How many times does the oscillation period of a mathematical pendulum on Earth differ from the oscillation

How many times does the oscillation period of a mathematical pendulum on Earth differ from the oscillation period of a pendulum placed on the Moon, if the acceleration of gravity on the Moon is g = 1.62 m / s2

The oscillation period of a mathematical pendulum can be calculated using the formula:

Т = 2 * π * √ (l / g), where

T is the period of oscillations, l is the length of the pendulum, g is the acceleration of gravity.

In Earth conditions, the acceleration of gravity is:

g1 = 9.82 (m / s2), respectively the period:

T1 = 2 * π * √ (l / g1).

On the moon

T2 = 2 * π * √ (l / g2),

where g2 = 1.62 (m / s2).

Let’s find the ratio of the periods T2 / T1.

T2 / T1 = (2 * π * √ (l / g2)) / (2 * π * √ (l / g1)) = √ g1 / g2 = √ 9.82 / 1.62 = √ 6.06 = 2 , 46

Answer: T2 / T1 = 2.46.