The Problem
How can I calculate a stiffness (k) and dampening (b) coefficient for an under-damped oscillation such that the system settles by a given time (t)?
The Environment
Given the under-damped oscillation formula:
$$ x(t) = e^{\frac{-bt}{2}} (\cos(\lambda t) + \frac{v}{\lambda} sin(\lambda t))$$
And the following known values:
$$ t = \text{duration} $$ $$ v = \text{velocity} $$ $$ \lambda = \sqrt{k - \frac{b^2}{4k}} $$ $$ b \leq k $$
I am attempting to determine a k and b value such that the system "settles" by a specific duration, t, e.g. 3 seconds.
$$ k = \text{stiffness} $$ $$ b = \text{damping} $$
To simplify things, k and b are directly related to one another via the following formula:
$$ 2d\sqrt{k} = b $$
where d is some value from (0...1]. A d value of 1 (critically damped) will mean there is no oscillation, while a value closer to 0 will increase the amount of oscillation.
What I've Tried
I know that I can calculate some b value by solving $\epsilon = e^{\frac{-bt}{2}}$:
$$ b = \frac{2\ln{\epsilon}}{t} $$
But this doesn't seem to take into account velocity which doesn't intuitively make sense. For example, given two velocities, one much larger than the other, we would have the same b value which would imply that by t the system with the larger velocity will not have settled as much as the other.
I have also tried solving the following equation:
$$ 0 = (\cos(\lambda t) + \frac{v}{\lambda} sin(\lambda t))$$
But this ends up with a formula that can't be solved without using a root finder and I'm not sure how to make a good guess (maybe plugging in the estimated b value above could work?).
Will post back here as I develop this problem and attempt to find a solution.
We can start with the relations $$ d^2 = \frac{b^2}{4k} \quad \text{and} \quad \lambda^2 = k - \frac{b^2}{4k} = k - d^2 \,. $$ For simplicity, consider the case where $d = 1$, i.e., $\lambda^2 = k - 1$ and $b^2 = 4k$.
The expression for the displacement in this case is $$ x(t) = \exp\left(-\sqrt{1+\lambda^2}\,t\right)\left[\cos(\lambda\,t) + \frac{v_0}{\lambda}\,\sin(\lambda\,t)\right] $$ When the system has settled, $dx/dt, d^2x/dt^2$ are also both simultaneously close to zero, i.e., less than $\varepsilon_1$.
For example, if $\varepsilon_1 = 10^{-20}$ and $v_0 = 10^{10}$, the exponential term will have to dominate over the velocity term. So you are correct when you use your approach to estimate $t$ at which all three quantities (displacement, velocity, acceleration) are close to zero: $$ t \sim -\frac{\log\left(\frac{\varepsilon_1\lambda}{v_0}\right)}{\sqrt{1+\lambda^2}} \sim -\frac{\log\varepsilon}{\sqrt{1+\lambda^2}}\,. $$ For physically reasonable values of $\lambda$ and $v$, $\varepsilon = \varepsilon_1\lambda/v_0$ will be a small number. You don't have to worry about $v_0$ and $\lambda$ inside the $\log$ term as long as you make sure that $\varepsilon$ is small. Asymptotics are the devil's invention (see paper here), particularly when the function you are exploring does not have a true asymptote but fluctuates around zero.