Jump to estion 11 of 16 Sapling Learning An undamped 1.35-kg horizontal spring o

Question

Jump to estion 11 of 16 Sapling Learning An undamped 1.35-kg horizontal spring oscillator has a spring constant of 39.6 N/m. While oscillating, it found to have a speed of 3.44 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? Number What is the oscllator's total mechanical energy as it passes through a position that is 0.680 of the amplitude away from the equilibrium position? Number O Previous Give Up & View Solution O Check Answer 0 Next Exit 0 F4 FS F6

Explanation / Answer

Ek = m v² / 2
Ek = 1.35 * 3.44² / 2
Ek = 7.98768 J

Ep = Ek

Ep = k x² / 2
7.98768 = 39.6 * x² / 2
x = 0.635 m....(answer a)
The total mechanical energy remains the same throughout the cycle, so
Em = Epmax = Ekmax = 7.98768 J

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