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Damping values

  1. May 25, 2008 #1
    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!
  2. jcsd
  3. May 25, 2008 #2


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

    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.
  4. May 25, 2008 #3
    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'
    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.
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