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

`g=3.76\ m/s^2`

Step-by-step 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{(l)/(g)} \\\\T^2=4\pi ^2* (l)/(g)\\\\g=(4\pi ^2 l)/(T^2)`

Put all the values,

`g=(4\pi ^2 * 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`.