Why does a critically damped oscillator undergo a quicker decay than an overdamped one? Starting with the initial conditions, $x(0)=x_0$ and $v(0)=0$, the solution of a damped oscillator:
$$\ddot{x}+2\gamma\dot{x}+\omega_0^2x=0,$$
for the overdamped case $(\gamma\gg \omega_0)$ is given by
$$x(t)=\frac{x_0}{2}e^{-\gamma t}\left[\left(1+\frac{\gamma}{\sqrt{\gamma^2-\omega_0^2}}\right)e^{\sqrt{\gamma^2-\omega_0^2}t}+
\left(1-\frac{\gamma}{\sqrt{\gamma^2-\omega_0^2}}\right)e^{-\sqrt{\gamma^2-\omega_0^2}t}
 \right]$$
and for the case of critically damped oscillation  $(\gamma\to\omega_0)$ is given by
$$x(t)=x_0(1+\omega_0t)e^{-\omega_0t}.$$
Without graphing these functions, can we analytically understand from these solutions why the decay of amplitude in the second case is quicker than in the first case?
 A: 
Without graphing these functions, can we analytically understand from these solutions why the decay of amplitude in the second case is quicker than in the first case?

There are two decaying modes in overdamped oscillations
$$e^{-\bigl(\gamma - \sqrt{\gamma^2 - \omega_0^2} \bigr) t} \qquad \text{and} \qquad e^{-\bigl(\gamma + \sqrt{\gamma^2 - \omega_0^2} \bigr) t}$$
With $\gamma \gg \omega_0$, both terms next to time $t$ in the above exponentials are positive. The exponential with smaller (positive) term will decay more slowly
$$e^{-\bigl(\gamma - \sqrt{\gamma^2 - \omega_0^2} \bigr) t} \qquad \rightarrow \qquad \text{decays slower} \tag 1$$
where $\sqrt{\gamma^2 - \omega_0^2} < \gamma$.
In critically damped oscillations, $\gamma = \omega_0$ and the slower decaying mode is
$$t e^{-\omega_0 t} \tag 2$$

*This is first time I do this analysis, I might be wrong about some stuff. I have to recheck everything.
Let's find ratio between the two decaying modes in Eqs. (1) and (2)
$$r(t) = \frac{e^{-\bigl(\gamma - \sqrt{\gamma^2 - \omega_0^2} \bigr) t}}{t e^{-\omega_0 t}} = \dots = \frac{1}{t} e^{\frac{2\omega_0 t}{1 + \sqrt{\frac{1 + \omega_0/\gamma}{1 - \omega_0/\gamma}}}}$$
For $\gamma \gg \omega_0$ the above ratio is simplified into
$$r(t) \approx \frac{1}{t} e^{+\omega_0 t} \tag 3$$
The mode from Eq. (1) decays when
$$t^\star = \frac{5}{\gamma - \sqrt{\gamma^2 - \omega_0^2}}$$
and the ratio from Eq. (3) at $t = t^\star$ is
$$r(t^\star) = \frac{\gamma}{5} \bigl(1 - \sqrt{1 - (\omega_0/\gamma)^2} \bigr) e^{\frac{5 \omega_0/\gamma}{1 - \sqrt{1 - (\omega_0/\gamma)^2}}}$$
If we introduce new variable $z = \omega_0/\gamma$, the above expression can be written as
$$r(t^\star) = \frac{\gamma}{5} \bigl(1 - \sqrt{1 - z^2} \bigr) e^{\frac{5 z}{1 - \sqrt{1 - z^2}}}$$
A: You got the math, but for the physical intuition...
"Critically damped" means that there is just enough damping to prevent oscillations. It is the smallest amount of damping you can have without oscillations. "Overdamped" has more damping than this, and as such means the decay will be slower.
Imagine a mass on a spring in syrup. If the syrup is thicker (more damping) then the mass will move slower through it to the final equilibrium position than in a thinner syrup.
A: The overdamped solution is so damped it moves more slowly towards equilibrium than the critical solution does.
