Physical Interpretation of an Overdamped Pendulum Consider a damped pendulum whose equation of motion is given in general by $$m\ddot{x}=-\mu\dot{x}-kx$$ where $\mu,k>0$
Rewrite this equation as 
$$\ddot{x}+2\gamma\dot{x}+\omega^2x=0,$$
where $2\gamma = \frac{\mu}{m}$ and $\omega^2 = \frac{k}{m}$.
If $\gamma>\omega$, the roots of this equation are real and distinct. Define $\beta^2=\gamma^2- \omega^2$. Then the roots are $-\gamma \pm\beta$.
I have no issue with this, the solution is of the form $$x(t)=e^{-\gamma t}\left(Ae^{\beta t}+Be^{-\beta t}\right).$$
Given some initial conditions $$x(0)=x_0 \quad \text{and} \quad \dot{x}(0)=v_0 $$
Does the system return to rest in a finite time?
I have no issue here as well. I can rewrite the solution to read as $$x(t)=Ae^{(-\gamma+\beta)t}\left[1+\frac{B}{A}e^{- 2\beta t}\right]$$
where the next time, say $t_1$ such that $x(t_1)=0$ is given by $$t_1 = \frac{1}{2\beta}\ln\left(-\frac{A}{B}\right)$$
Here is where I struggle though. I can picture what happens if $$\frac{A}{B}<-1$$
I can choose some arbitrary number, say -2, so that $\ln(-(-2))=\ln(2)$ and thus $t_1$ is finite. In other words given some initial velocity the pendulum will move from its initial position and without oscillating return to its equilibrium position after time $t_1$.
However, now I am stuck.
If $\dfrac{A}{B}\to -\infty$ then $t_1 \to \infty$ I can make peace with. Am I right in saying that the system is so over damped that this pendulum almost comes to a complete stop and sloooooooooooooowly over an infinitely long period returns to equilibrium? If so, my brain can picture that, and I am happy.
But now the cases I can't make peace with.


*

*If $A=-B$, i.e. $x(0)=0$, so we start at equilibrium, regardless of my initial velocity $\dot{x}(0)=v_0$, it is true that $t_1$ is always zero as $\ln(1)=0$. In my brain, I can picture a pendulum starting at equilibrium position, flying with some crazy initial velocity, and that equation is telling me, yeah.. The next time it's at $0$ is now instantaneously at $0$. 
So what happened? It's impossible that the system is so overdamped that it's almost not damped at all and the pendulum flies infinitesimally fast to the equilibrium and then stops suddenly without shooting past. I kind of almost want to say it's negatively overdamped.. and that can't make sense.. can it?

*Similarly, If $-1 <\frac{A}{B} - <0$, then $t_1$ is negative.

*Similarly, If $\frac{A}{B} = 0$, or rather as $\frac{A}{B} \to -0$ then $t_1 \to -\infty$ is problematic.

*Self-explanatory ... If $\frac{A}{B} > 0$ then we have a ln of a negative number which is also problematic.
So my question is really this. Are there physical relationships to the other 4 cases or is it one of those "the maths is possible but the physics is meaningless" scenarios. i.e. If $\frac{A}{B} > 0$ then that equation cannot physically ever be representing an over damped oscillator... or at least not in this universe. I am really hoping that's the case.
TL;DR - What is happening with over damped pendulums? Or does the equation of motion become meaningless under certain initial conditions?
PS: I am putting this under homework because my brain might be missing something obvious that someone might illuminate. My background is in applied mathematics and fluid mechanics. For the life of me I can't picture this mentally or find literature online that addresses these issues and unluckily I am curious to know things that bother me.
 A: Let us solve the differential equation to see what we get. (spoiler: it is not what you originally posted).
You have $$\ddot{x}+2\gamma\dot x+\omega^2x=0$$
The characteristic equation of the DE is,
$$\lambda^2+2\gamma\lambda+\omega^2=0$$
And solving for $\lambda$ you get:
$$\lambda_{1,2}=\frac{-2\gamma\pm \sqrt{4\gamma^2-4\omega^2}}{2}=-\gamma\pm\sqrt{\gamma^2-\omega^2}$$
And the usual way to express this is,
$$\lambda_{1,2}=-\gamma\pm i\sqrt{\omega^2-\gamma^2}$$
The physical parameters here are $\omega$, the natural frequency of the system (the frequency of oscillation of the undampened system) and $\gamma$, which is related to the damping factor $\zeta:=\gamma/\omega$. The value of $\zeta$ tells you whether the system is underdamped ($\zeta<1$), critically damped ($\zeta=1$) or overdamped ($\zeta>1$). You should be able to find these concepts in more details in a System Modelling book.
Basically, the only kind of system for which you will observe oscillatory motion is $\zeta<1$, because there you have an imaginary component on the roots, and you know that translates to sines and cosines.
You can see the prototype response for these different kind of systems in the picture I attached.
Regarding your question, you can see that the general solution of the equation is:


