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?

  • $\begingroup$ Your tags are in the category of oscillators and you only mentioned an exponential functional form. Have you considered heavily damped sinusoidal functional forms? $\endgroup$ Jan 2, 2022 at 16:49
  • $\begingroup$ @DavidWhite a sinus IS an exp function, one with an imaginary exponent. In the above formulas you see omega-squared, meaning that you won't see the i, because i squared is -1 is real. By squaring, you look at the amplitude. $\endgroup$
    – Roland
    Jan 4, 2022 at 14:31

8 Answers 8


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.

  • 5
    $\begingroup$ It's not intuitively obvious that you couldn't save time by having the sphere fall a bit faster, and have the resulting tiny bounce take less than the time saved. But it's easy to accept that that's how the math turns out, i.e. that getting turned around after bouncing takes a long time, longer than you could save by going a bit faster. $\endgroup$ Jan 3, 2022 at 21:16
  • 2
    $\begingroup$ @peter - now that I think about it, yeah, it won't be very intuitive. a lot of people can think that a slightly faster-moving ball with a few smaller rebounds can be faster as compared to a slow-moving ball that touches the bottom and sits there. When writing the answer, I deliberately asked for an assumption that viscosity is too low so that people visualize it as an extreme case of rebounds that will take a lot of time. However, at boundary conditions, that intuition can go for a toss. Thinking about it, I feel maths is the only thing that will be able to clear the confusion in that case. $\endgroup$ Jan 4, 2022 at 4:48
  • $\begingroup$ This is hardly a second order system. You mention mass and viscosity. Unless you add that the bottom is a spring-like object. You mention that the ball may bounce, but only in passing. A block of concrete will also bounce when falling on a concrete floor, but that is hardly the same as a mass-spring system with little resistance and a lot of "oscillation". $\endgroup$
    – Roland
    Jan 4, 2022 at 14:45
  • $\begingroup$ "no rebound"???? with critical damping there is usually a slight rebound, if I remember right some 7 percent. Looking at a moving coil volt meter that is critically damped, you will clearly see this slight "overshoot". $\endgroup$
    – Roland
    Jan 4, 2022 at 14:48
  • $\begingroup$ @Roland Moving coil meters are slightly under damped to try and ensure that they do not "stick" before reaching the correct reading. $\endgroup$
    – Farcher
    Jan 5, 2022 at 11:30

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to compare the heavy damping path and the critical damping path you have to adjust the initial conditions you obtain

$$x_C(t)=\left( c_{{1}}+c_{{2}}t \right) {{\rm e}^{-\beta\,t}}$$ and $$x_H(t)=w_1\,e^{\mu_1\,t}+w_2\,e^{\mu_2\,t}\\ w_1=\frac 12\,{\frac {c_{{1}}{\omega_{{0}}}^{2}-c_{{1}}{\beta}^{2}-\sqrt {-{ \omega_{{0}}}^{2}+{\beta}^{2}}c_{{2}}}{{\omega_{{0}}}^{2}-{\beta}^{2}} } \\ w_2=\frac 12\,{\frac {\sqrt {-{\omega_{{0}}}^{2}+{\beta}^{2}} \left( -\sqrt {-{ \omega_{{0}}}^{2}+{\beta}^{2}}c_{{1}}+c_{{2}} \right) }{{\omega_{{0}}} ^{2}-{\beta}^{2}}} $$


hence: critical damping return to equilibrium fastest because the area under the "critical path" $~\int_0^{T} x_C(t)\,dt~$ is smaller then the area under the "heavy path" $~\int_0^{T} x_H(t)\,dt~$

the initial conditions:

$$\mu_1=-\beta+\sqrt {-{\omega_{{0}}}^{2}+{\beta}^{2}}\\ \mu_2=-\beta-\sqrt {-{\omega_{{0}}}^{2}+{\beta}^{2}}$$

$$x_H(0)=x_C(0)\quad\Rightarrow \\w_{{1}}+w_{{2}}=c_{{1}}\\ \dot x_H(0)=\dot x_C(0)\quad\Rightarrow \\-w_{{1}}\beta+w_{{1}}\sqrt {-{\omega_{{0}}}^{2}+{\beta}^{2}}-w_{{2}} \beta-w_{{2}}\sqrt {-{\omega_{{0}}}^{2}+{\beta}^{2}}=c_{{2}}-\beta\,c_ {{1}}$$

