# Damping values

forty
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!

Homework Helper
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.

forty
Woah i was doing this at about 2am in the morning and realized what a stuff up I've 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 can't work out how to calculate the values of k that lead to under damped, critically damped, over damped.