| Topic: |
Science > Physics |
| User: |
"JS Groot" |
| Date: |
12 Apr 2005 06:08:19 AM |
| Object: |
Critical damping of pendulum |
The motion of an ordinary pendulum with viscous damping is
described by the non-linear differential equation
theta\dotdot + a * theta\dot + b * sin(theta) = 0
It is linearised by assuming theta is small:
theta\dotdot + a * theta\dot + b * theta = 0
In this case the pendulum is critically damped if a^2= 4b.
My question is: what is the equation for critical damping in
the non-linearised case? Is an analytical expression available?
Jos Groot
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| User: "tadchem" |
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| Title: Re: Critical damping of pendulum |
12 Apr 2005 08:48:38 AM |
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Try this:
http://mathworld.wolfram.com/DampedSimpleHarmonicMotionCriticalDamping.html
TOm Davidson
Richmond, VA
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| User: "JS Groot" |
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| Title: Re: Critical damping of pendulum |
12 Apr 2005 10:15:34 AM |
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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.
Jos Groot
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| User: "tadchem" |
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| Title: Re: Critical damping of pendulum |
12 Apr 2005 10:40:46 AM |
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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)
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)
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.
As the idea of "critical damping" is to reduce the amplitude to 0 in no
more than half a cycle, the non-linearity of the dependence of
frequency upon various parameters does not enter into the problem.
It does not matter whether a large amplitude oscillation takes more
time than a smaller amplitude oscillation, because there is only time
for a half cycle oscillation at best.
What is the 'frequency' of something that is not permitted to repeat?
Tom Davidson
Richmond, VA
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| User: "Gregory L. Hansen" |
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| Title: Re: Critical damping of pendulum |
13 Apr 2005 09:50:11 AM |
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In article <1113320446.324279.289460@o13g2000cwo.googlegroups.com>,
tadchem <thomas.davidson@dla.mil> 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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| User: "JS Groot" |
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| Title: Re: Critical damping of pendulum |
13 Apr 2005 10:25:20 AM |
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"Gregory L. Hansen" wrote:
[...]
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.
I'm not after the analytical solution of the amplitude as a function of
time - which does not exist, as you state correctly. Instead I
was guessing that maybe an analytical expression exists for the
critical damping. I can imagine that for certain quantities
it is not necessary to solve the equation for theta(t) exactly.
I cannot give an example at the moment, though.
Jos Groot
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| User: "JS Groot" |
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| Title: Re: Critical damping of pendulum |
13 Apr 2005 05:49:45 AM |
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tadchem 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:
[...]
The definition I use for critical damping is: the minimum amount of damping
that keeps the system from oscillating.
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.
I solved my two equations numerically for the case
of theta(0)=pi/2 and theta\dot(0)= 0, i.e., a pendulum
which start swinging from rest, from the horizontal position
(0 rad= the vertical, rest position). The critical damping was
different for the linear and non-linear equation for these
large amplitude oscillation. That's the reason for my original
question.
As the idea of "critical damping" is to reduce the amplitude to 0 in no
more than half a cycle, the non-linearity of the dependence of
frequency upon various parameters does not enter into the problem.
It does not matter whether a large amplitude oscillation takes more
time than a smaller amplitude oscillation, because there is only time
for a half cycle oscillation at best.
I agree on this, but my definition simply does not incorporate the
notion of "half a cycle". It also isn't in any of the definitions you quoted.
Is this a common definition? I've never come across it.
What is the 'frequency' of something that is not permitted to repeat?
0 Hz, I guess?
Jos Groot
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| User: "JS Groot" |
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| Title: Re: Critical damping of pendulum |
18 Apr 2005 06:01:20 AM |
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tadchem wrote:
[...]
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.
You are right. If the linearised equation has (or has not) an oscillatory
solution for small amplitudes the non-linear equation has (or has not). So
the critical damping, the boundary between oscillatory and non-oscillatory
solutions, is the same.
Jos Groot
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