# At what speed does the weight of a spring pendulum, having a mass of 0.1 kg, pass through the equilibrium position

At what speed does the weight of a spring pendulum, having a mass of 0.1 kg, pass through the equilibrium position if the stiffness of the spring is 40H \ m, and the amplitude of oscillation is 2 cm?

Given:
m = 0.1kg,
k = 40N / m,
X = 2cm = 0.02m;
Find: v -?
According to the law of conservation of energy, the potential energy of the spring at the maximum deviation of the load from equilibrium is equal to the kinetic energy of the load when passing through the equilibrium point:
Ep = Ek,
where
Ep = 1/2 * k * X ^ 2,
Ek = 1/2 * m * v ^ 2;
Substitute these expressions into the equation:
1/2 * m * v ^ 2 = 1/2 * k * X ^ 2;
Let’s find the required speed of the load:
v = X * sqrt (k / m) = 0.02 * sqrt (400) = 0.4m / s. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.