Calculating the Period of a Pendulum with Varying Acceleration

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Homework Help Overview

The problem involves calculating the period of a simple pendulum under the influence of varying acceleration due to gravity, specifically when the pendulum is accelerating downwards by 0.59 g. The original poster expresses confusion regarding the application of this acceleration to the formula for the period of the pendulum.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why they can substitute (9.8 - 0.59g) for the new acceleration in the period formula. Some participants question the terminology used, suggesting that "accelerates" may be more appropriate than "oscillates." Others explore the implications of the downward acceleration on the effective gravity experienced by the pendulum.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the relationship between the downward acceleration and the effective gravitational force acting on the pendulum. There is a recognition of the need to understand the concept of effective weight in an accelerating frame, but no consensus has been reached yet.

Contextual Notes

Participants are navigating the implications of the problem's wording and the physics of acceleration in a gravitational context. The original poster's confusion stems from interpreting the net acceleration and its effect on the pendulum's period.

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Homework Statement




A simple pendulum has a period T. What is the period if the entire pendulum oscillates 0.59 g downwards? (give answer as a ratio of Tnew/T)


Homework Equations



T=2[tex]\pi[/tex]([tex]\sqrt{L/g}[/tex]

The Attempt at a Solution



I'm so confused about this problem. I actually got the answer in the textbook by putting in (9.8-0.59g) for the new acceleration but I really don't understand why.. it was just trial and error. I don't understand why you can just plug in 0.59g for the new acceleration- isn't this the net acceleration anyways? Would subtracting it result in a net acceleration of 0.41g?
 
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the entire pendulum oscillates 0.59 g downwards
Could that be "accelerates" instead of "oscillates"?
If so, it simply reduces the acceleration due to gravity felt by the pendulum. Like riding in an elevator accelerating downward.
 
Yes, accelerates, sorry my mistake. Okay thank you that helps, but I'm still confused as to why though you can't just use 0.59 g if that is what it is accelerating at?
 
On an elevator standing still your weight would be mg or 9.81*m.
If the elevator was accelerating downward at 1 m/s², you would feel a bit lighter: (9.81 - 1)*m. The acceleration takes away from the regular gravity. On a spacecraft falling at 9.81 m/s² you would feel weightless. The pendulum period is a measure of this remaining gravity.
 
Ohh okay I was approaching the problem completely wrong.. I didn't realize it said the entire pendulum is accelerated downwards. Thanks for the help!
 

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