A pendulum has a period of 0.90 s on Earth. What is its period on a planet where the acceleration of gravity is about 0.29 that on Earth?

Answer 1

use T = 2π√(L/g)

to get L. Now using that L, replace g with 0.29g to get the new T.

.90 = 2π√(L/9.8)
L = 0.201

.29*9.8 = 2.842

So, the new
T = 2π√(.201/2.842) = 1.671 s
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Or, noting that T varies inversely as the √g, using .29g will cause the period to increase by a factor of 1/√.29 = 1.857 or 1.671 s.