Consider a spring device whose equation is given by$$my""+\mu y"+ky=0$$ and also let $D=\mu^2-4mk$. Now there are three cases and I to be considering the cases that $D=0$ and $D>0$:

When $D=0$, the solution is that the kind $y=(a+bt)e^rt$. (Critically damped)When $D>0$, the systems is of the form $y=c_1e^r_1t+c_2e^r_2t$. (Overdamped)

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).

Question

Why room the graphs that the solution in this two cases so similar (while in the other case $DHow to tell indigenous the graph whether we have actually a critically damped device or an overdamped system?


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

$$ \bbox \chrischona2015.orgop \lim \limits_\omega \, \to \,0 f(t) = \left( f(0) + \left( f"(0) - \,\,\rho f(0) \right)t \right)e^\,\rho \,t $$

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.