A pendulum is dropped 0.50 m from
the equilibrium position as shown below. 
What is the speed of the pendulum bob as it passes through the
equilibrium position?

-- When the ball is at the top, before it's dropped, it has potential energy above the equilibrium position.Potential energy = (mass) x (gravity) x (height) = (mass) x (G) x (0.5)As it passes through the equilibrium position, it has kinetic energy.Kinetic energy = (1/2) x (mass) x (speed)²How much kinetic energy does it have at the bottom ?EXACTLY the potential energy that it started out with at the top !THAT's where the kinetic energy came from.So the two expressions for energy are equal.K.E. at the bottom = P.E. at the top.(1/2) x (mass) x (speed)² = (mass) x (G) x (0.5)Divide each side by (mass) . . . (the mass of the ball goes away, and has no effect on the answer !)(1/2) x (speed)² = (G) x (0.5)Multiply each side by 2 :(speed)² = Gspeed = √G = √9.8 = 3.13 meters per second, regardless of the mass of the ball !