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October 31, 2014

October 31, 2014

Posted by **Chris** on Sunday, October 30, 2011 at 3:06am.

(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?

- Engineering -
**drwls**, Sunday, October 30, 2011 at 6:36amI 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.

- Engineering -
**MathMate**, Sunday, October 30, 2011 at 8:03amIf 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.

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