# find frequency eqaution and then compute for the non zero natural frequencies

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DP el 13 de Mayo de 2024
Editada: Sam Chak el 17 de Mayo de 2024
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Sam Chak el 13 de Mayo de 2024
Editada: Sam Chak el 17 de Mayo de 2024
Hi @DP
A 5-coupled mass-spring-damper network constitutes a 10th-order system. If you are referring to the 'frequency equation' as the Laplace transfer function, we can utilize the 'charpoly' and 'tf' functions. Additionally, the 'damp()' command can provide you with information about the system's natural frequencies.
Update: Check if you want to find the natural frequencies of the mass-damper-spring system using this approach:
m1 = 1;
m2 = m1;
m3 = m1;
m4 = m1;
m5 = m1;
c2 = 2*m1/100;
c3 = c2;
c4 = c2;
c5 = c2;
k2 = m1/1.1486;
k3 = k2;
k4 = k2;
k5 = k2;
% Construct mass matrix
M = [m1, 0, 0, 0, 0;
0, m2, 0, 0, 0;
0, 0, m3, 0, 0;
0, 0, 0, m4, 0;
0, 0, 0, 0, m5];
% Construct damping matrix
C = [c2, -c2, 0, 0, 0;
-c2, c2+c3, -c3, 0, 0;
0, -c3, c3+c4, -c4, 0;
0, 0, -c4, c4+c5, -c5;
0, 0, 0, -c5, c5];
C = 1000*C;
% Construct stiffness matrix
K = [k2, -k2, 0, 0, 0;
-k2, k2+k3, -k3, 0, 0;
0, -k3, k3+k4, k4, 0;
0, 0, -k4, k4+k5, -k5;
0, 0, 0, -k5, k5];
%% state matrix
% A = [zeros(size(M)), eye(size(M));
% -M\K, -M\C]
% CE = charpoly(A);
% Gp = tf(1, CE)
% damp(Gp)
%% Eigenvalues
lambda = eig(M\K)
lambda =
2.2793 + 0.0000i 2.2793 - 0.0000i 1.7413 + 0.0000i 0.3325 + 0.0000i 0.3325 + 0.0000i
%% Natural frequencies of the system
omega = sqrt(lambda)
omega =
1.5097 + 0.0000i 1.5097 - 0.0000i 1.3196 + 0.0000i 0.5767 + 0.0000i 0.5767 + 0.0000i
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DP el 13 de Mayo de 2024
Editada: Sam Chak el 13 de Mayo de 2024
i m refering to the frequency eqaution as deriving the deteminant function as below for the spring damper system
where
syms omega;
M = [
1.8000 0 0 0 0
0 6.3000 0 0 0
0 0 5.4000 0 0
0 0 0 22.5000 0
0 0 0 0 54.0000];
C = [
10000 -10000 0 0 0
-10000 10500 -500 0 0
0 -500 2000 -1500 0
0 0 -1500 2600 -1100
0 0 0 -1100 1100];
K = [
100000000 -100000000 0 0 0
-100000000 100050000 -50000 0 0
0 -50000 125000 75000 0
0 0 -75000 85000 -10000
0 0 0 -10000 10000];
det(K - omega^2 * M - omega * C)
ans =
and then getting the ouputs for non zero natural frequencies , is my approach correct?
Sam Chak el 13 de Mayo de 2024
Hi @DP
I am not aware of another term called the "determinant function." In my understanding, the general term for such an equation is referred to as the "characteristic polynomial" or "characteristic equation." That's why I initially used the 'charpoly()' function. However, numerically solving the 10th-degree polynomial will indeed provide you with ten eigenfrequencies, as demonstrated in my example.
syms omega;
M = [
1.8000 0 0 0 0
0 6.3000 0 0 0
0 0 5.4000 0 0
0 0 0 22.5000 0
0 0 0 0 54.0000];
C = [
10000 -10000 0 0 0
-10000 10500 -500 0 0
0 -500 2000 -1500 0
0 0 -1500 2600 -1100
0 0 0 -1100 1100];
K = [
100000000 -100000000 0 0 0
-100000000 100050000 -50000 0 0
0 -50000 125000 75000 0
0 0 -75000 85000 -10000
0 0 0 -10000 10000];
eqn = det(K - omega^2 * M - omega * C)
eqn =
lambda = vpasolve(eqn == 0)
lambda =

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