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PhoenixFire Group Title

Given the equation of damped shm \[\frac{d^2x}{dt^2} +B\frac{dx}{dt}+Cx=0\] And the solution \[x=Ae^{-\frac{t}{T}} cos(wt+d)\] Where T is the time it takes for the amplitude to fall a factor of 1/e. How do I find T in terms of B and C?

  • 4 months ago
  • 4 months ago

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  1. theEric Group Title
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    Do you have to solve the differential equation maybe? If not, there might be a formula... I'm sorry, I might have known this once, but I don't now.

    • 4 months ago
  2. PhoenixFire Group Title
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    @theEric Don't need to solve the differential as the solution to the differential is given. The question is asking for the time period it takes for the amplitude of the solution to fall by a factor of 1/e based on the values of B and C from the differential.

    • 4 months ago
  3. theEric Group Title
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    Yeah, I guess you wouldn't solve that...

    • 4 months ago
  4. PhoenixFire Group Title
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    @theEric Nevermind I figured it out. The general solution to damped shm is: \[x=Ae^{-\alpha t} cos(wt+d)\] where \(\alpha\) is the coefficient of exponential decay given by \(\alpha=\frac{b}{2m}\) and m is the constant for the second order term of the diff. so in this case m=1. Now the in my question \(\large e^{-\frac{t}{T}}=e^{-\alpha t}\) when \(\alpha = \frac{1}{T}\) or \(T = \frac{1}{\alpha}=\frac{2m}{b}\) for my question \(T=\frac{2}{B}\)

    • 4 months ago
  5. theEric Group Title
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    Cool! Congrats! I was just looking through my book, but I didn't arrive at any result.

    • 4 months ago
  6. theEric Group Title
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    My book has that the amplitude at any point will be \(Ae^{-\gamma t}\) where \(\gamma=\frac12B\) adapted to correspond to your equation* So the amplitude is generally \(\Large Ae^{-\frac{Bt}2}\) If \(\Large Ae^{-\frac{Bt}2}=Ae^{-\frac tT}\) \(-\dfrac{Bt}2=-\dfrac tT\) Then \(T=\dfrac 2B\) Again :)

    • 4 months ago
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