Answer to Question #25135 in Mechanics | Relativity for Romulo Bermillo

Question

A simple pendulum 99.5 cm long is observed to have a period of 2.05 s at a certain place. Determine the acceleration of gravity at the place.

Answer 1

For simple pendulum:
T = 2*pi*Sqrt [L/g],
T - pendulum period
pi = 3.14
L - pendulum length
g - acceleration of gravity
Therefore:
g = L * (2*pi/T)^2 = 0.995m* ( 2*3.14/2.05 s)^2 = 9.35 m/s^2
Answer: g = 9.35 m/s^2