The period of oscillation of the load suspended from the spring is 10 seconds. Determine the stiffness

The period of oscillation of the load suspended from the spring is 10 seconds. Determine the stiffness of the spring, if the mass of the load is 200 g. Find the total energy of the oscillatory process, if the amplitude of oscillations is 4 cm.

T = 10 s.

m = 200 g = 0.2 kg.

A = 4 cm = 0.04 m.

k -?

E -?

The period of the spring pendulum T is determined by the formula: T = 2 * P * √m / √k, where P is the number pi, m is the mass of the load, k is the stiffness of the spring.

√k = 2 * P * √m / T.

k = 4 * P ^ 2 * m / T2.

k = 4 * (3.14) ^ 2 * 0.2 kg / (10 s) 2 = 0.079 N / m.

In the position with maximum deflection, the total energy T consists only of the potential energy En: E = En = k * x ^ 2/2 = k * A ^ 2/2.

E = 0.079 N / m * (0.04 m) ^ 2/2 = 0.00006 J.

Answer: the stiffness of the spring is k = 0.079 N / m, the total energy is E = 0.00006 J.