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how to extract roots of equation

조회 수: 2 (최근 30일)
saman ahmadi
saman ahmadi 2020년 9월 17일
답변: VBBV 2023년 12월 8일
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)

답변 (1개)

VBBV
VBBV 2023년 12월 8일
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 = 

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