Calculating the Period of a Pendulum with Varying Acceleration

[maccha]

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?

 

[Delphi51]

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.

 

[maccha]

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?

 

[Delphi51]

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.

 

[maccha]

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!