Question

A simple pendulum of length of 1.37 m and mass of 6.66 kg is given an initial speed of 2.85 m/s at its equilibrium position. Determine its period (assuming the pendulum undergoes simple harmonic motion). The acceleration due to gravity is 9.8 m/s 2 . Answer in units of s.

Answer 1

Answer:

2.35 s

Explanation:

The period of a simple pendulum is expressed as;

                                T = 2π$\sqrt{\frac{L}{g} }$

Where

T is the period in seconds

L is the length in metres

g is acceleration due to gravity

                                 T = 2π$\sqrt{\frac{1.37}{9.8}}$

                                 T = 2.349 s

                                 T = 2.35 s