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

2.35 s

Step-by-step explanation:

The period of a simple pendulum is expressed as;

T = 2π
`\sqrt{(L)/(g) }`

Where

T is the period in seconds

L is the length in metres

g is acceleration due to gravity

T = 2π
`\sqrt{(1.37)/(9.8)}`

T = 2.349 s

T = 2.35 s