A body of mass m = 0.4 kg, suspended on a spring with a stiffness k = 0.1 N / m, performs small harmonic oscillations

| February 18, 2021 | education

A body of mass m = 0.4 kg, suspended on a spring with a stiffness k = 0.1 N / m, performs small harmonic oscillations, if the minimum speed of the body is 8 cm / s, then the amplitude of oscillations is?

m = 0.4 kg.
k = 0.1 N / m.
V = 8 cm / s = 0.08 m / s.
x -?
According to the law of conservation of energy, the maximum deflection of the pendulum will be when all the kinetic energy Ek is converted into potential energy En.
The kinetic energy Ek is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the body, V is the speed of the body.
The potential energy of a spring pendulum, Ep, is determined by the formula: Ep = k * x ^ 2/2, where k is the stiffness of the spring, x is the elongation of the spring.
m * V ^ 2/2 = k * x ^ 2/2.
x = √ (m * V ^ 2 / k).
x = √ (0.4 kg * (0.08 m / s) ^ 2 / 0.1 N / m) = 0.16 m.
Answer: the vibration amplitude is x = 0.16 m.



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.