A weight of 200 g vibrates on a spring. At what speed does the load pass the equilibrium position

A weight of 200 g vibrates on a spring. At what speed does the load pass the equilibrium position if its maximum potential energy is 0.02 J?

m = 200 g = 0.2 kg.
Ep = 0.02 J.
V -?
According to the law of conservation of total mechanical energy, during oscillations, the potential energy En is converted into kinetic Ek and vice versa.
En = k * x ^ 2/2.
Ek = m * V ^ 2/2.
The maximum potential energy En will be in the extreme position when the body stops, that is, Ek = 0.
The maximum kinetic energy will be when passing the equilibrium position, when En = 0.
En = Ek.
En = m * V ^ 2/2.
V = √ (2 * En / m).
V = √ (2 * 0.02 J / 0.2 kg) = 0.44 m / s.
Answer: the maximum speed of movement when passing the equilibrium position is V = 0.44 m / s.