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Re: Re: Non-Linear pendulum

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  • Subject: [mg105011] Re: Re: [mg104874] Non-Linear pendulum
  • From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
  • Date: Tue, 17 Nov 2009 05:17:25 -0500 (EST)

Dear David,

in your answer to this question (as well as in some other answers) I met the construction:

\[Theta][t_] = \[Theta][t] /. s

Previously I met constructions like \[Theta][t_] with SetDelayed, rather than with Set. Could you please 
comment a bit on the use of such a construct with Set?

Thank you in advance and best regards, Alexei


 

pendulum = {Line[{{0, 0}, {0, -1}}], Circle[{0, -1.3}, 0.3]};
l = 20;
g = 9.81;
Clear[\[Theta]];
s = First@
  NDSolve[{\[Theta]''[t] == -g/l Sin[\[Theta][t]], \[Theta][0] == 
     Pi/2, \[Theta]'[0] == 0}, \[Theta], {t, 0, 30}]
\[Theta][t_] = \[Theta][t] /. S

Animate[
 Graphics[Rotate[pendulum, \[Theta][t], {0, 0}],
  PlotRange -> {{-2, 2}, {-2, .5}}],
 {t, 0, 30},
 AnimationRunning -> False]


David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/ <http://home.comcast.net/%7Edjmpark/>  


From: Allamarein [mailto:matteo.diplomacy at gmail.com] 

I'm getting to know Mathematica. I want to compile a code to see the
non-linear pendulum behavior.

pendulum= {Line[{{0, 0}, {0, -1}}], Circle[{0, -1.3}, 0.3]};
l = 20;
g = 9.81;
s = NDSolve[
   { \[Theta]''[t] == -g /l Sin[\[Theta][t]],
    \[Theta][0] == Pi/2,
    \[Theta]'[0] == 0}, \[Theta],
   {t, 0, 30}];
Animate[
 Graphics[Rotate[pendulum, \[Theta[]t], {0, 0}],
  PlotRange -> {{-2, 2}, {0, -2}}],
 {t, 0, 30}, AnimationRunning -> False]

This code doesn't work. I realized my error is in Rotate argument. If
I change this line with:

Graphics[Rotate[pendulum, Sin[t], {0, 0}]

code runs, but it's not the result (obviously).
How can I correct my code, to see the pendulum oscillates with \[Theta]
[t] law?



-- 
Alexei Boulbitch, Dr., habil.
Senior Scientist

IEE S.A.
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