Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm . The explorer finds that the pendulum completes 92.0 full swing cycles in a time of 143 s. What is the magnitude of the gravitational acceleration on this planet?

Answer 1

P = 2*pi*sqrt(L/g).
Solving this for g:
g = (4*pi^2*L)/P^2.
the pendulum as a length of 54.0 cm and that the pendulum completes 92 full swings in 143 s.
P = 143/92 s.
Then, given that L = 54.0 cm:
g = (4*pi^2*L)/P^2
= [4*3.14^2*(0.54 m)]/(143/92 s)^2
= 8.814 m/s^2.