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 spe ed of 3.90 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation
1 answer:
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
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