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please i need it ASAP. thak you very much.

2006-09-13 16:29:34 · 1 answers · asked by Raymond 1 in Science & Mathematics Engineering

1 answers

In the general case, where the vibration(s) may be analyzed using a 2'nd order DE of the form
mx'' + rx' + kx = 0 then the auxiliary equation
mΩ² + rΩ + k = 0 will lead to solutions in Ω being real and distinct, real and single-valued, or complex conjugates.

If Ω is real and distinct it will be of the form (a±b) from the quadratic equation. Then the general solution for x is of the form
x = c1*e^(-(a-b)t) + c2*e^(-(a+b)t) and the x ->0 as t-->∞
This is called the overdamped case and x will return to its' equilibrium point without making a full orbit about the equilibrium point in the phase plane.

If Ω is complex valued, it will have the form -α ± iß where
α = r/2m and ß = √((k/m) - (r/2m)²) and the general solution is
x = (c1*cos(ßt) + c2*sin(ßt)*e^(-αt)
Again, due to the e^(-αt) term, the oscillations will die out as t-->∞
This is called the underdamped case and x will return to its' equilibrium point after making multiple orbits about the equilibrium point in the phase plane.

In the case Ω is single-valued, the general solution is
x = (c1 + c2t)*e^((-r/2m)t)
This is called the critically damped case and x will return to its' equilibrium point after making at most one complete orbit about the equilibrium point in the phase plane.

One thing to keep in mind is that, in the general case of a complex oscillating system or a complex forcing function, this analysis must be done for each of the Fourier components of the oscillatory motion.

Hope that helps.


Doug

2006-09-13 18:26:58 · answer #1 · answered by doug_donaghue 7 · 0 0

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