Question

A 5.00kg mass is attached to a 2.00m long pendulum near the surface of the planet Earth. Calculate the period of oscillation.

Answer 1

T = 2.84 s

Step-by-step explanation:

Time period of simple pendulum is given as

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

here we know that

`L = 2 m`

`g = 9.81 m/s^2`

now plug in all data in above equation

`T = 2\pi \sqrt{(2)/(9.81)}`

`T = 2.84 s`

so the time period of simple pendulum above the earth surface is approximately 2.84 s