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PhoenixFire
 one year ago
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?
PhoenixFire
 one year ago
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?

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theEric
 one year ago
Best ResponseYou've already chosen the best response.1Do 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.

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1@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.

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, I guess you wouldn't solve that...

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.1@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 \(T=\frac{2}{B}\)

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Cool! Congrats! I was just looking through my book, but I didn't arrive at any result.

theEric
 one year ago
Best ResponseYou've already chosen the best response.1My 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 :)
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