Newtonian Mechanics – Why Critical Damping Returns to Equilibrium Faster than Overdamping

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How can one prove that the critical damping regime returns to equilibrium fastest? Is it true that there are cases where the heavy damped solution will return faster?

Consider a free harmonic oscillator $$x''(t) + \gamma x'(t)+ \omega_0^2 x(t) = 0.$$ This has solutions of the following:

Heavy damping: $\gamma > 2 \omega_0$
$$
x(t) = A_1 e^{(-\gamma/2-\sqrt{\gamma^2/4-\omega_0^2})t} + A_2 e^{(-\gamma/2+\sqrt{\gamma^2/4-\omega_0^2})t} = A_1 e^{-\mu_1t} + A_2 e^{-\mu_2t}, \tag{1}
$$

where
$$
\mu_1 = \frac{\gamma}{2}+\sqrt{\frac{\gamma^2}{4}-\omega_0^2} \text{ , } \mu_2 = \frac{\gamma}{2}-\sqrt{\frac{\gamma^2}{4}-\omega_0^2}.
$$

Clearly, $\mu_1 > \gamma/2 > \omega_0$ (since heavy damping configuration) and $\mu_1 \mu_2 = \omega_0^2$ so therefore $\mu_2 < \omega_0$.

Critical damping: $\gamma = 2 \omega_0$
$$
x(t) = (B_1+B_2t)e^{-\gamma/2t} = (B_1+B_2t)e^{-\omega_ot}. \tag{2}
$$

In $(1)$, the second term is the slowest to decay since $\mu_1 > \mu_2$. Furthermore, the second term in $(1)$ decays more slowly than the exponential term in $(2)$ since $\mu_2 < \omega_0$. Hence, critical damping returns to equilibrium faster than heavy damping.

However, what if the constant $A_2$ is zero? Surely then the heavy damping scenario returns to equilibrium more quickly than critical damping?

Best Answer

There are good mathematical answers to this question. I'd like to give an intuitive one.

Consider the following situation: a sphere falling through a viscous fluid:

sphere falling through a viscous fluid

Equilibrium is when the sphere sits at the bottom without any movement.

Damping can be increased by making the fluid more viscous.

An underdamped situation is when the fluid's viscosity is too low. When we drop the sphere, it will bounce back and forth before settling down.

In a very overdamped situation, the sphere will take a long time to fall through the fluid to reach the bottom.

In the perfectly balanced situation (critical damping), the sphere falls as fast as it possibly could and there is no rebound. It just touches the bottom surface and sits there. This will be the fastest as compared to previous two cases.

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