Ax=Cos(wt')Cos(dx+Deltax)+Sinwt' Sin(dx+Deltax) that has to stay only X/Ax=Cos(wt') how to come to this conclution ? How to get red of Cos(dx+Deltax))+Sinwt' Sin(dx+Deltax) I dont know :((((((
    any idea?
     
  5. Feb 26, 2006 #4
    I figured it out

    thanks again
     
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