Time for critically damped oscillator to reach equilibrium? The title says it all. With my limited knowledge of physics and math, I have no idea where to begin, as the position function I have for a critically damped oscillator, $x=e^{-\omega_0t}[x_0+(v_0+\omega_0x_0)t]$ where $\omega_0$ is the undamped frequency of the oscillator, does not have an analytical solution for $t$.  By equilibrium, I mean within a few decimal places of equilibrium, as the oscillator only approaches $0$ as $t$ goes to infinity.
 A: I think you could use the Newton-Raphson method to solve this for $x=a$, where $a$ is some constant close to zero that you choose (e.g. 0.01).
Let's make a function $f(t)$ from your function:
$$f(t) = \mathrm{e}^{-\omega_0 t}\left[x_0 + (v_0+w_0 x_0)t\right] - a$$
We would like this to be equal to zero, since that's the case in your original equation, with your choice of $x=a$.
Calculate the derivative of $f(t)$ with respect to $t$.  After some messing about I think (you must check!) that you get
$$f'(t) = \mathrm{e}^{-\omega_0 t}\left[v_0-\omega_0(v_0+\omega_0 x_0)t\right]$$
Start with some initial guess for $t$, which we will call $t_0$.  Then calculate a new guess $t_1$ as follows
$$t_1 = t_0 - \frac{f(t_0)}{f'(t_0)}$$
Then another guess:
$$t_2 = t_1 - \frac{f(t_1)}{f'(t_1)}$$
and keep going like that until your guesses get close enough together that you don't care about the difference.
A problem with this method in general is that the guesses do not always converge.  I think it should work in your case, but you'll need to be a bit careful about your first guess.  If you want to be more rigorous, you can calculate the second derivative (check!):
$$f''(t) = \omega_o \mathrm{e}^{-\omega_0 t}\left[-(2v_o+\omega_0 x_0) + \omega_0(v_0 + \omega_0 x_0)t\right]$$
then choose $t_0$ so that:

*

*$f'(t_0) < 0$ and $f''(t_0) > 0$ if $f(t_0) > 0$

*$f'(t_0) > 0$ and $f''(t_0) < 0$ if $f(t_0) < 0$
In that case I suspect your guesses should go nicely to the right place.
