Elevetor and acceleration as elevator goes up - algebra involved?

[riseofphoenix]

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?

 

[riseofphoenix]

Wait...the period doesn't change if it's moving upward at constant acceleration...

So the answer would be 2*T₀ right?

 

[haruspex]

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.