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If you have a mass-spring simple harmonic oscillator, with a mass of $m$, spring coefficient $k$ and damping coefficient $c$, the critical damping condition as shown here and here is defined by:

$$ω = \sqrt{\frac{k}{m}}$$

$$c = \sqrt{4mk}$$

Let's say you have a hypothetical such critically damped oscillator of a known $m$, $k$ and $c$ which is released with zero velocity from an initial displacement of $x$.

It will eventually settle out to a final displacement of $0$. Can you calculate the maximum velocity that will be achieved along the way? If so how?

I could "solve" this by running it through a stepwise simulation with a given sampling rate, but I am wondering if there is an actual solution mathematically for this problem so it is not so inefficient and I can be more precise.

Is there an equation that describes the displacement curve of a critically damped oscillator released from zero velocity at a certain starting point? If so then I presume I would just have to find the derivative of that and get the maximum absolute value.

Any thoughts? Thanks.

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2 Answers 2

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the ODE is:

$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$

with $$c=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$

you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$

$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$

hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$

you can obtain the time where the velocity is maximum by solving this equation

$$\frac{d}{dt} v(t)=0$$

for t, $~\Rightarrow~t_m=t$ and

$~|v_m|=\frac{\omega\,x_0}{e}$

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    $\begingroup$ Wow that simplifies so much! What an easy equation to use then. Thanks. $\endgroup$
    – mike
    May 31 at 16:34
  • $\begingroup$ I am looking at it in more detail and I am curious to see how you got the $x(t)$ equation. Would you be willing to add one or two more steps in there to show how you got it from the lines above it? I presume $v(t)$ was then found by taking derivative of $x(t)$. Thanks for any clarification. I appreciate your reply already though so no worries if it's too much work. Thanks again. $\endgroup$
    – mike
    Jun 1 at 0:32
  • $\begingroup$ If the eigenvalues are equal then you make this solution ansatz $x\left( t\right) =c_{1}e^{\lambda t}+c_{2}te^{\lambda t}$ you got it? $\endgroup$
    – Eli
    Jun 1 at 6:34
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You can find the equation of motion for the critical damping regime with the given initial conditions and then find the maximum veocity, which corresponds to zero acceleration. You can find the general results for th simpler case (damping proportional to speed) in many places. Here is on example: https://byjus.com/jee/damped-oscillation/#:~:text=The%20effect%20of%20radiation%20by,are%20said%20to%20be%20damped You will have to find the equation of motion for your initial conditions (vo=0, xo). For other types of damping you may have to do numerical calculations.

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