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

Question #25135

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.

Expert's answer

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

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

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