Re: [Phys-l] Queston: simple pendulum lengthen suspension.

[Bernard Cleyet <bernardcleyet@redshift.com>]

For the case you (PK) discuss specifically, IIRC, there is universal 
agreement. 

Perhaps I should have begun w/ my motivation and the method I tried to 
use to answer my question.  OTOH it might have "coloured" the replies.

About the turn + of the century.  Horologists designed rather period 
constant pendulua clocks.  Very high Q (well over an hour for A to 
decrease to 1/e), minimal interference w/ the period by the addition and 
extraction of energy.  [Gravity impulse at 30 s interval and at that 
time stepping the clock -- master and slave(s) w/ the master in < one mm 
Hg vacuum etc.]   A most important manufacturer sold slaves as use as 
masters for non Astronomical and naval use.  In that case they often 
have a "weight" tray in which compensating weights  are added (or 
subtracted) rather than use the thumb screw to vary the bob's distance.  
It takes a very long time for the pendulum to reach equilibrium ("settle 
down") after such adjustment.  After learning this from a Horologist, I 
posed the question, "have you noticed a change in the amplitude?"  Very 
little examination of amplitude has been done 'till the introduction of 
the "Microset" to the "Horological" community, so no answer.  Since the 
usual addition is a few grams w/ a 15 kg bob, probably not noticeable, 
and no one has reported such w/ the Microset.  so I, as reported, tried 
numerically modelling. 

The replies have led me to understand my model and whether it conforms 
to the question.  It is as Jack U., a sudden (The time steps are 0.1 
milli s.) introduction of a new initial condition w/ the others the same 
(theta dot and theta).  So it doesn't necessarily conform to the 
"gedanken...".  For the l at amplitude (max. displacement), I think it 
conforms.  Adding mass below the bob lengthens l, or D and G^2  [omega^2 
= gD/G^2],  essentially the same tho different amounts between the 
simple and physical pendulua).   Equivalence ensures that the A remains 
the same, if added in a manner that doesn't change E(k).  So 
thoughtlessly (naively). I got this correct, ... I pray.

For the addition at "bottom dead centre" My model is wildly wrong.  
Adding mass will subtract E because it is an impulse (energy required to 
accelerate the mass to the same as the bob.  It is an inelastic 
collision.  So not only will A be reduced for the previous reason, but 
also to conserve E.  E (k)max. = mgh max. = mg l[1- cos(theta)], as JS 
agreed.

My model kept theta and theta dot (now at max.) the same and increased 
l.  This effectively increased the moment of momentum and and the E(k) 
angular.  Since U is proportional to l and E(k) to l^2, it's no wonder 
the amplitude increased! 

bc



PHILIP KELLER wrote:

> Would you get at the same concept by reframing the original question as a
> comparison question:
>
> Given two simple pendulums of lengths L and L+delta, both passing through
> their equilibrium positions at the same speed.  Which has greater
> amplitude?
>
> This way, we don't haave to worry about the mechanism used to lengthen the
> string.
>
> As for the case where the lengthening occurs at the extreme point (where v
> = 0), I don't see why the angular amplitude would be affected.  It would
> be like replacing the string with a longer string and then restarting the
> oscillations, but still from the same angle.  But the equilibrium speed
> would increase.
>
> Phil Keller
> 1st post to list -- hope it isn't gibberish :)
>
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