Question

On the planet xenos, an astronaut observes that a 1.59 m long pendulum has a period of 1.57 s. what is the free-fall acceleration on xenos?

Answer 1

The period T of a pendulum of length L is given by

`T=2 \pi \sqrt{ (L)/(g) }`
where g is the free-fall acceleration value.
Given that on Xenos T=1.57 s and L=1.59 m, we can re-arrange the formula and replacing these numbers to find the value of g on Xenos:

`g=L ((2 \pi)^2)/(T^2)=(1.59m) ((2 \pi)^2)/((1.57s)^2)=25.44 m/s^2`