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Amplitude of damped mass-spring system

  1. Jan 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the IVP
    [tex]
    \ddot{r}+\dot{r}+20r=0\\
    r(0) = 0.8\\
    \dot{r}(0) = 0
    [/tex]
    for the length of an oscillating spring (damped), we find that the general solution is
    [tex]
    r=e^{-0.5t}[0.8\cos(\sqrt{19.75}t)+\frac{0.4}{\sqrt{19.75}}\sin(\sqrt{19.75}t)]
    [/tex]
    and I wish to find the curve bounding the upper part of the solution, i.e. the amplitude of the oscillations.
    2. Relevant equations


    3. The attempt at a solution
    I understand that if
    [tex]
    r=e^{-0.5t}0.8\cos(\sqrt{19.75}t)
    [/tex]
    for example, then the amplitude would be obtained by removing the cosine factor, since [itex]e^{-0.5t}0.8[/itex] correspond to the amplitude and [itex]\cos(\sqrt{19.75}t)[/itex] corresponds to how fast the spring is oscillating. But how does one do when there are two oscillating terms as above?
     
  2. jcsd
  3. Jan 15, 2016 #2

    LCKurtz

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    Use$$A\cos\theta + B\sin\theta = \sqrt{A^2+B^2}\left(\frac A {\sqrt{A^2+B^2}}\cos\theta + \frac B {\sqrt{A^2+B^2}}\sin\theta \right)$$ $$=\sqrt{A^2+B^2}\cos(\theta-\alpha)$$
     
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