Under the action of a force of 0.3 N, the spring is lengthened by 1 cm.

Under the action of a force of 0.3 N, the spring is lengthened by 1 cm. Determine the vibration energy of the load suspended on this spring if the load vibrates with an amplitude of 3 cm.

F1 = 0.3 N.

x1 = 1 cm = 0.01 m.

A = 3 cm = 0.03 m.

E -?

The energy of vibrations of the load on the spring E is found by the formula: E = k * A ^ 2/2, where k is the coefficient of stiffness of the spring, A is the greatest deviation from the equilibrium position, that is, the amplitude.

According to Hooke’s law, the elastic force F1 is directly proportional to the elongation of the spring: F1 = k * x1.

Let us express the stiffness of the spring: k = F1 / x1.

The formula for determining the vibration energy will take the form: E = F1 * A ^ 2/2 * x1.

E = 0.3 H * (0.03 m) ^ 2/2 * 0.01 m = 0.0135 J.

Answer: the vibration energy of the load is E = 0.0135 J.