Initial conditions for horizantal pendulum
Pioneer1
the following question regarding a differential equation.
I have been working on an analysis of the equation of motion of the
Cavendish experiment. I asked a similar question at sci.math
and Robert Israel kindly supplied a numerical solution for the non-
linear equation. My goal was to compare the linear and non-linear
solutions. This is almost done, but as I was looking at the
differential equation I realized that there may be a problem with the
initial condition y = 0 at t = 0. Here y=theta = excursion angle.
The equation of motion is given by
I y'' + ky = 2GMmd/(a - yd)^2
Primes are time derivatives.
I = Moment of inertia
k = Torsion constant
G = Gravity constant
M,m = weights
d = moment arm
a = distance between weights
I ignored the damping term. I also linearize the equation by setting
yd = 0.
Iy'' + ky = 2GMmd/a^2
So the initial condition is y = 0 at t = 0. Then,
0 = 2GMmd/a^2
But the only way 0 = 2GMmd/a^2 can be true is if G = 0. But G is never
zero. It is unphysical to assume that G is zero.
Does this reasoning make sense?
I have more details here:
Similar discussion at sci.physics.research
http://groups.google.com/group/sci.physics.research/browse_thread/thread/d391940cc173f9dc/eed90e6c3fee0edc#eed90e6c3fee0edc
Many thanks for helping.
Robert Israel
> I would greatly appreciate if readers in this group could comment on
> the following question regarding a differential equation.
>
> I have been working on an analysis of the equation of motion of the
> Cavendish experiment. I asked a similar question at sci.math
>
>
>http://groups.google.com/group/sci.math/browse_thread/thread/a6ee2f782df09625
>/53cf5573d354a3ab#53cf5573d354a3ab
>
> and Robert Israel kindly supplied a numerical solution for the non-
> linear equation. My goal was to compare the linear and non-linear
> solutions. This is almost done, but as I was looking at the
> differential equation I realized that there may be a problem with the
> initial condition y = 0 at t = 0. Here y=theta = excursion angle.
>
> The equation of motion is given by
>
> I y'' + ky = 2GMmd/(a - yd)^2
>
> Primes are time derivatives.
>
> I = Moment of inertia
> k = Torsion constant
> G = Gravity constant
> M,m = weights
> d = moment arm
> a = distance between weights
>
> I ignored the damping term. I also linearize the equation by setting
> yd = 0.
>
> Iy'' + ky = 2GMmd/a^2
That's incorrect. The correct linear approximation of 2GMmd/(a - yd)^2 for
small y is
2GMmd/a^2 + 4GMmd^2/a^3 y
> So the initial condition is y = 0 at t = 0. Then,
>
> 0 = 2GMmd/a^2
No. y = 0 at t = 0 but y'' is not 0.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Pioneer1
<isr...@math.MyUniversitysInitials.ca> wrote:
> Pioneer1 <1pione...@gmail.com> writes:
> That's incorrect. The correct linear approximation of 2GMmd/(a - yd)^2 for
> small y is
>
> 2GMmd/a^2 + 4GMmd^2/a^3 y
Thanks I'll change the equation.
>
> No. y = 0 at t = 0 but y'' is not 0.
This is a bit counterintutive for me. I was imagining that at t=0 the
pendulum arm is stationary and has not yet moved. According to this
assumption y=0, y'=0 and y''=0. Does this make sense?
If we assume y''(0)=0, will this change the numerical solution you
posted in sci.math?
Also, I'd like to mention that I used your numerical solution and the
analytical solution given by Jean-Marc Gulliet in Maple group
to compare the linear and non-linear solutions in excel. A plot of it
is here
As you can see there is a discrepency in the periods. The linear
solution has amplitude 0.08 but the non-linear has amplitude 0.01. I
think I made a mistake about the period somewhere but I couldn't
locate it as yet. I would appreciate comments. Thanks again.
Robert Israel
> On Oct 7, 3:49 am, Robert Israel
> <isr...@math.MyUniversitysInitials.ca> wrote:
> > Pioneer1 <1pione...@gmail.com> writes:
> > That's incorrect. The correct linear approximation of 2GMmd/(a - yd)^2
> > for
> > small y is
> >
> > 2GMmd/a^2 + 4GMmd^2/a^3 y
>
> Thanks I'll change the equation.
> >
> > No. y = 0 at t = 0 but y'' is not 0.
>
> This is a bit counterintutive for me. I was imagining that at t=0 the
> pendulum arm is stationary and has not yet moved. According to this
> assumption y=0, y'=0 and y''=0. Does this make sense?
No. Well, only a little bit of sense. For t < 0, y'' = 0 if the
arm is held stationary. For t > 0, y'' is what the DE says it is, which has a
nonzero limit as t -> 0+. Strictly speaking, y''(0) doesn't exist for the
actual arm, but the solution for t > 0 (which is what you're really
concerned about) has an analytic continuation in which y''(0) is nonzero.