Question

A 1.2 kg object on a vertical spring with k+ 500 N/m is compressed from its...

A 1.2 kg object on a vertical spring with k+ 500 N/m is compressed from its equilibrium position. Once it is released, the mass is seen to have a velocity of 1.4 m/s at a position of 0.1 m from the equilibrium.

a.) Determine the equation for this oscillation.

b.) What is the maximum speed?

c.) What is the maximum acceleration?

d.) At what position from the equilibrium does the potential energy equal the kinetic energy?

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Answer #1

A> Velocity VEWA A> Velocity = v= wrA282 3 K = 500N/m m = 1.2 1-2 kg. n = 0.1m, v=1.4 m/s 1.4 ben A²-12 (1.4) K (A²-x2) A 1•2

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