1. , Free vibration Eigenvalue equation EIGENVALUE EQUATION
Free vibration occurs in a mechanical system under initial conditions when F(t) equals to zero. So, external forces equals to zero at the RHS of the differential equation of a multi-degree of freedom system.
, Free vibration
Assuming
and substituting it in the differential equation, we get the equation.
Dividing the equation by est, we find the equation below. To obtain a solution, A must not equal to zero and, the determinant of the matrix must be zero, so the resulting determinat is called as eigenvalue equation.
Eigenvalue equation

2. Example 1.1 (Continue)
We reconsider Example 1.1 for this lecture.
Remembering the equation of motion for Example 1.1.
Recall the eigenvalue equation. Multiplying the mass matrix by s2, multiplying the damping matrix by s, taking the stiffness matrix itself and then summing the matrices, the resulting eigenvalue equation is found as follows.
Eigenvalue Equation

3. a=[0.0001828,0.05,4.46,182.96,5241.18];p=roots(a);vpa(p,4)
Substituting the numerical values given in the example, the determinant of the resulting matrix given below is calculated.
Eigenvalue Equation :
m=0.184 kg, L =0.24 m, k=201.1 N/m, c=2.34 Ns/m
The determinant gives a polynomial equation depending on s.
We can solve the polynomial equation with MATLAB. Defining the coefficients of polynomial equation in a vector defining “a” in MATLAB, we can find the eigenvalues of the mechanical system by “roots” commands in MATLAB.
MatLAB Code:
a=[ ,0.05,4.46,182.96, ];p=roots(a);vpa(p,4)

4. Eigenvalues: -18.5+42.1i, -18.5-42.1i, -97.4, -139
We can also find the eigenvalues of the system by writing an alternative code in MATLAB.
Eigenvalue Equation
m=0.184 kg, L =0.24 m, k=201.1 N/m, c=2.34 Ns/m
MatLAB Code:
clc;clear
m0=0.184;l0=0.24;k0=201.1;c0=2.34;
m=[m0*l0^2/8,0;0,3*m0/4];
c=[27*c0*l0^2/16,-9*c0*l0/4;-9*c0*l0/4,6*c0];
k=[9*k0*l0^2/8,-3*k0*l0/2;-3*k0*l0/2,4*k0];
syms s;p=solve(det(m*s^2+c*s+k));vpa(p,4)
Using any programs given above, we can obtain the eigenvalues of the system.
Eigenvalues: i, i, -97.4, -139
We have four eigenvalues, the two of them are complex roots, and the others are real roots.

5. Eigenvalues : s1=-18.5+42.1i, s2= -18.5-42.1i, s3= -97.4, s4= -139
The form of free vibration response :
The form of free vibration is determined from the eigenvalues. A pair of eigenvalues having complex conjugate roots gives an exponential-harmonic response. Each eigenvalue having a real root gives a pure exponential response.
Ai is the amplitude and φ is the phase angle for the response. The values of A1, φ1, A2 and A3 are determined from initial conditions.
When t goes to infinity, the free vibration response goes to zero. So, the steady state response becomes zero. A system is stable when its whole eigenvalues have negative real parts
Another definition of stability is that the free vibration response goes to zero when t goes to infinity.

6. For the system response: Δt=0.0023 , t∞=0.34
From the eigenvalues having complex conjugate roots, we can calculate the parameters such as the undamped natural frequency, the damping ratio, the time step, and time at which a response reaches steady state.
rad/s ω0 iω -σ φ
p=-σ+iω
f0=1/T0
From the eigenvalues having real roots, we can calculate the parameters such as the time constant, the time step, and time at which a response reaches steady state.
p=-σ
For s3=-97.4, Δt=0.0033, t∞=0.0645
For s4=-139 Δt=0.0023, t∞=0.0452
The smallest value of the time steps and the biggest value of steady-state times are chosen as time parameters for the system to plot a dynamic response.
For the system response: Δt= , t∞=0.34

7. The plots of exponential-harmonic responses for different damping ratios and pure exponential responses for different time constants are shown below.
x(t) t 5 3 1
0.2
0.5
ξ=0.1 t
x(t)