We solved an example problem last time, but let's compare it with the general solution for forced oscillations described in the literature.
The equation of motion when an external force Fsinωt acts on mass m is:
mx"＋Cx'＋kx＝Fsinωt ・・・（21）
x"＋2ζωox'＋ωo^2x＝fsinωt ・・・（22）
However, ωo=√(k/m), f=F/m, ζ=C/Cc=C/(2√(mk))
This solution is
x＝Ce^(-ζωot)cos(√(1-ζ^2)ωot−Φ)
＋A/√[{(1-(ω/ωo)^2}^2＋(2ζω/ωo)^2]* sin(ωt-θ) ・・ (23)
A＝f/(ωo^2)＝F/k
θ＝tan-1[2ζ(ω/ωo)/{1-(ω/ωo)^2}]
C and Φ are constants determined from the initial conditions.
The first term's amplitude decreases over time due to attenuation. Therefore, when considering the response due to continuous forced oscillation, only the second term needs to be considered.
The first term shows the transient response.
x＝Ce^(-ζωot)cos(√(1-ζ^2)ωot−Φ) ・・・ （24）
The second term shows the steady-state response.
x＝ A/√[{(1-(ω/ωo)^2}^2＋(2ζω/ωo)^2]* sin(ωt-θ) ・・・(25)
If we list equations (3) and (22) from the previous example together,
x''＋2ζωox'＋ωo^2x＝fsinωt ・・・（22）
x''＋12x'＋400x＝20cos10t ・・・ （3）
When comparing the coefficients
10→ω、20→ωo、0.3→ζ、20→f
This can be understood. If we replace the numerical values with symbols in the previous calculation process, we obtain equation (25).
Since sinωt is the waveform of cos(ωt+π/2) which is 90 degrees ahead of cos, we can consider sin and cos in the same way if we take the phase difference into account.
Now, let's focus on the amplitude in equation (25) for the steady-state response. The amplitude at frequency ω is
x/A＝1/√[{(1-(ω/ωo)^2}^2＋(2ζω/ωo)^2]
However, A = f/(ωo^2) ... (26)
Equation (26) is known as the amplitude ratio or response ratio.
In the literature, the x-axis is plotted as ω/ωo, and the graph is used to explain various values of ζ.
Coefficient representing damping
Here, I have drawn a graph of the response magnification for example equation (27).
x''＋12x'＋400x＝20cosωt ・・・ （27）
Substitute ωo = 20 and ζ = 0.3 into equations (26) and (23).
x/A＝1/√[{1-(ω/20)^2}^2＋{0.6ω/ωo}^2] ・・・（28）
A＝f/ωo^2
θ＝tan-1[(0.6ω/20)/{1-(ω/20)^2}] ・・・ （29）
The figure below shows a graph of equations (28) and (29).
Figure 1: Example of a one-degree-of-freedom forced vibration model
Figure 2: Response magnification
Figure 3: Response magnification plotted logarithmically on the X and Y axes.
Figure 4: Phase difference
When ω is ωo = 20, that is, when the frequency of forced oscillation coincides with the natural frequency
x/A≒1/(2ζ) ・・・ （30）
This is what it will be.
The value of ωm at which the response magnification is maximized is slightly smaller than ωo.
Differentiating equation (25) gives the value of ω for which the derivative is zero. (See supplement)
ωm/ωo＝√(1-2ζ^2) ・・・ （31）
Substituting this into equation (24), the maximum response is
(x/A)max＝1/{2ζ√(1-ζ^2)} ・・・ (32)
For ζ≪1, it will be as follows:
ωm＝ωo
(x/A)max＝1/(2ζ) ・・・ （33）
As ζ increases, ωm becomes smaller than the natural frequency ωo.
Figure 1 shows this model, and Figure 2 shows equation (28) as a graph. The maximum value is 1.75 times when ωm is approximately 18 (rad/s).
Figure 3 is a Bode plot, which is a graph with the X and Y axes plotted on a logarithmic scale.
For example, the value of the Y axis is the maximum value.
(x/A)max (dB)=20Log1.75=4.86 (dB)
As can be seen in Figure 3, where ω is large, doubling ω results in a linear attenuation with a slope of approximately -12 dB (-12 dB/octave, and a tenfold increase results in a slope of -40 dB (-40 dB/decade)).
Figure 4 is a graph of the phase of equation (29), plotted with a logarithmic scale for the x-axis. At ω = ωo, the phase is 90 degrees, and as ω increases, it approaches 180 degrees.
As you may have noticed, Figures 3 and 4 are graphs commonly seen in FFT analyzers. Using an FFT analyzer, Figure 3 can be experimentally obtained as acceleration/force, or displacement/force obtained by double-integrating acceleration.
We have explained the vibration response when an external force acts at a single frequency, but even when multiple external forces of different frequencies act in combination, we can simply superimpose the responses to each external force at each frequency. Considering that complex waveforms of external forces can be Fourier transformed, this is where the significance of using an FFT analyzer lies.
<Supplement>
We will find ωm using differential formulas.
y＝f(g(x))
dy/dη＝dy/dz*dz/dη＝f'(g(η))g'(η)
We will use this.
In equation (26), set η = ω/ωo
y＝1/√[{(1-η^2}^2＋(2ζη)^2]
＝[(1-η^2)^2＋(2ζη)^2]^(1/2)
The condition for an extreme value is dy/dη = 0.
dy/dη＝1/2*{(1-η^2)^2+(2ζη)^2}^(-1/2)*{-4η+4η^3+8ηζ^2}
=1/2*{(1-η^2)^2+(2ζη)^2}^(-1/2)*{η(-4+8ζ^2)+4η^3}
= 0
Also, considering η≠1 and η>0
1/2*{(1-η^2)^2+(2ζη)^2}^(-1/2)≠0
If we divide both sides by this then
{η(-4+8ζ^2)+4η^3}＝0
Therefore
η＝√(1-2ζ^2)
References: Practical Mechanical Vibration Engineering, by Masaharu Kunieda, published by Rikogakusha.
(Excerpt from the email newsletter issued on June 22, 2006)
