Question

The acceleration due to gravity on the surface of moon is 1.7 m s-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s?

Answer 1

Acceleration due to gravity on the surface of moon, \( g’ = 1.7\ m/s^2 \)

Acceleration due to gravity on the surface of earth, \( g = 9.8\ m/s^2 \)

Time period of a simple pendulum on earth, T = 3.5 s

\( T = 2 π \sqrt{ l / g} \)

Where,

l is the length of the pendulum

Therefore,

\( l = {T^2 / (2 π)^2} \times 2 \)

On substituting, we get,

\( l = (3.5)2 / {4 \times (3.14)2} \times 9.8 m \)

The length of the pendulum remains constant

On moon’s surface, time period, \( T’ = 2 π\sqrt {l / g’} \)

\( = 2 π \sqrt{[(3.5)2 / 4 \times (3.14)2 \times 9.8 / 1.7}] \)

We get,

= 8.4\ s=8.4 s

Therefore, the time period of the simple pendulum on the surface of the moon is 8.4 s.

The time period of the simple pendulum on the surface of the moon is 8.4 sec.