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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.

Example graph with some arbitrary values of b and k

featherless
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  • Could you give us an idea of the units of each of your quantities? A superficial look suggests that $\lambda, b$ have units $T^{-1}$, $v$ has units of $T^{-1}$, and $k$ has units $T^{-2}$. Also, is $v$ an initial velocity or is it $dx/dt$? – Biswajit Banerjee Apr 04 '14 at 00:55
  • Sure! t is seconds. x(t) starts at 1 for t = 0 and approaches 0 as t approaches infinity. v is the initial velocity. Just added a graph as well. – featherless Apr 04 '14 at 01:17
  • $\lambda$, $b$, and $k$ are each scalar values. In the graph attached $k = 339.3215$, $b = 18.4207$, and $\lambda = 15.9528$ ($d = 0.5$ was used to calculate $k$ after estimating $b$ using the first equation I noted for solving $b$ with a $t$ of 1). – featherless Apr 04 '14 at 01:25

1 Answers1

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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.

Biswajit Banerjee
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  • The problem is that as the epsilon gets smaller the dampening curve gets shallower. I could choose an infinitely small epsilon in order to ensure that the initial velocity is never a problem, but this would lose detail in the curve. I'm trying to find a way to calculate $k$ and $b$ values such that the system comes very close to rest at a given $t$ duration without dampening the curve so much that I lose detail. – featherless Apr 04 '14 at 02:51
  • Ideally the model should have physically realistic initial conditions that respect the assumption of linear superposition. That implies that the init vel is small. If you want high accuracy there is no option other than numerical solutions. Another rough estimate can be found by trying to fit a bessel fn to the curve and picking one of the zeros. – Biswajit Banerjee Apr 04 '14 at 07:26
  • Given the last equation that you posted, how would I solve for $\lambda$, given its presence twice in the equation? Is approximation via a root finder the only method? Also it's important to note that assuming d=1 trivializes the equation to the critically-damped case and won't necessarily work for d<1 (underdamped) unless I'm misunderstanding the simplification (quite likely). – featherless Apr 09 '14 at 14:51