*

*For $\gamma<\omega$
$$e^{-\gamma t}\left[A\text{ }\mathrm{sin}\left(t\sqrt{\omega^2-\gamma^2}\right)+B\text{ }\mathrm{cos}\left(t\sqrt{\omega^2-\gamma^2}\right)\right]$$


*

*For $\gamma=\omega$
$$A\text{ }e^{-\gamma t}+B\text{ }te^{-\gamma t}$$


*

*And for $\gamma>\omega$
$$A\text{ }\exp\left[-\gamma{\left(1+\sqrt{1-\frac{\omega^2}{\gamma^2}}\right)}t\right]+B\text{ }\exp\left[-\gamma{\left(1-\sqrt{1-\frac{\omega^2}{\gamma^2}}\right)t}\right]$$
Where I rearranged thing a little in the last equation to make it clearer that the exponents are negative. I graphed this with $\omega =10, A=B=\frac{1}{2}$ for the three cases, with $\zeta=\frac{1}{2},1,2$, and I get the following:

The red is underdamped, the blue is critically damped and the green is overdamped. They all decay exponentially to zero.

Does the system return to rest in a finite time?

No, it does not (if, by returning to rest you mean that the system returns to the equilibrium point with $v=0$). As you can see in the three equations for the solutions, the amplitude of $x$ decreases exponentially. This means that you asymptotically approach the rest state ($x=p=0$), but never reach it.

Edit: I will add, per the request of the OP, another graph. In this case with $-A=B=\frac{1}{2}$, $\omega=10$ and $\zeta=2$.

We can think of this system as one where the oscillator starts at the equilibrium position but with a high velocity (we push the mass very hard and reach desired velocity exactly at the equilibrium point of the oscillator and we start recording the position at that instant ($t=0$)).
A: (Answer written for v2 of question with given stable system ODE)
First, for the case of real, distinct roots, I believe there are just two cases to consider given a non-zero initial position and $t\ge 0$:


*

*The initial velocity is such that the system asymptotically approaches $x=0$

*The initial velocity is such that the system goes through $x=0$ for some $t>0$ and then asymptotically approaches $x=0$
To see this, stipulate that the general solution is
$$x(t) = Ae^{-\alpha t}+Be^{-(\alpha + \beta)t}$$
where the decay constants $\alpha,\,\beta$ are positive real numbers. The initial conditions are then
$$x(0) = A + B$$
$$\dot x(0) = -\alpha x(0) - \beta B$$
Now, solve for $t$ such that $x(t) = 0$:
$$t = -\frac{1}{\beta}\ln\left(-\frac{A}{B}\right)$$
Clearly, this equation can only be satisfied for finite $t\gt0$ when $B\lt -A$
For $-A \lt B \lt 0$, there are solutions for finite $t\lt 0$, and for $B\ge 0$, there are no solutions for finite $t$.
Since it is evident by inspection that $x(t)\rightarrow 0$ as $t\rightarrow\infty$, the two cases I mention above follow.
Note: the $B=-A$ (zero initial position) case is the case that is 'on the boundary' between the solutions with $x=0$ for $t\gt 0$ and $x=0$ for $t\lt 0$.
Here are plots of four sets of initial conditions for the general solution $x(t) = Ae^{-t}+Be^{-2t}$ that illustrate the cases above:

