A weight of 4 kg performs harmonic vibrations under the action of a spring with a stiffness of 75 N / m.

A weight of 4 kg performs harmonic vibrations under the action of a spring with a stiffness of 75 N / m. At what displacement of the load from the equilibrium position is the modulus of its velocity equal to 5 m / s, if in the equilibrium position the modulus of its velocity is 10 m / s?

m = 4 kg.

k = 75 N / m.

V = 5 m / s.

Vmax = 10 m / s.

x -?

Let’s use the law of conservation of total mechanical energy.

The total mechanical energy of the body E is the sum of the potential En and the kinetic energy Ek of the body.

En = k * x ^ 2/2.

Ek = m * V ^ 2/2.

At the moment of passing the equilibrium position, the total mechanical energy of the body E consists only of the kinetic energy Ek = m * Vmax ^ 2/2.

At another moment in time En: E = k * x ^ 2/2 + m * V ^ 2/2.

m * Vmax ^ 2/2 = k * x ^ 2/2 + m * V ^ 2/2.

k * x ^ 2/2 = m * Vmax ^ 2/2 – m * V ^ 2/2.

x ^ 2 = m * (Vmax ^ 2 – V ^ 2) / k.

x = √ (m * (Vmax ^ 2 – V ^ 2) / k).

x = √ (4 kg * ((10 m / s) ^ 2 – (5 m / s) ^ 2) / 75 N / m) = 2 m.

Answer: the displacement of the load from the equilibrium position is x = 2 m.