An object with mass 8.2 kg is attached to a spring with spring stiffness constant 370 N/m and is executing simple harmonic motion. When the object is 0.16 m from its equilibrium position, it is moving with a speed of 0.31 m/s. Calculate the maximum velocity attained by the object.

Answer 1

(1/2)MV^2 + (1/2)kX^2 = Total energy
= 0.394 + 4.736 = 5.13 Joules

For the maximum velocity (when X = 0) solve
(1/2) M V^2 = 5.13 J

V = 1.119 m/s