Re: Critical damping of pendulum

In article <1113320446.324279.289460@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
tadchem <thomas.davidson@xxxxxxx> wrote:
>
>JS Groot wrote:
>> tadchem wrote:
>> >
>> > Try this:
>> >
>http://mathworld.wolfram.com/DampedSimpleHarmonicMotionCriticalDamping.html
>>
>> Thanks for the link. However, the solution presented there is also
>> the one for the linearised differential equation (the 2nd equation in
>> my posting), resulting in simple harmonic motion.
>
>"Critical damping" is defined various ways:
>
>Critical Damping: Critical damping is the smallest amount of damping at
>which a given system is able to respond to a step function without
>overshoot.
>(from http://www.flw.com/define_c.htm)
>
>critical damping - damping that just prevents oscillations: damping of
>a system that is just enough to prevent oscillations occurring
>(from
>http://encarta.msn.com/encnet/features/dictionary/DictionaryResults.aspx?refid=561531742)
>
>critical damping - The minimum damping that will prevent or stop
>oscillation in the shortest amount of time, typically associated with
>oscillatory systems like geophones.
>(from
>http://www.glossary.oilfield.slb.com/Display.cfm?Term=critical%20damping)

Those are all the same thing.

>
>Critical damping is a special case of damped simple harmonic motion in
>which (beta)^2 - 4*(omega-0)^2 = 0
>(from
>http://mathworld.wolfram.com/DampedSimpleHarmonicMotionCriticalDamping.html)

That's a special case of the definitions above.

>
>These are all independent of whether the oscillator is linear or
>non-linear to the extent that for all non-linear oscillators, small
>amplitude oscillations approach linearity so the frequency approaches
>that of the linear oscillator.

(beta)^2 - 4*(omega_0)^2 = 0 assumes the oscillator is simple harmonic.
It's valid for a pendulum only when the oscillation amplitude is small
enough that it can be treated as harmonic.

Relative to a displaced and released harmonic oscillator, a non-linear
oscillator displaced the same amount will have a speed surplus or deficit
as it approaches zero. The pendulum in particular deviates from Hooke's
law and has less oomph at the extreme displacement, and I'd expect it will
be critically damped with a smaller amount of damping than is required for
the SHO, but with a damping that depends on the initial condition. Since
the equation of motion for even an undamped pendulum involves an elliptic
integral, I can't think of a way to solve it except numerically and
preparing a graph of critical damping versus initial displacement.
--
"The polhode rolls without slipping on the herpolhode lying in the
invariable plane." -- Goldstein, Classical Mechanics 2nd. ed., p207.
.

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