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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?
 6 months ago
 6 months ago
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?
 6 months ago
 6 months ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
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.
 6 months ago

PhoenixFire Group TitleBest 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.
 6 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Yeah, I guess you wouldn't solve that...
 6 months ago

PhoenixFire Group TitleBest 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 \(T=\frac{2}{B}\)
 6 months ago

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

theEric Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 6 months ago
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