A 1.0 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.0 cm/s and the initial displacement is 2.0 cm,then what is the maximum spring potential energy of the simple harmonic motion?
A) 0.0026 J
B) 0.0059 J
C) 0.0035 J
D) 0.0067 J
E) 0.0060 J

Question

Quizplus · Accepted Answer

The maximum spring potential energy of the simple harmonic motion is equal to the initial kinetic energy of the mass. The initial kinetic energy is given by:$$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1.0 	ext{ kg})(0.04 	ext{ m/s})^2 = 0.0008 	ext{ J}$$Therefore, the maximum spring potential energy is also 0.0008 J.

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