# The amplitude of the harmonic vibrations of the point is xm = 2 cm, the total vibration energy is W = 3 * 10 ^ -7 J.

**The amplitude of the harmonic vibrations of the point is xm = 2 cm, the total vibration energy is W = 3 * 10 ^ -7 J. At what displacement from the equilibrium position does the force F = 2.25 * 10 ^ -5 N act on the vibrating point?**

Xm = 2 cm = 0.02 m.

W = 3 * 10 ^ -7 J.

F = 2.25 * 10 ^ -5 N.

NS – ?

The total mechanical energy of the body W at the point with the maximum amplitude consists only of the potential energy Wp.

Let us write down the potential energy of a point at maximum deflection: Wp = k * Xm ^ 2/2, where k is the stiffness of the spring, Xm is the vibration amplitude.

k = 2 * Wn / Xm ^ 2.

Let’s write down Hooke’s law: F = k * X, where F is the elastic force, k is the stiffness of the spring, X is the displacement from the equilibrium position.

X = F / k = F * Xm ^ 2/2 * Wp.

X = 2.25 * 10 ^ -5 H * (0.02 m) ^ 2/2 * 3 * 10 ^ -7 J = 0.015 m.

Answer: the displacement from the equilibrium position is X = 0.015 m.