A: 1) the system will never get to $0$ in a finite time. Therefore, it makes sense that the only time to can find where $x=0$ is at $t=0$. It isn't saying that the object starts moving but then instantly snaps back to $0$. Your solution is showing you that you are only at equilibrium at $t=0$, which is true.
For all other cases you have picked initial conditions such that the object never reaches the origin. Or in other words, you would need to have been at the origin initially in the past (hence the negative time).

Keep in mind that this is an over damped system, so the last time the object will "reach equilibrium" will be due to an essentially exponential decay where the origin is only arrived at as $t\to\infty$. For the cases where $t_1>0$ you have just set the initial conditions such that the system overshoots equilibrium before this exponential decay occurs.
A: I think something has gone wrong very early in your analysis. For example:

1: If $A=-B$, i.e. $x(0)=0$, so we start at equilibrium, regardless of my initial velocity $\dot{x}(0)=v_0$, it is true that $t_1$ is always zero as $\ln(1)=0$. 

Yep! There is only one root and it is at $t=0$. But then you say, 

In my brain, I can picture a pendulum starting at equilibrium position, flying with some crazy initial velocity and that equation is telling me, yeah.. the next time it's at 0 is now instantaneously at 0. So what happened?

And the key is, nothing happened.
It started at $x = 0$ with $\dot x \ne 0.$ That is where it started.
That is also the only place it is ever at zero. Looking really abstract, if you draw out the factor of $B$ rather than the factor of $A$, you have this equation $(A/B) + e^{-2\beta t} = 0$ and it only has at most one root, namely it only has one if $A/B < 0$. It had no other roots it could possibly have. You started it at $x=0$ and it will never be at zero again.
It didn’t “return back to zero after no time”... it never returned back to 0. How could it? The graph is $A~\sinh(\beta t)~e^{-\gamma t}$ and for $t > 0$ that is the product of three positive terms! How are we going to multiply positives to get anything that is not a positive number? (OK so $A$ could be negative but, y'know, “without loss of generality” etc. we can assume that we orient our axes so that either $x(0) > 0$ or if $x=0$ then so that $\dot x(0) > 0.$)
More to the point the single root only exists for positive $t$ if you fire the particle at the $x=0$ position with a certain negative velocity. If you do not fire it hard enough, then the damping just gets rid of this initial momentum. If you do fire it hard enough, it crosses over with some finite remaining velocity and then it sits at $x < 0$ until the infinity of time.
“Aha, but I have good physical intuitions and there must be some boundary between these where I give the particle just enough energy to hit zero but not cross over.” That is a beautiful idea which unfortunately also does not work... because the time when the crossing-over happens happens to be $t=\infty$ at this boundary case; the root “comes in from infinity” as you increase $v_0.$
There is only one root, and it is never the end of the dynamics, except in the trivial case where $x_0 = v_0 = 0.$
A: 
I have a feeling this ODE behaves like a damped oscillator (for
  certain values)

(sigh) Link to the version of this question I wrote this answer to
There are a family of initial conditions that yield an $x(t)$ that goes to zero as $t\rightarrow\infty$: 
$$\dot x(0) = -(3 + \sqrt{19})\cdot x(0)$$
This corresponds to $A=x(0),\, B=0$
But the essential problem with this system (in the context of your lengthy analysis of it) is that it isn't stable which is to say that the slightest perturbation away from these initial conditions will yield an $x(t)$ that goes to $\pm\infty$ as $t\rightarrow\infty$
Here's a graph of a $A=1, B=0$ solution as well as a $A/B=-2$ solution that you mention early in your post:

This is really all there is to it. There's a family of solutions that look suspiciously like a (decaying) first order system (no damped oscillation evident), and then the rest - all of which 'run away' from $x=0$.  I just don't see any evidence damped oscillator behavior in any of the solutions.

