Question

The period of vibration of a body of mass 25gm attached to a spring, vibrating on a smooth horizontal surface, when it is displaced 10cm to the right of its extreme position, the period of vibration is 1.57sec and the velocity at the end of this displacement is 0.4m/s. Determine the i) Spring constant ii) Total energy iii) Amplitude

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.

Answer 2

Answer:

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