Question

An undamped 2.87 kg horizontal spring oscillator has a spring constant of 26.5 N/m. While oscillating, it is found to have a speed of 3.02 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? amplitude of oscillation: What is the oscillator's total mechanical energy as it passes through a position that is 0.629 of the amplitude away from the equilibrium position? total mechanical energy:

Answer 1

Answers the kinetic Enersy in the spring is enal to the mechanical work done by the spring & the kinetic enersy in the spring is, E = // KA² the mechanical work done by the spring is E = 1 m ² compare the above two Equations to find the amplitude of oscillations x = 72 m2 - usmik 3.02 mis 2.87kg 26.5P/m 1A=0.9938 m the mechanical work done by the spring is. E = my - 1 2 4 2.87* (3:02) 2 [E = 13.087 J]