The amplitude of free oscillations of a spring pendulum is 0.1 m. The weight of the load of this pendulum is 0.1 kg

The amplitude of free oscillations of a spring pendulum is 0.1 m. The weight of the load of this pendulum is 0.1 kg and the spring rate is 40 h / m. Determine the speed of the load at the moment it passes the equilibrium position.

A = 0.1 m.

m = 0.1 kg.

k = 40 N / m.

V -?

When passing the equilibrium position, the pendulum will have only kinetic energy Ek = m * V2 / 2.

In the position with the greatest deflection, that is, in the position when the deflection of the body is equal to the amplitude, the pendulum has only potential energy En = k * A ^ 2/2.

According to the law of conservation of total mechanical energy: Ek = En.

m * V ^ 2/2 = k * A ^ 2/2.

m * V ^ 2 = k * A ^ 2.

V ^ 2 = k * A ^ 2 / m.

V = √ (k * A ^ 2 / m).

V = √ (40 N / m * (0.1 m) ^ 2 / 0.1 kg) = 2 m / s.

Answer: when passing the equilibrium position, the load has a speed of V = 2 m / s.