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Physics, 07.04.2021 23:20 willowcollins1256

The motion of a 3.0 kg mass on the end of a spring is described by
x(t) = (0.20 m)*Cos(2t/3.0s).
What is a time when the system is at the equilibrium position?

Please explain how to find this answer


The motion of a 3.0

kg mass on the end of a spring is described by
x(t) = (0.20 m)*Cos(2t/3.0s).

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Answers: 1

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The motion of a 3.0 kg mass on the end of a spring is described by
x(t) = (0.20 m)*Cos(2t/3....
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