While I understand that this two instances are very different and also the role $y=(a+bt)e^rt$ is very different native the role $y=c_1e^r_1t+c_2e^r_2t$, it likewise seems come me that the two features have very comparable graphs (in particular, very comparable end behaviours).

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You are watching: Difference between overdamped and critically damped

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When you compose the solution of the overdamped mechanism as$$\eqalign & f(t) = c_\,1 e^\,\rho \,t + \omega \,t + c_\,2 e^\,\rho \,t - \omega \,t = \left( c_\,1 e^\,\omega \,t + c_\,2 e^\, - \omega \,t \right)e^\,\rho \,t = \cr & = \left( a\cosh \,\left( \omega \,t \right) + b\sinh \,\left( \omega \,t \right) \right)e^\,\rho \,t \cr $$

and i charged the initial conditions, for instance for $f(0)$ and $f"(0)$, girlfriend get$$\left\{ \matrix f(0) = a \hfill \cr f"(0) = \,\,\rho a + b\omega \quad \Rightarrow \quad b = \;1 \over w\left( f"(0) - \,\,\rho f(0) \right) \hfill \cr \right.$$

so$$f(t) = \left( f(0)\cosh \,\left( \omega \,t \right) + 1 \over w\left( f"(0) - \,\,\rho f(0) \right)\sinh \,\left( \omega \,t \right) \right)e^\,\rho \,t $$

Now, if the damping ideologies the vital value, that is $\omega \to 0$, then

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And similarly, beginning from an under-damped system, whereby you have actually normal $\sin $ and also $\cos$ rather of the hyperbolic version.

So the critically-damped solution is in ~ the frontier in between the two, chrischona2015.orgematically and physically, and also not easy distinguishable at an initial sight when really near to the crucial value.In fact, the under-damping instance will always be evidenced by more or much less visible oscillations.In case of overdamping instead, because for passive system $\rho$ is negative, the exponential degeneration is prevalentand masking in the long time. In the quick time instead, the difference between $\cosh (\omega t)$ and $1$ and also $\sinh (\omega t)$ and $\omega t$is no appreciable.