A load suspended on a spring with a coefficient of elasticity of 25 N / m oscillates with an amplitude of 12 cm.

A load suspended on a spring with a coefficient of elasticity of 25 N / m oscillates with an amplitude of 12 cm. The speed of the load at the moment of equilibrium is 0.4. Find the mass of the load.

k = 25 N / m.

A = 12 cm = 0.12 m.

V = 0.4 m / s.

m -?

With harmonic vibrations, there is a transition from one type of mechanical energy to another. At the moment with the greatest deflection, the spring pendulum has only potential energy, which is determined by the formula: En = k * A ^ 2/2, where k is the stiffness of the spring, A is the greatest deflection, that is, the amplitude.

At the moment of passing the equilibrium position, it has only kinetic energy: Ek = m * V ^ 2/2, where m is the mass of the load, V is the speed of its movement.

En = Ek.

k * A ^ 2/2 = m * V ^ 2/2.

k * A ^ 2 = m * V ^ 2.

The formula for determining the mass of the cargo will be: m = k * A ^ 2 / V ^ 2.

m = 25 N / m * (0.12 m) ^ 2 / (0.4 m / s) ^ 2 = 2.25 kg.

Answer: the mass of the cargo is m = 2.25 kg.