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Problem simplifying the solution of an ODE.

  1. Feb 21, 2013 #1
    So, I was following the derivation in my physics book of:
    [itex]x(t) = c_1e^-(\frac{\gamma t}{2})\cos(\omega_d t)+c_2e^-(\frac{\gamma t}{2})\sin(\omega_d t)[/itex]

    Until they simply get to this in one step:
    [itex]Ae^-(\frac{\gamma t}{2})\cos(\omega_d t + \phi)[/itex]

    I've tried reading many other sources for this derivation of the underdamped oscillator, and I follow up until this last critical step and they don't hint at the omitted steps.
     
  2. jcsd
  3. Feb 21, 2013 #2

    HallsofIvy

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    That's a pretty standard step. Use the identity cos(A+ B)= cos(A)cos(B)- sin(A)sin(B).

    Comparing that to [itex]\alpha cos(\theta)+ \beta sin(\theta)[/itex]
    ([itex]\alpha= c_1e^{-\gamma t/2}[/itex], [itex]\beta= c_2e^{-\gamma t/2}[/itex] and [itex]\theta= \omega_d t[/itex].)
    We need [itex]cos(A)= \alpha[/itex] and [itex]sin(A)= \beta[/itex]. Of course, that is not possible unless [itex]\alpha^2+ \beta^2= 1[/itex]. If that is not true, then we multiply and divide by [itex]\alpha^2+ \beta^2[/itex]:
    [tex](\alpha^2+ \beta^2)\left(\frac{\alpha}{\alpha^2+ \beta^2}cos(\theta)+ \frac{\beta}{\alpha^2+ \beta^2}sin(\theta)\right)[/tex]
     
  4. Feb 21, 2013 #3

    LCKurtz

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    I think Halls meant$$
    \sqrt{\alpha^2+ \beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+ \beta^2}}cos(\theta)+ \frac{\beta}{\sqrt{\alpha^2+ \beta^2}}sin(\theta)\right)$$
     
  5. Feb 21, 2013 #4
    Got it, thanks everyone.
     
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