Calculating the Period of a Pendulum with Varying Acceleration

AI Thread Summary
The discussion centers on calculating the period of a pendulum when it experiences a downward acceleration of 0.59 g. The correct approach involves adjusting the effective gravitational acceleration by subtracting the downward acceleration from the standard gravity (9.8 m/s²). This results in a new acceleration that affects the pendulum's period, as the period formula T=2π(√(L/g)) relies on the effective gravitational force. Participants clarify that the term "oscillates" should be understood as "accelerates," which helps in grasping the concept of reduced effective gravity. Understanding this adjustment is crucial for accurately determining the pendulum's new 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\pi(\sqrt{L/g}

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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