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F&C sections 3.4 and 3.6
With a damping term linear in the velocity, the equation of motion is:
mx" + cx' + kx = 0.
Let g = c/2m, w02 = k/m, and divide through by m to get:
x" + 2gx' + w02x = 0 or using the D notation (D2 + 2gD + w02)x = 0.
This can be factored to give:
[D + g - (g2 - w02)1/2] [D + g + (g2 - w02)1/2] x = 0
Let q2 = g2 - w02 . Then we have 3 cases to consider:
q2 > 0 This situation is called overdamped.
q2 = 0 This situation is called critically damped.
q2 < 0 This situation is called underdamped.
This is mathematically the easiest case. The solution is:
x(t) = A1e-(g-q)t + A2e-(g+q)t
v(t) = -(g-q)A1e-(g-q)t - (g+q)A2e-(g+q)t
This solution corresponds to the position returning exponentially to the equilibrium position. Despite the name "damped harmonic oscillation", there is no oscillation! Well, if you arrange things right there might be one oscillation.
And note that q is always less than g, so it is never possible for g-q to become negative and therefore turn the A1 exponential into a positive exponential.
When q=0, the two terms of the DE are identical:
[D + g] [D + g] x = 0
So the solution x1 = Ae-gt has only one arbitrary constant. But this is a second order DE -- this means there should be TWO arbitrary constants. Physically this corresponds to being able to independently specify the initial position and velocity for the oscillator. The solution to this dilema is to ask if there is a function x2(t) that yields a result of the form Ae-gt after being acted on by [D + g]. Then the second [D + g] guarantees that the result is 0. Mathematically this means:
[D + g] x2(t) = Ae-gt
The solution, which can be verified by substitution, is
x2(t) = Ate-gt
So, our general solution for the critically damped case is the sum of x1 and x2:
x(t) = Ate-gt + Be-gt = (At + B)e-gt .
v(t) = (A - gB - gAt)e-gt .
When q2 < 0, the solution to the DE is an exponential with a complex exponent. As shown in Appendix D, eiq = cosq + isinq. I will skip the derivation (you can find it in the text), and jump to the solution:
x(t) = e-gt [Csin(wdt) + Dcos(wdt)] = e-gt Acos(wdt + f)
v(t) = -Ae-gt [gsin(wdt + f) - wdcos(wdt + f)]
These two forms of the solution are equivalent. In the first case, C and D are the unknown constants while in the second case they are A and f. The frequency wd = iq = (w02 - g2)1/2 = (k/m - c2/4m2)1/2 and is called the damped frequency (w0 is the undamped or natural frequency).
This solution is almost the same as the solution of the undamped oscillator. There are two differences:
The amplitude decays exponentially to zero, and
the frequency of oscillation is wd rather than w0 .
As when there is no damping, the energy of the oscillator at any instant in time is given by:
E = 1/2 mx'2 + 1/2 kx2 .
When there is no damping the energy is constant. With damping, we expect that the energy will tend to decrease with time and we find that:
dE/dt = mx'x" + kxx' = (mx" + kx)x' = -cx'2
since mx" + cx' + kx = 0 is the equation of motion. The frictional constant c is positive as well as the square of the velocity, so the change of energy with time is always negative, that is, the energy is always decreasing. The instantaneous rate of energy loss varies with the velocity, but the fractional energy loss averaged over one period of oscillation is approximately constant.
DE = ò0Td dE/dt dt = -gmw02A2e-2gt Td
where Td = 2p/wd. This result valid when the damping is small -- the weak damping limit. The energy of the oscillator at time t is E = 1/2 mA2w02e-2gt, so we find that the fraction energy loss is
|DE/E| = 2gTd .
We identify 2g to be an inverse time such that the product is unitless. Calling the time t, then t=(2g)-1, and
|DE/E| = Td / t .
The quality factor, Q, is defined as 2p times the energy stored in the oscillator divided by the energy lost in a single period of oscillation. Therefore we get immediately that, in the case of weak damping,
Q = 2pt / Td .
© 31 March 1999 R. Harr