Apendulum 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)²

How 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 = G

speed = √G = √9.8 = 3.13 meters per second, regardless of the mass of the ball !

2. they have negligible mass.