The length of a simple pendulum is 0.94 m and the mass of the particle )the Bob) at the end of the cable is 0.

Question

The length of a simple pendulum is 0.94 m and the mass of the particle )the Bob) at the end of the cable is 0.77 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.6 degrees and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion.
A) What is the angular frequency of the motion
B) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth.
C) What is the bob's speed as it passes through the lowest point of the swing

Answer 1

A. Frequency = sqrt(g/L) rad/s. This is distinct from the angular rate of travel of the pendulum at any point in the swing.
B. Total mechanical energy = initial potential energy PE = mgLsin(theta)
C. At the lowest point, all energy is kinetic. KE = PE = mv^2/2, so v = sqrt(2PE/m) = sqrt(2gLsin(theta))