An undamped 1.48 kg horizontal spring oscillator has a spring constant of 35.4 N/m. While oscillating, it is found to have a speed of 3.90 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation

Answer: 0.798 m

Explanation:

Given

Mass of the spring oscillator, m = 1.48 kg

Force constant of the spring, k = 35.4 N/m

Speed of oscillation, v = 3.9 m/s

Kinetic Energy = 1/2 mv²

Kinetic Energy = 1/2 * 1.48 * 3.9²

KE = 0.5 * 22.5108

KE = 11.26 J

Using the law of conservation of Energy. The Potential Energy of the system is equal to Kinetic Energy of the system

KE = PE

PE = 1/2kx²

11.26 = 1/2 * 35.4 * x²

11.26 = 17.7x²

x² = 11.26 / 17.7

x² = 0.6362

x = √0.6362

x = 0.798 m