Question

A simple pendulum performsSHM in a vertical plane with an amplitude of 5.00cm.At time zero,the bob has a displacement of 5.00cm from its equilibrium position 0.500s later, determine the period of the motion?

Answer 1

The period of a simple pendulum is determined by
its length and the acceleration due to gravity. In this case, with a displacement of 5.00cm from the equilibrium position, the period of the motion is approximately 0.448s.

Step-by-step explanation:

The period of a simple pendulum depends on its length and the acceleration due to gravity. The formula for the period (T) of a simple pendulum is given by:

T = 2π√(L/g)

Where L is the length of the pendulum and g is the acceleration due to gravity. In this case, the displacement of the bob from its equilibrium position is 5.00cm. Since the bob is displaced by the same distance as its amplitude, the length of the pendulum can be taken as the displacement from the equilibrium position. Therefore, L = 5.00cm. The acceleration due to gravity is approximately 9.8 m/s². Substituting these values into the equation:

T = 2π√(0.05m / 9.8m/s²) = 2π√(0.0051s²/m) = 2π(0.071s) =0.448s