# The amplitude of oscillations of a spring pendulum with a mass of 100 g and a stiffness of 40

The amplitude of oscillations of a spring pendulum with a mass of 100 g and a stiffness of 40 N \ m is 5 cm. Find the speed of a body when displaced by 3 cm.

m = 100 g = 0.1 kg.

k = 40 N / m.

A = 5cm = 0.05m.

x = 3 cm = 0.03 m.

V -?

To solve the problem, we will use the law of conservation of total mechanical energy: En = En1 + Ek1.

En = k * A ^ 2/2.

En1 = k * x ^ 2/2.

Ek1 = m * V ^ 2/2.

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

V2 = k * A ^ 2 / m – k * x2 / m.

The speed of the load will be determined by the formula: V = √ (k * A ^ 2 / m – k * x ^ 2 / m).

V = √ (40 N / m * (0.05 m) ^ 2 / 0.1 kg – 40 N / m * (0.03 m) ^ 2 / 0.1 kg) = 0.8 m / s.

Answer: the weight of the spring pendulum will have a speed of V = 0.8 m / 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.