Question

A second’s pendulum is taken on the surface of the moon where the acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain the same or increase or decrease? Give a reason.

Answer 1

We know the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity

i.e. T α `1/sqrt("g")`

As acceleration due to gravity on the surface of the moon decreases as compared to that of the earth
∵ g_moon < g_earth
∴ T_moon > T_earth

⇒ The time period of the second’s pendulum increases when it is taken to the surface of the moon.