Taylor 5.4-5.5 (today) and 5.5-5.6 (Monday).
where q² = β² - ω0². Then we have 4 cases to consider:
q² > 0 This case is called overdamped.
q² = 0 This case is called critically damped.
q² < 0 This case is called underdamped.
q² = -ω0² This case corresponds to β=0 and is called undamped. We have already solved this case and will skip it.
This is mathematically the easiest case. The solution is:
This solution is the transition from overdamped to underdamped. It is of interest because for some devices (shock absorbers on a vehicle) this is the optimal regime.
When q² < 0, the solution to the DE is an exponential with a complex exponent. Recall that eiθ = cosθ + isinθ. The solution is then:
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 δ. The frequency ωd = iq = (ω0² - β²)½ = (k/m - b²/4m²)½ and is called the damped frequency (ω0 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 ωd rather than ω0.
As when there is no damping, the energy of the oscillator at any instant in time is given by:
When there is no damping the energy is constant. With damping, we expect that the energy will tend to decrease with time and as you should find in your homework problem,
The frictional constant b 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.
where τd = 2π/ωd. The approximation is valid when the exponential changes little during one cycle so that it can be treated as a constant in the integral. This result is valid when the damping is small, the so-called weak damping limit. The energy of the oscillator at time t is E(t) = ½mA²ω0²e-2βt, so we find that the fraction of energy lost is
We identify 2β to be an inverse time such that the product is unitless. Calling the time &tau, then τ=(2β)-1, and
The quality factor, Q, is defined as 2π 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,
The following table summarizes relations between the constants used in the damped oscillator results.
| quantity | relations | comments |
|---|---|---|
| ω0 | ω02 = k/m | positive |
| ωd | ωd2 = ω02 - β2 | positive |
| β | β = b / 2m | positive |
Four cases to consider: no damping, underdamped, critically damped, and overdamped. These differ in the amount of damping as determined by the relative values of β and ω0.
| damping | β | decay parameter |
|---|---|---|
| none | β=0 | 0 |
| under | β<ω0 | γ |
| critical | β=ω0 | β |
| over | β>ω0 | β - sqrt(β² - ω0²) |
For an oscillator with a natural (undamped) frequency of ω0, let's look at how quickly the system returns to its equilibrium state as a function of the damping characterized by β. This is determined by how quickly the exponential factor decreases to zero, and is shown in Fig. 5.13. To construct this graph, we imagine a system with a fixed natural frequency ω. Now we arrange things so that we can continuously vary the damping as denoted by the value of β = b/2m. When β = 0 there is no damping. When β < ω0, the damping goes like e-βt. When β = ω0, the damping factor is e-βt = e-ω0t. When β > ω0, there are two damping factors, and, to consider the most general situation where the initial conditions are random, the time to return to equilibrium is determined by the slowest damping factor. The slowest damping factor is e-[β - (β² - ω0²)½]t, and for β sufficiently large, this is approximately e-(ω0/β)t. This is the result shown in Fig. 5.13. The system returns to equilibrium most quickly at critical damping, making this the optimal regime for shock absorbers on vehicles.