Question

The acceleration due to gravity on the surface of moon is $1.7m/s^2$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $3.5s$ ? ($g$ on the surface of earth is $9.8m/s^2)$

Answer 1

Given, Acceleration due to gravity on the surface of earth $=9.8m/s^2$
Time period of the pendulum on the surface of the earth $=3.5s$
$T=2\pi \sqrt{\frac{l}{g}}$
Where, $l=\text{length of the pendulum}$
$l=\frac{T^{2}g}{4{\pi }^2}$
$l=\frac{3.5\times 3.5\times 9.8}{4\times 3.14\times 3.14}$
$l=3.043m$

Given, Acceleration due to gravity on the surface of moon $=1.7m/s^2$
As length of the pendulum remains constant,$l=3.043m$
Time period of the pendulum on the surface of the moon,$T=2\pi \sqrt{\frac{l}{g}}$
$T=2\times 3.14\sqrt{\frac{3.043}{1.7}}$
$T=2\times 3.14\times 1.338$
$T=8.4s$.