The vertical spring pendulum performs harmonic oscillations with an amplitude of A = 2.4 cm.

The vertical spring pendulum performs harmonic oscillations with an amplitude of A = 2.4 cm. determine the displacement x1 of the load from the equilibrium position at the point at which the restoring force acts on it, the modulus of which is F = 30 mN, if the total mechanical energy of the pendulum is W = 3.6 mJ.

A = 2.4 cm = 0.024 m.

F1 = 30 mN = 0.03 N.

W = 3.6 mJ = 0.0036 J.

x1 -?

According to Hooke’s law, the elastic force F, which occurs when the spring is deformed, is directly proportional to the spring elongation x: F = k * x, where k is the spring stiffness.

F1 = k * x1.

x1 = F1 / k.

At the moment with the greatest deviation x = A, the total mechanical energy of the pendulum consists only of the potential energy of the deformed spring: W = k * A ^ 2/2.

k = 2 * W / A ^ 2.

The elongation of the spring x1 will be determined by the formula: x1 = F1 * A ^ 2/2 * W.

x1 = 0.03 H * (0.024 m) ^ 2/2 * 0.0036 J = 0.0024 m.

Answer: the load will have a displacement x1 = 0.0024 m.