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.
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.
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