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.(adsbygoogle = window.adsbygoogle || []).push({});

(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!

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# Homework Help: Damping values

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