Question: A mass of 2kg is suspended from a spring with spring constant k = 4N/m and natural length L = 2m. The mass is also subject to an external downwards force F = 2sin(2t) Newtons. Initially the mass is released at a height 50cm above the equilibrium position. Assume the mass is subject to a damping force proportional to its velocity. The acceleration due to gravity is 9.8m/s/s. (A) Whats is the extension of the spring when it is as equilibrium 0 = mg - T 0 = 2*9.8 -4x x = 4.9m (B) Using Newton's law of motion, derive the differential equation for the displacement of the mass y(t) below the equilibrium position. Use the downward direction as the positive direction. ma = weight + external force - T - air resistance(ky') i worked this out to be: y'' + (k/9.8)x' + (4/9.8)x = 2sin(2t) (C) For what values of the damping constant is the motion: under damped, critically damped, over damped. This is where I'm stuck, i thought that it might have something to do with the discriminant being = 0, greater then 0 or less then 0. Any help would be greatly appreciated!