(2) When no external force is acting
Solving the equations of motion involves solving differential equations. This time, let's solve an example. I believe it will be easier to understand if we compare it with the terminology used in specialized textbooks, so I will proceed with the explanation in that manner.
2-1 General formulas explained in specialized books, etc.
From the previous issue, the equation of motion is expressed as follows:
mx''＋cx'＋kx＝0 ・・・(1)
Divide by m
x‘’＋(c/m)x‘＋（k/m）x＝0 ・・・（2）
Here
k/m=ωo^2 (ωo: natural frequency)
Furthermore, the critical damping coefficient Cc
Cc/2m＝√k/m＝ωo
ζ=c/Cc (ζ: damping constant)
In that case, (2) is
x‘’＋2ζωox‘＋ωo^2x＝0 ・・・（3）
This solution is
x＝Ae^（-ζωot）cos(√1-ζ^2*ωot−Φ） ・・（4）
or
x＝Ae^（-ζωot）sin(√1-ζ^2*ωot＋Ψ）
ωn = √1 - ζ^2*ωo (damped free vibration frequency)
A, Φ, and Ψ are values determined by the initial conditions.
ζ≪1
ωn＝ωo
x＝Ae^（-ζωot）cos(ωot-Φ） ・・・（4‘）
This is how it is explained.
We will focus on what ωo and ζ mean.
2-2 Example
(1) Let's take the following example as an equation.
x‘’＋2x‘＋400x＝0 ・・・（5）
This is the equation for when the mass is shifted to the right in the diagram from the previous issue and then gently released.
A differential equation with zero on the right-hand side is called a homogeneous equation.
When solving this equation, damping and oscillation are expressed as complex exponential functions, so the form of the solution is
x＝e^λt ・・・（6）
Let's set this and find λ. λ is a complex number.
When you differentiate (6)
x'＝λe^λt、 X''＝λ^2*e^λt
Substitute this into equation (5)
λ^2*e^λt＋2λ*e^λt＋400e^λt＝0
（λ^2＋2λ＋400）e^λt＝0 ・・・（7）
Since e^λt ≠ 0
λ^2＋2λ＋400＝0 ・・・(8）
Equation (8) represents the vibration characteristics of the system, and is therefore called the characteristic equation.
Using the formula for finding roots,
λ＝［｛-b±√b^2-4ac｝/2a］
＝｛-2±√4-4*400｝/2
＝-1±√1-400
＝-1±j√399 ・・・（9‘）
≒-1±j√400
≒-1±j20 ・・・（9）
We obtained two solutions for x.
x1＝e^-1+j20、 x2＝e^-1-j20
[Theorem 1]
When two solutions, x1 and x2, are found for a homogeneous equation, the solutions are
x＝Ax1+Bx2
Thus, A and B are integration constants determined by the initial conditions and other factors.
By Theorem 1, and also by using Auler's formula (10)
e^jθ＝cosθ＋jsinθ
e^-jθ＝cosθ−jsinθ ・・・（10）
x＝Ae^(-1+j20)t＋Be^(-1-j20)t
＝e^-t｛Ae^j20t＋Be^-j20t｝
＝e^-t｛Acos20t＋jAsin20t＋Bcos20t-jBsin20t｝
＝e^-t｛(A+B)cos20t＋j(A-B)sin20t｝
Here, if we define new constants as C = A + B and D = j(AB),
x＝e^-t｛Ccos20t＋Dsin20t｝ ・・・（11）
And we obtained the general solution.
Differentiating equation (10)
x'＝−e^-t｛Ccos20t＋Dsin20t｝
＋e^-t｛-20Csin20t＋20Dcos20t｝ ・・・（12）
To determine C and D, if we shift the displacement by 1m and gently release it, the initial conditions are:
At t=0, x=1, x'=0
Substituting this condition into equations (11) and (12) gives
1＝C
0＝−C＋20D
twist,
C＝1
D＝1/20
Therefore,
x＝e^-t｛cos20t＋1/20*sin20t｝ ・・・（13）
Furthermore, using the formula for the composition of trigonometric functions,
x＝e^-t√（1^2+1/20^2）cos（20t-Φ）
＝1.0012e^-tcos（20t-Φ） ・・・（13‘）
Φ＝tan^-1（D/C）＝tan-1（1/20）
We were able to determine the vibrational displacement x when mass m is shifted by 1m and then gently released.
Figure 1 shows a graph of equation (13').
We were able to determine the vibrational displacement x when mass m is shifted by 1m and then gently released.
Figure 1 shows a graph of equation (13').
Figure 1 Attenuation waveform
(Excerpt from the email newsletter issued on February 23, 2006)
