Question

A simple pendulum having a length of 1.02 m and a mass of 6.66 kg undergoes simple harmonic motion when given an initial speed of 0.874 m/s at its equilibrium position. Determine its period. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of s

Answer 1

Time period of pendulum is 2.02 s.

Step-by-step explanation:

A simple pendulum is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.

The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.

By Huygens law of period of pendulum,

T = 2π
`\sqrt{(L)/(g) }` eqn 1

where L is the length of pendulum,

g is acceleration due to gravity

Period of pendulum is independent of the mass of pendulum,

Substituting values in eqn 1

T = 2π
`\sqrt{(1.02)/(9.8\\) }`

T = 2.02 s

Time period of pendulum is 2.02 s.