An undamped 2.86-kg horizontal spring oscillator has a spring constant of 36.1 N

Question

An undamped 2.86-kg horizontal spring oscillator has a spring constant of 36.1 N/m. While oscillating, it is found to have a speed of 2.96 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? b) What is the oscillator's total mechanical energy as it passes through a position that is 0.684 of the amplitude away from the equilibrium position?

Explanation / Answer

1) at equilibrium, velocity is maximum => entirely Kinetic energy...........let the amplitude be "A".........balancing energy: 0.5kA^2 = 0.5mv^2....A = [(2.86*2.96^2) / 36.1]^0.5 = 0.69 m.....answer......

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