Question

The acceleration due to gravity on the surface of moon is 1.7 ms^-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? (g on the surface of earth is 9.8 ms^-2

Answer 1

Acceleration due to gravity on the surface of moon, g'=1.7 ms^-2
Acceleration due to gravity on the surface of earth, g=9.8 ms^-2
Time period of a simple pendulum on earth, T=3.5 s
\(T=2π\sqrt\frac{l}{g}\)
Where,

l is the length of the pendulum

\(∴l=\frac{T^2}{(2π)^2}×g\)
\(=\frac{(3.5)^2}{4×(3.14)^2}×9.8 m\)

The length of the pendulum remains constant.

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

\(=2π\sqrt\frac{\frac{(3.5)^2}{4×(3.14)^2}×9.8}{1.7}=8.4 s\)

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