A weight of 350 g falls from a height of 2 m onto the cup, suspended on a spring with a stiffness of 500 N / m

A weight of 350 g falls from a height of 2 m onto the cup, suspended on a spring with a stiffness of 500 N / m, and remains on the cup. Determine the amplitude of steady-state vibrations if the mass of the cup and spring can be neglected

k = 500 N / m.

h = 2 m.

g = 10 N / kg.

m = 350 g = 0.35 kg.

A -?

The amplitude of oscillations A is the greatest deviation of the cup from the equilibrium position.

When a load falls on a porridge, the law of conservation of total mechanical energy is valid. The potential energy of the body Еп1 is converted into the potential energy of the compressed spring Еп2.

En1 = m * g * h, where m is the mass of the body, g is the acceleration of gravity, h is the height from which the load falls.

En2 = k * A ^ 2/2, where k is the stiffness of the spring, A is the greatest deviation from the equilibrium position.

m * g * h = k * A ^ 2/2.

A2 = 2 * m * g * h / k.

A = √ (2 * m * g * h / k).

A = √ (2 * 0.35 kg * 10 N / kg * 2 m / 500 N / m) = 0.167 m = 17 cm.

Answer: the amplitude of steady-state oscillations will be A = 17 cm.