Re: Initial conditions for horizantal pendulum

On Oct 7, 3:49 am, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Pioneer1 <1pione...@xxxxxxxxx> 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

http://groups.google.com/group/comp.soft-sys.math.maple/browse_thread/thread/271f9ccc3afe2184/9c7e4022ca5d5115?hl=en#9c7e4022ca5d5115

to compare the linear and non-linear solutions in excel. A plot of it
is here

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_%28equation_of_motion%29#Comparison_of_linear_and_nonlinear_solutions

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.

.