Question

A body of mas 4 kg attached to a spring is displaced through 0.04m from its
equilibrium position and then released. If the spring constant is 400 N/m. find the
time period of vibration?​

Answer 1

Answer:

Mass of the body attached to the spring, m=0.025 kg.

Initial displacement from the equilibrium position, x=0.1 m.

Spring constant, k=0.4 N/m.

Velocity at the end of displacement, v=0.4 m/s. I presume that the meaning of this is that the body is displaced by a distance of 0.1 m initially and then given a velocity of =0.4 m/s at this position.

At the initial point, the velocity is 0.4 m/s.

⇒ The kinetic energy is 12mv2=12×0.025×(0.4)2=2×10−3 J.

At the initial position, the displacement is 0.1 m from the equilibrium position.

⇒ The potential energy is 12kx2=12×0.4×(0.1)2=2×10−3 J.

⇒ The total energy is 2×10−3+2×10−3=4×10−3 J.

Hope it helps you

Pls mark me as Brainliest