A 0.20-kg object mass attached to a spring whose spring constant is 500 N/m executes simple harmonic motion.If its maximum speed is 5.0 m/s, the amplitude of its oscillation is:
A)0.0020 m
B)0.10 m
C)0.20 m
D)25 m
E)250 m

Question

Quizplus · Accepted Answer

The maximum speed of the object, v_max = 5.0 m/s, occurs at the equilibrium point (where the displacement is 0) of the oscillation, where all the potential energy of the spring is converted into kinetic energy of the object. Therefore, the total energy, E_total, of the system is given by
E_total = (1/2)kA^2,
where k is the spring constant and A is the amplitude of the oscillation.

At the maximum displacement from equilibrium, the potential energy of the spring is also maximum and equal in magnitude to the kinetic energy of the object at the equilibrium point. Therefore, we can relate the maximum speed to the amplitude by conservation of energy as follows:
(1/2)kA^2 = (1/2)mv_max^2,
where m is the mass of the object.

Solving for A, we get
A = sqrt[(mv_max^2)/k] = sqrt[(0.20 kg x (5.0 m/s)^2)/500 N/m] = 0.10 m.

Therefore, the amplitude of the oscillation is 0.10 m.

A 020-Kg Object Mass Attached to a Spring Whose Spring Constant