Question

A 2.2 m long simple pendulum oscillates with a period of 4.8 s on the surface of unknown planet. What is the surface gravity of the planet?

Answer 1

Answer:

$g=3.76\ m/s^2$

Explanation:

Given that,

The length of a simple pendulum, l = 2.2 m

The time period of oscillations, T = 4.8 s

We need to find the surface gravity of the planet. The time period of the planet is given by the relation as follows :

$T=2\pi \sqrt{\dfrac{l}{g}} \\\\T^2=4\pi ^2\times \dfrac{l}{g}\\\\g=\dfrac{4\pi ^2 l}{T^2}$

Put all the values,

$g=\dfrac{4\pi ^2 \times 2.2}{(4.8)^2}\\\\=3.76\ m/s^2$

So, the value of the surface gravity of the planet is equal to $3.76\ m/s^2$.