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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 = T_{0}

Elevator moves upward with constant acceleration a = 3g

T = 2π√(L/9.8+3g)

So I have to solve for T_{0}

T_{0}= [STRIKE]2π[/STRIKE]√(L/9.8+a) = [STRIKE]2π[/STRIKE]√(L/9.8+3g)

T_{0}= (√(L/9.8+a))^{2}= (√(L/9.8+3g))^{2}

T_{0}= L/(9.8+a) = L/(9.8+3g)

T_{0}= [ L/9.8+a ] / [ L/9.8+3g ]

T_{0}= [STRIKE]L[/STRIKE](9.8 + a) / [STRIKE]L[/STRIKE](9.8 + 3g)

T_{0}= (9.8 + a) / (9.8 + 3g)

I'm not sure I'm doing this right... Help?

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# Homework Help: Elevetor and acceleration as elevator goes up - algebra involved?

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