I do not see why the determinant must be zero. Your final equation is a statement that it IS zero, except you seem to have left out an "l" term after the first k in equation 2.
You final equation is of the form
a*cosA*sinB -sinA*cosB = 0
a = tanA/tanB
where a = 1/sqrt2
A = k*l/sqrt2
B = k*l
You may have to solve for kl by iteration. I don't see an easy way.
-From the following two linear homogeneous algebraic equations:(sqr= square root)
(1) B*sin(kl/sqr2) = D*sin(kl)
(2) (k/sqr2)*B*cos(kl/sqr2) = (k)*D*cos(kl)
-Form matrix of these 2 equations and solving the determinant=0 will lead to: (1/sqr2)*cos(kl/sqr2)*sin(kl) - sin(kl/sqr2)*cos(kl)= 0
-How do i solve this?
2 answers
If it's of any help, here's a plot of the function
f(x)=sqrt(2)*cos(x)*sin((x)/sqrt(2))-sin(x)*cos((x)/sqrt(2))
http://imageshack.us/photo/my-images/196/1319958400.png/
Since there is only one variable, and the graph is harmonic, there's probably a way to solve the equation, but I cannot think of one for now. Use iteration as drwls suggested for the moment.
f(x)=sqrt(2)*cos(x)*sin((x)/sqrt(2))-sin(x)*cos((x)/sqrt(2))
http://imageshack.us/photo/my-images/196/1319958400.png/
Since there is only one variable, and the graph is harmonic, there's probably a way to solve the equation, but I cannot think of one for now. Use iteration as drwls suggested for the moment.
0 answers