A spring pendulum performs harmonic oscillations with an amplitude of 0.05 m. When the displacement

A spring pendulum performs harmonic oscillations with an amplitude of 0.05 m. When the displacement is x = 0.02 m, the elastic force is Fel = 6 * 10 ^ -5 H. What are the potential and kinetic energies corresponding to this displacement, and the total energy of the pendulum.

A = 0.05 m.

x = 0.02 m.

Fupr = 6 * 10-5 N.

Ek -?

Ep -?

E -?

According to Hooke’s law: Fcont = k * x, where k is the stiffness of the spring, x is the displacement from the equilibrium position.

k = Fcont / x.

k = 6 * 10-5 N / 0.02 m = 0.003 N / m.

Potential energy of a spring pendulum Ep is determined by the formula: Ep = k * x2 / 2, where k is the stiffness of the spring, x is the displacement of the spring from the equilibrium position.

Ep = 0.003 N / m * (0.02 m) 2/2 = 0.6 * 10-6 J.

Since the amplitude A is the maximum deviation from the equilibrium position, then in this position the total mechanical energy of the spring E consists only of the potential En = k * A2 / 2.

E = 0.003 N / m * (0.05 m) 2/2 = 3.75 * 10-6 J.

E = En + Ek.

Ek = E – En.

Ek = 3.75 * 10-6 J – 0.6 * 10-6 J = 3.15 * 10-6 J.

Answer: E = 3.75 * 10-6 J, En = 0.6 * 10-6 J, Ek = 3.15 * 10-6 J.