Question

A pendulum is placed on a distant planet. The length is one meter, and the measured period is 1.4 seconds, what is the acceleration of gravity on that planet?

Answer 1

`a=20.14\ m/s^2`

Step-by-step explanation:

The time period of the simple pendulum is given by :

`T=2\pi \sqrt{(l)/(g)}`

l is the length of the pendulum

g is the acceleration due to gravity

We have,

T = 1.4 s, l = 1 m

So,

`T^2=(4\pi^2 l)/(g)\\\\g= (4\pi^2 l)/(T^2)\\\\g= (4\pi^2 * 1)/((1.4)^2)\\\\g=20.14\ m/s^2`

So, the acceleration due to gravity of that planet is
`20.14\ m/s^2`.