Question

An undamped 2.74-kg horizontal spring oscillator has a spring constant of 32.2 N/m. While oscillating, it is found to have a speed of 3.01 m/s as it passes through its equilibrium position.

A. What is its amplitude of oscillation?

B.What is the oscillator's total mechanical energy as it passes through a position that is 0.527 of the amplitude away from the equilibrium position?

Answer 1

a)Kinetic energy is given as:

E_k = m ∙ v² / 2
E_k = 2.74 ∙ 3.01² / 2
E_k = 12.41 J

Now (1/2)kA² = 12.41 J

(1/2) 32.2 x A² = 12.41

A = 0.87 m (Required amplitude of oscillation)

b)

Since mechanical energy remains constant hence energy at position that is 0.527 of the amplitude away from the equilibrium position = 12.41 J