from here you obtain $~w_1,~w_2~ $ as a function of $c_1~,c_2,\omega_0~,\beta$

  • $\begingroup$ Correct me if I have made a mistake, but I'm assuming you have used $\mu_1 = -\beta + \sqrt{\beta^2-\omega_0^2}$ and $\mu_2 = -\beta - \sqrt{\beta^2-\omega_0^2}$. Now consider the initial conditions such that $w_1 = 0$. We find that $c_2 = - c_1 \sqrt{\beta^2-\omega_0^2}$ and $w_2 = c_1$. Hence $$ x_c(t) = c_1(1 - \sqrt{\beta^2-\omega_0^2}t)e^{-\beta t} $$ $$ x_H(t) = c_1e^{(-\beta - \sqrt{\beta^2-\omega_0^2})t} $$ If this is correct, clearly the heavy damped case returns to equilibrium more quickly (e.g. plot with $\beta = 1, \, \omega = 0.1, \, c_1 = 10$. So is this an exception? $\endgroup$
    – user246795
    Jan 2, 2022 at 21:59
  • $\begingroup$ @user246795 why is $~w_1=0~$ ? i put more information see edit "initial conditions " $\endgroup$
    – Eli
    Jan 2, 2022 at 22:16
  • $\begingroup$ I'm asking is the case when $w_1 = 0$ an exception to the general statement "critically damped systems return to equilibrium fastest"? I accept that for general non-zero coefficients $w_1, \,w_2$ it is true. $\endgroup$
    – user246795
    Jan 3, 2022 at 8:45
  • 2
    $\begingroup$ @user246795 if you think you have found an exception to a rule, it should be possible to give an explicit example with "real world" values (i.e. in this case: a specific initial position and velocity which lead to a specific damping factor being more optimal than the calculated critical damping for those conditions). Personally I think that Joe Natal is correct that you have made a free variable of something which is not actually free, but in these cases it is easier to discuss potential misunderstandings (or prove correctness) with a concrete example. $\endgroup$
    – Dave
    Jan 3, 2022 at 15:28
  • $\begingroup$ @Dave do you agree that if we can find initial conditions such that $w_1 =0$, then we will a specific case where the heavy damping case returns to equilibrium more quickly? I get that any $\mu_2 x(0) = \dot{x}(0)$ will give $w_1 = 0$, where $x(0)$ and $\dot{x}(0)$are initial position and velocity respectively. $\endgroup$
    – user246795
    Jan 3, 2022 at 15:57

It is only reasonable to compare the overdampled and critically damped case given the same initial conditions. Equivalency of both the initial velocity and position gives two equations.

Equating positions at $t = 0$ gives

$$ A_1 + A_2 = B_1 $$

and equating initial velocities:

$$ \mu_1 A_1 + \mu_2 A_2 = \omega_0 B_1 - B_2 $$

This means $A_2$ is not necessarily a free parameter. Plug in initial conditions (e.g. $x(0) = 1$ and $x'(0) = 0$) and then compare the behvior of the two system responses.

  • $\begingroup$ Yes, I had forgotten to mention this. But considering a case where $A_2$ was zero (and non-trivially $A_1 \neq 0$) then we are still comparing simply the two exponential terms, for which the over damped case is going to approach equilibrium faster, right? $\endgroup$
    – user246795
    Jan 2, 2022 at 16:31

If your initial condition is such that $A_2$ is exactly zero, then indeed, in the absence of the second term in the rightmost object in (1), the first one would decrease faster than both terms rightmost object in (2).

But exactitude does not exist in the real world. Even if initially $A_2$ is as small as you can arrange it to be, it will never be exactly zero. And since both terms in (2) decrease exponentially faster than $e^{-\mu_2 t}$, however small a non-exactly zero value of $A_2$ is present in the very carefully prepared initial state you use, eventually the term that damps at the slowest rate, namely $A_2 e^{-\mu_2 t}$ will dominate.

Of course, if you prepare your initial state very carefully, so $A_2$ is extremely small, by the time $A_2 e^{-\mu_2 t}$ dominates the motion everything will be below the precision of your measuring apparatus. Then the experimental curve would show a faster decrease for the "heavy damping" up to the point where it is lost in the thickness of the ink of the line.

