how to extract roots of equation

3 Ansichten (letzte 30 Tage)
saman ahmadi
saman ahmadi am 17 Sep. 2020
Beantwortet: VBBV am 8 Dez. 2023
Hi. I want to extract roots(w) of below two equation(equations are detA and detB). The roots of equation are with respect to N, how can i do this? thank you
syms w N
k1=70;
k2=200;
m1=0.1;
m2=0.064;
m3=0.04;
r=0.25
M1=(m2+m3)/m1;
M2=m2/m1;
K=k2/k1;
wn1=(sqrt(k1/m1))/(2*pi);
wn2=(sqrt(2*k2/m3))/(2*pi);
A=[-(1+M1)*(w/wn1)^2+2-2*cos(pi) -M2*(w/wn1)^2 -(M1-M2)*(w/wn1)^2;-(w/wn1)^2 -N*(w/wn1)^2+(2*K)/M2 -K/M2;-(w/wn2)^2 -0.5 1-(w/wn2)^2];
detA=det(A)
B=[-(1+M1)*(w/wn1)^2+2-2*cos(0) -M2*(w/wn1)^2 -(M1-M2)*(w/wn1)^2;-(w/wn1)^2 -N*(w/wn1)^2+(2*K)/M2 -K/M2;-(w/wn2)^2 -0.5 1-(w/wn2)^2];
detB=det(B)

Antworten (1)

VBBV
VBBV am 8 Dez. 2023
syms w N
k1=70;
k2=200;
m1=0.1;
m2=0.064;
m3=0.04;
r=0.25
r = 0.2500
M1=(m2+m3)/m1;
M2=m2/m1;
K=k2/k1;
wn1=(sqrt(k1/m1))/(2*pi);
wn2=(sqrt(2*k2/m3))/(2*pi);
A=[-(1+M1)*(w/wn1)^2+2-2*cos(pi) -M2*(w/wn1)^2 -(M1-M2)*(w/wn1)^2;-(w/wn1)^2 -N*(w/wn1)^2+(2*K)/M2 -K/M2;-(w/wn2)^2 -0.5 1-(w/wn2)^2];
detA=det(A)
detA = 
B=[-(1+M1)*(w/wn1)^2+2-2*cos(0) -M2*(w/wn1)^2 -(M1-M2)*(w/wn1)^2;-(w/wn1)^2 -N*(w/wn1)^2+(2*K)/M2 -K/M2;-(w/wn2)^2 -0.5 1-(w/wn2)^2];
detB=det(B)
detB = 
sol = solve([detA,detB],[N,w])
sol = struct with fields:
N: [4×1 sym] w: [4×1 sym]
vpa(sol.N,4)
ans = 
vpa(sol.w,4)
ans = 

Kategorien

Mehr zu Mathematics finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by