Elevetor and acceleration as elevator goes up - algebra involved?

AI Thread Summary
The discussion revolves around the period of a pendulum in an elevator experiencing upward acceleration. The formula for the period is initially given as T = 2π√(L/g), and adjustments are made for the elevator's acceleration. When the elevator accelerates upward at 3g, the modified period becomes T = 2π√(L/(9.8 + 3g)). Participants clarify that the period does not change with constant upward acceleration, contrasting it with constant velocity scenarios. The need for correctly writing the equations and understanding the implications of acceleration on the pendulum's period is emphasized.
riseofphoenix
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Period for pendulum is T = 2π√(L/g)

When a pendulum is in an elevator and it goes up

T = 2π√(L/9.8+a)

At rest:

T = T0

Elevator moves upward with constant acceleration a = 3g

T = 2π√(L/9.8+3g)

So I have to solve for T0

T0 = [STRIKE]2π[/STRIKE]√(L/9.8+a) = [STRIKE]2π[/STRIKE]√(L/9.8+3g)
T0 = (√(L/9.8+a))2 = (√(L/9.8+3g))2
T0 = L/(9.8+a) = L/(9.8+3g)
T0 = [ L/9.8+a ] / [ L/9.8+3g ]
T0 = [STRIKE]L[/STRIKE](9.8 + a) / [STRIKE]L[/STRIKE](9.8 + 3g)
T0 = (9.8 + a) / (9.8 + 3g)

I'm not sure I'm doing this right... Help?
 
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Wait...the period doesn't change if it's moving upward at constant acceleration...

So the answer would be 2*T0 right?
 
riseofphoenix said:
Wait...the period doesn't change if it's moving upward at constant acceleration...
No, that would be true at constant velocity
Period for pendulum is T = 2π√(L/g)
When a pendulum is in an elevator and it goes up
T = 2π√(L/9.8+a)
T = 2π√(L/9.8+3g)
It helps to get the parentheses right. Can you write that equation correctly?
T0 = [STRIKE]2π[/STRIKE]√(L/9.8+a) = [STRIKE]2π[/STRIKE]√(L/9.8+3g)
No, T0 is the period when a = 0.
Write out the correct equation for that and the equation for T when a = 3g.
 

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