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The problem is much simpler if you just think "differental equations".
At any instant of time you are allowed to constants to determine the behavior of the undisturbed system for all subsequent times. So the physical question is limited to: How does the lengthening process affect the angle, and how does it affect the velocity of the bob.
Regards,
Jack
On Sat, 17 Mar 2007, Bernard Cleyet wrote:
>
>
> Brian Whatcott wrote:
>
>> At 03:41 PM 3/17/2007, you wrote:
>>
>>
>>
>>> What happens to the amplitude if one suddenly lengthens the suspension
>>> at equilibrium position or at max. displacement?
>>>
>>> bc, puzzled
>>>
>>>
>>
>> Sticking to the idea of an oscillator, which constantly switches energy
>> between modes: in this case kinetic and potential energy: I see that
>> lengthening the pendulum at the center position, while holding its
>> sidewards kinetic energy constant allows the pendulum's amplitude
>> to increase, so that it gains the same potential energy as before at
>> the end of stroke.
>>
>>
> U = mgh = mgl(1- cos A)
>
> If l is longer, then A should be smaller???
>
> what is my problem?
>
>> If the pendulum's length is increased at the end of stroke, the
>> ensuing exchange to kinetic energy involves a smaller transaction,
>> because less potential energy is converted by the reduced drop,
>> though the former amplitude is maintained.
>>
>> These model manipulations are simplified by considering that the
>> change of length is taken at the pivot.
>>
>>
> yes. (latter)
>
> This case is intuitive, as long as the reduction in potential E by
> dropping the bob is not converted into bob KE, which seems the case;
> where it goes is not clear to me.
>
> I modelled this numerically and obtained the BW result. Curiously, (At
> least bc is curious.) The increase in amplitude is directly proportional
> to the period increase.
>
> bc, still puzzled.
>
>
>>
>>
>> Brian Whatcott Altus OK Eureka!
>>
>>
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>>
>>
>>
>>
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