I suppose that, for cases where $\gamma$ is just barely larger than $2\omega_0$ it could even be done experimentally. At the price of an enormous effort.

But who cares ?


What the OP asks for is indeed possible but only for very specific initial conditions.

There are a few things to discuss, the first being what does "returning to equilibrium faster" mean. First, we are trying to compare the behavior of two different systems (I will assume they have the same damping coefficient $\gamma$ but different resonant frequencies $\omega_{0,1}$, $\omega_{0,2}$), so it makes sense to only compare solutions to the two systems with the same initial conditions.

Second, unless the system is at rest with $x(t) = 0$ forever, the system is never actually in equilibrium and takes an infinite amount of time to reach equilibrium. One possible way to define "returning to equilibrium faster" would be that, given some fixed initial conditions $x_0, v_0$ , the solution $x_1(t)$ to system 1 approaches $0$ asymptotically faster than the solution $x_2(t)$ to system 2, both satisfying $x_1(0) = x_2(0) = x_0$ and $x_1'(0) = x_2'(0) = v_0$.

Another possible way to define it would be to compare the times $T_1, T_2$ it takes for $x_1(t), x_2(t)$ to get below a certain fixed value $M$, i.e. $T_1$ is the smallest $t \geq 0$ such that $\left|x_1(t)\right| < M$ for all $t > T_1$, and similarly for $T_2$. Then $x_1$ returns to equilibrium faster than $x_2$ if $T_1 < T_2$.

I will work with the first definition in terms of the asymptotic behavior.

In that case as you already mentioned, you are looking for initial conditions for which $A_2 = 0$. In that case, since $\mu_1 > \frac{\gamma}{2}$ and $\mu = \frac{\gamma}{2}$, the solution to system 1 would approach $0$ asymptotically faster than the solution to system 2.

This means that our solution in the overdamped system is $x_1(t) = A_1e^{-\mu_1t}$. Looking at $t = 0$, we have $x_1(0) = A_1$ and $x_1'(0) = -\mu_1 A_1$. Therefore, it is possible for a solution to system 1 to reach equilibrium faster than a solution to system 2 if the initial conditions satisfy the property that $x'(0) = -\mu_1 x(0)$. As $\mu_1$ is positive, and assuming we start from a positive value of $x$, then this just means that you need to give an initial velocity with just the right amount in the negative direction to achieve this. However, if we are starting at rest (zero initial velocity), it is impossible unless we are already at equilibrium.

One explicit numerical example for such a situation: let $\gamma = 4$ and let $\omega_0 = \sqrt{3}$ in the overdamped case and $\omega_0 = 2$ in the critically damped case. Let the initial conditions be $x(0) = 1$ and $x'(0) = -3$.

The differential equation in the overdamped case is $x''(t) + 4x'(t) + 3x(t) = 0$. The solution satisfying $x(0) = 1$, $x'(0) = -3$ is $x_1(t) = e^{-3t}$ as can be easily checked.

The differential equation in the critically damped case is $x''(t) + 4x'(t) + 4x(t) = 0$. The solution satisfying $x(0) = 1$, $x'(0) = -3$ is $x_2(t) = (1 - t)e^{-2t}$ as can also easily be checked.

And indeed $e^{-3t}$ returns to $0$ asymptotically faster than $(1 - t)e^{-2t}$ so this is an example of the phenomenon the OP is asking for.

As mentioned before, we had to pick the initial conditions just right to make this work. In practice there will never be an exact precision in the initial velocity that is given, so there will always be some component decaying at the slower rate. However, it may still be possible to make the magnitude of this component small enough for short times, and then we can use the second definition of "returning to equilibrium faster" to produce more realistic examples of the phenomenon.

Another thought: the previous discussion was based on the assumption that we keep the damping coefficient $\gamma$ the same and vary the resonant frequency $\omega_0$ to achieve overdamping or critical damping. However it might be more interesting to keep $\omega_0$ constant and vary $\gamma$ instead. So we have damping coefficients $\gamma_1 > \gamma_2 = 2\omega_0$ for the overdamped system and the critically damped system, respectively.

