The amplitude of small free vibrations of the spring pendulum is 9 cm, the weight of the load is 100 g, and the spring

The amplitude of small free vibrations of the spring pendulum is 9 cm, the weight of the load is 100 g, and the spring rate is 40 N / m. The maximum speed of the oscillating weight is.

A = 9 cm = 0.09 m.

m = 100 g = 0.1 kg.

k = 40 N / m.

Vmax -?

According to the law of conservation of total mechanical energy, when the spring pendulum oscillates, the kinetic energy Ek is transformed into potential En and vice versa.

In the position of maximum deflection, the pendulum has only potential energy En. At the moment of passing the equilibrium position, the pendulum has only kinetic energy Ek.

En = k * x ^ 2/2, where k is the stiffness of the spring, x is the deviation from the equilibrium position. In the position of the greatest deviation x = A.

Ek = m * Vmax ^ 2/2.

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

We express the maximum speed by the formula: Vmax = A * √k / √m.

Vmax = 0.09 m * √40 N / m / √0.1 kg = 1.8 m / s.

Answer: Vmax = 1.8 m / s.