Difference between critically damped systems and overdamped systems Consider a spring system whose equation is given by
$$my''+\mu y'+ky=0$$ and let $D=\mu^2-4mk$. Now there are three cases and I am considering the cases that $D=0$ and $D>0$:


*

*When $D=0$, the solution is of the form $y=(a+bt)e^{rt}$. (Critically damped)

*When $D>0$, the solution is of the form $y=c_1e^{r_1t}+c_2e^{r_2t}$. (Overdamped)


While I understand that these two cases are very different and the function $y=(a+bt)e^{rt}$ is very different from the function $y=c_1e^{r_1t}+c_2e^{r_2t}$, it also seems to me that the two functions have very similar graphs (in particular, very similar end behaviours). 
Question


*

*Why are the graphs of the solution in these two cases so similar (while in the other case $D<0$, the graph is very different)?

*How to tell from the graph whether we have a critically damped system or an overdamped system?

 A: When you write the solution of the overdamped system 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 impose the initial conditions, for instance for  $f(0)$ and $f'(0)$, you 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 approaches the critical value, that is $\omega \to 0$, then
$$ \bbox[lightyellow] {  
\mathop {\lim }\limits_{\omega \, \to \,0} f(t) = \left( {f(0) + \left( {f'(0) - \,\,\rho f(0)} \right)t} \right)e^{\,\rho \,t} 
}$$
And similarly, starting from an under-damped system, where you have normal $\sin $ and $\cos$ instead of the hyperbolic version.
So the critically-damped response is at the frontier between the two, mathematically and physically, 
and not easy distinguishable at first sight when very near to the critical value.
In fact, the under-damping case will always be evidenced by more or less visible oscillations.
In case of overdamping instead, since for passive system $\rho$ is negative, the exponential decay is prevalent
and masking in the long time. In the short time instead, the difference between $\cosh (\omega t)$ and $1$ and $\sinh (\omega t)$ and $\omega t$
is not appreciable.
