A simple pendulum of length l executes simple harmonic motion with a small angular amplitude ₀ . What is the linear speed of the bob when its kinetic energy is equal to its potential energy? (Assume potential energy is zero at the mean position and g is the acceleration due to gravity)

Step-by-step solution

For a simple pendulum of length l executing SHM with angular amplitude ₀ , the linear amplitude is: A = l ₀ The angular frequency of the simple pendulum is: = g l The total mechanical energy E of the pendulum bob of mass m is: E = 1 2 m ^2 A^2 = 1 2 m ( g l ) (l ₀)^2 = 1 2 m g l ₀^2 When the kinetic energy K is equal to the potential energy U , the kinetic energy is exactly half of the total energy: K = E 2 Substituting the expressions for K and E : 1 2 m v^2 = 1 2 ( 1 2 m g l ₀^2 ) 1 2 m v^2 = 1 4 m g l ₀^2 v^2 =