How long is a simple pendulum with a period of 5 seconds? How long would this pendulum have to be if it were to operate on the moon

Answer 1

The period, T, of a pendulum is given by

T = 2pi √[L / g];

where L is the length of the pendulum and g is the acceleration of the gravitational field the pendulum is in. Solving for L, we obtain

L = (T^2)(g) / (4pi^2)

On earth, take g ≈ 9.8 m/s^2. For a period of 5 seconds, we would need a length of about 6 meters. On the moon, take g ≈ 1.6 m/s^2. For a period of 5 seconds, we would need a length of about 1 meter.

Answer 2

Period T = 2π√l/g; l = T²g/(4π²) = 25s²·9.81m/s²/(4π²) = 6.21 m

On the Moon g = 1.625 m/s² and because l = 25s²·1.625m/s²/(4π²) = 1.03 m