Question

An undamped 2.19 kg horizontal spring oscillator has a spring constant of 34.5 Nm. While oscillating, it is found to have a speed of 2.50 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 0.584 of the amplitude away from the equilibrium position? total mechanical energy:

Answer 1

Solution 1)

We know,

0.5 kA^2 = 0.5mv^2

34.5*A^2=2.19 kg*2.5^2

So, A = 0.629 m (Ans)

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Mechanical energy = 0.5 mV^2 = 6.84 J (Ans)

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Good luck!:)