1 answer

An undamped 1.02-kg horizontal spring oscillator has a spring constant of 29.9 N/m. While oscillating, it...

Question:

An undamped 1.02-kg horizontal spring oscillator has a spring constant of 29.9 N/m. While oscillating, it is found to have a speed of 2.21 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? What is the oscillator\'s total mechanical energy as it passes through a position that is 0.701 of the amplitude away from the equilibrium position?


Answers

According to the given problem,

Ek = m ∙ v² / 2
Ek = 1.02*2.21² / 2
Ek = 2.4901 J

Ep = Ek

Ep = k ∙ x² / 2
2.4901 = 29.9 ∙ x² / 2
x = 0.4082 m.

The total mechanical energy remains the same throughout the cycle, so
Em = Epmax = Ekmax = 2.49.01 J

.

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