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A simple pendulum has a period of 3.65 s on the surface of the Earth. If this pendulum were placed on the surface of the Moon, where the gravitational acceleration is 1.62 m/s2m/s2 , what would its period be?

Question

A simple pendulum has a period of 3.65 s on the surface of the Earth. If this pendulum were placed on the surface of the Moon, where the gravitational acceleration is 1.62 m/s2m/s2 , what would its period be?

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Solution

The period of a simple pendulum is given by the formula:

T = 2π √(L/g)

where: T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

On Earth, we know that T = 3.65 s and g = 9.81 m/s². We can rearrange the formula to solve for L:

L = (T² * g) / (4π²)

Substituting the known values:

L = (3.65² * 9.81) / (4π²) ≈ 1.08 m

Now, if we take this pendulum to the Moon where g = 1.62 m/s², we can use the same formula to find the new period T':

T' = 2π √(L/g')

Substituting the known values:

T' = 2π √(1.08 / 1.62) ≈ 5.17 s

So, the period of the pendulum on the Moon would be approximately 5.17 seconds.

This problem has been solved

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