Then $\mu_1 = \frac{\gamma_1}{2} + \sqrt{\frac{\gamma_1^2}{4} - \omega_0^2}$, $\mu_2 = \frac{\gamma_1}{2} - \sqrt{\frac{\gamma_1^2}{4} - \omega_0^2}$, $\mu = \omega_0$

Since $\mu_1$ is a strictly increasing function of $\gamma_1$ and $\mu_1 = \omega_0$ when $\gamma_1 = 2\omega_0$, it is again true that $\mu_1 > \mu$. Similarly $\mu_2 < \mu$. So we can proceed exactly as in the previous discussion to find initial conditions for which the term involving $\mu_2$ is zero, and keep the term involving $\mu_1$ which decays faster than critically damped, for a fixed $\omega_0$.


Critical damping is the condition which transitions between overshoot of the static equilibrium position (under damped) and an exponential decay to the static equilibrium position (over damped).

With less damping than critical damping the system will reach the static equilibrium position quicker than one which was critically damped but will overshoot and continue oscillating about the static equilibrium position with an exponentially decaying (never ending) amplitude.

With more damping than critical damping the system will always lag behind a critically damped system as it will lose mechanical energy faster and reach the static equilibrium position exponentially (never ending).

Critical damping means that the system will reach the static equilibrium position with no overshoot and with zero velocity and hence will be the fastest.


Simplest answer is that it is just a matter of jargon. Obviously, with very little damping the system may vibrate increasingly (unstable), or the vibration may dampen out after a long time.

With a lot of damping it will get to the stable position but it may take long, very long, or "forever".

So somewhere in between is the amount of damping where the system gets stable in the shortest time. That amount of damping will be called "critical damping".


WLOG you can rescale x so that the initial displacement is 1 and rescale time so that $\omega _0 = 1$. For convenience you may also relabel $\gamma = 2q$ which is fine because we will check our result as $q$ varies. We take initial velocity to be zero because otherwise the result is not necessarily true.

$\ddot{x} + 2q \dot{x} + x = 0$

The solution to the equation using our initial conditions is: $$ x = e^{-t} [\frac{p+q}{2p} e^{(p+1-q)t} + \frac{p-q}{2p} e^{-(p + q-1)t}] $$ $$ p := \sqrt{q^2 -1} $$

If you want to verify then express it as

$$ x = \frac{p+q}{2p} e^{(-q+p)t} + \frac{p-q}{2p} e^{(-q-p)t} $$

which clearly matches the initial conditions and the auxillary equation $\lambda ^2 + 2q \lambda + 1 = 0$. Note that $q>1$ as we are over-damped and $p<q<p+1$.

The critically damped solution is $e^{-t}(1+t)$.

So all we need to compare is $(1+t)$ vs $[\frac{p+q}{2p} e^{(p+1-q)t} + \frac{p-q}{2p} e^{-(p + q-1)t}]$

The first derivative of the term in square brackets is $$ [\frac{p+q-1}{2p} e^{(p+1-q)t} + \frac{p-q+1}{2p} e^{-(p + q-1)t}] $$

which matches the value of the first derivative of (1+t) at $t=0$. Namely 1.

The second derivative of the term in square brackets is $$ \frac{(p+q-1)*(p-q+1)}{2p} e^{(p+1-q)t} - \frac{(p-q+1)(p+q-1)}{2p} e^{-(p + q-1)t} $$

which, for t>0 is strictly greater than

$$ \frac{(p+q-1)*(p-q+1)}{2p} e^{(p+1-q)t} - \frac{(p-q+1)(p+q-1)}{2p} e^{-(p+1 - q))t} $$

because p+1-q < p+q-1.

In turn, for t>0 this is strictly greater than 0. As opposed to the second derivative of (1+t) which is zero everywhere.

We may now use the Cauchy Mean Value Theorem to conlude the first derivative of the term in square brackets is greater than the first derivative of (1+t) for t>0.

We may now use the Cauchy Mean Value Thorem again to conclude the term in square brackets is greater than (1+t) for t>0 and now we're done.

Cauchy Mean Value Theorem: If $f$ and $g$ are continuous on [a,b] and differentiable on (a,b), then there is a number x in (a,b) such that $\frac{f(b)-f(a)}{g(b)-g(a)} = \frac {f'(x)}{g'(x)}$.

From the Cauchy Mean Value Theorem it is trivial to prove that if f(0) = g(0) and f'(x) > g'(x) for x>0 then f(x) > g(x) for x>0, which is what I actually used in my argument about the derivatives.

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