Question

A simple pendulum with a length of 2 m oscillates on the Earth's surface. What is the period of oscillations?

Answer 1

Answer:

Period, T = 27.84 seconds

Explanation:

Given the following data;

Length of pendulum = 2m

We know that acceleration due to gravity on earth is equal to 9.8 m/s²

To find the period of oscillation of the simple pendulum;

Mathematically, it is given by the formula;

$Period, T = 2 \pi \sqrt {lg}$

Substituting into the equation, we have;

$Period, T = 2 * 3.142 * \sqrt {2*9.8}$

$Period, T = 6.284 * \sqrt {19.6}$

$Period, T = 6.284 * 4.43$

Period, T = 27.84 seconds