# Damping values

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!

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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)
I didn't get that. What happened to g? And what is x here?

(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!
That's strange, I thought the concept of overdamping, underdamping and critical damping applied only to damped and unforced oscillations. The oscillation is clearly forced here.

Woah i was doing this at about 2am in the morning and realised what a stuff up ive made here.

for (B)

ma = weight + external force - T - air resistance(ky')
i worked this out to be:
my'' = mg + 2sin(2t) - 4(4.9+y) - ky'
rearranging
y'' + (k/2)y' + (2)y = 2sin(2t)

thats my new differential equation but i still cant work out how to calculate the values of k that lead to under damped, critically damped, over damped.