Re your edited question:  IIRC, for an overdamped (stable) 2nd order system with non-zero initial conditions, the trajectory for $t>0$ either asymptotically approaches $x=0$ or crosses $x=0$ once before asymptotically approaching $x=0$. Isn't that all there is to it?
A: In my work, I am dealing a lot with cases where A=-B in a somewhat related case of aroma release (It's not an overdamped pendulum but instead two coupled first-order differential equations that can be written as a single second-order differential equation like yours). 
The way that I look at is as a sum of two exponetial terms:
$$x(t) = A e^{-c_1t} + B e^{-c_2t} $$
with $c_2 > c_1 > 0$ relating to two decaying terms


*

*One term is a more slowly decaying term $A e^{-c_1t}$. 

*The other term $B e^{-c_2t}$ is a faster decaying term (or I actually see it as an increasing term when B is negative). 
This second term makes the end result, $x(t)$, a curve that is moving towards this slowly decaying term (the difference with it is decreasing exponentially).

Image example:

So you can see the four different situations depicted by the four black curves. In all cases they are moving towards this red curve which is that slowly decaying term $A e^{-c_1t}$. The curves have an asymptote in $x=0$ but differ in the point when they cross the point x=0, with the four cases: 
$$\begin{array}{rl}t>0 &\quad\text{for} \quad  B/A<-1\\
 t=0 &\quad\text{for} \quad B/A=-1\\
t<0  &\quad\text{for} \quad-1<B/A<0\\
\text{never} &\quad\text{for} \quad 0 \leq B/A \\
\end{array}$$

This last case $B/A > 0$ relates to a pendulum that initially has a fast speed, $\dot{x}$, toward $x=0$. This speed get's slowed down quickly due to the large damping force $-\mu\dot{x}$. This damping of the speed relates to the faster $B e^{-c_2t}$ (this term is in the same way a slowing down term when $B/A < 0$ but then the speed is away from $x=0$) . When you reach the slower red curve relating to the $A e^{-c_1t}$ term then the damping force almost equals the spring force and $m\ddot{x}=-\mu\dot{x}-kx \approx 0$.
A: The equation of motion of an harmonic oscillator can be written as:
$
\ddot{x}+2\xi\omega_n\dot{x}+\omega_n^2x=0
$
with initial conditions $x(0)=x_0$ and $\dot{x}(0)=\dot{x}_0$, where $\omega_n$ is the natural frequency of the system and $\xi$ is called the damping ratio, and can be calculated as
$
\xi=\frac{\mu}{2m\omega_n}
$
The system is said to be overdamped if $\xi>1$ and, for this case, the solution of the differential equation is:
$
x(t)=C_1e^{\left(-\xi+\sqrt{\xi^2-1}\right)\omega_nt}+C_2e^{\left(-\xi-\sqrt{\xi^2-1}\right)\omega_nt}
$
The constants $C_{1,2}$ are obtained from the initial conditions:
$
C_{1,2}=\frac{\pm x_0\omega_n\left( \xi \pm \sqrt{\xi^2-1} \right) \pm \dot{x}_0}{2\omega_n\sqrt{\xi^2-1}}
$
We notice from the response of this system: The motion is aperiodic, regardless of the initial condition. The motion diminishes exponentially with time.
When $\xi$ is appreciably greater than unity, one of the two decaying exponentials decreases much faster than the other, so the faster-decaying exponential term may be neglected. The solution can be approximately written as:
$
x(t)\approx C_2e^{\left(-\xi-\sqrt{\xi^2-1}\right)\omega_nt}
$
Now, the decaying exponential will never be exactly zero, but you can get an estimate from zero. Say, $\left| x(t) \right|<\epsilon$, so
$
C_2e^{\left(-\xi-\sqrt{\xi^2-1}\right)\omega_nt} \approx \epsilon
$
and you can solve for $t$.
