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.
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$\begingroup$ Presumably the equilibrium is where $x=0$? If so, $\mathrm{e}^{-\omega_0 t}$ will always dominate the term in $t$ in parentheses, and the expression will therefore tend towards equilibrium as $t$ increases, but it will take infinitely long to actually get there. Am I missing something? $\endgroup$– TonyCommented Feb 8, 2021 at 4:40
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$\begingroup$ @Tony You are correct, the oscillator will reach equilibrium at $t = \infty$. I meant when it "practically" reaches equilibrium, say, within 2 decimal places. I will amend my question. $\endgroup$– user288363Commented Feb 8, 2021 at 4:46
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$\begingroup$ Seems like you should numerically calculate, for a given set of parameters, when it reaches whatever threshold you set. Then you can sweep the parameters as you want, and see what happens. Might be good coding practice. Why do you want an expression if you already know a simple one can't be found? $\endgroup$– prolyxCommented Feb 8, 2021 at 5:01
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$\begingroup$ @JonathanJeffrey Sorry, but I'm having a hard time understanding your comment... I only know highschool-level math, and I don't even know what to look up to help me understand it... from what I do understand, you're telling me the only thing I can do is approximate the answer, but to do that, from what I can tell, I'd need to approximate the $-1$ branch of the Lambert $W$ function or something like that, and I don't know where to find that. $\endgroup$– user288363Commented Feb 8, 2021 at 5:22
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$\begingroup$ If you solve this equation $~e^{-\omega_0\,t}=c~$ for t you obtain $~t=-\frac{\ln(c)}{\omega_0}~$ for a very small value of c you get the solution $\endgroup$– EliCommented Feb 8, 2021 at 9:02
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1 Answer
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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.
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1$\begingroup$ Your first derivative is correct by my account, but your second derivative should be $\omega_0e^{-\omega_0t}[\omega_0(v_0+\omega_0x_0)t-\omega_0x_0-2v_0]$ $\endgroup$– user288363Commented Feb 8, 2021 at 14:36
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$\begingroup$ @Freddy Thank you, you are quite right. I have made a correction. $\endgroup$– TonyCommented Feb 8, 2021 at 21:38