in desperate need for help (Trigonometric equation)

Good people of the internet,

I really need your help in solving this equation:

>> al=(((10^4+g^4)^0.5 -10^2)^0.5)*L;
>> bl=(((10^4+g^4)^0.5 +10^2)^0.5)*L;
>> char=2*al*bl*(1-cos(al)*cosh(bl))+(bl^2-al^2)*sin(al)*sinh(bl)
eqn=char==0
- sin(L*((g^4 + 10000)^(1/2) - 100)^(1/2))*sinh(L*((g^4 + 10000)^(1/2) + 100)^(1/2))*(L^2*((g^4 + 10000)^(1/2) - 100) - L^2*((g^4 + 10000)^(1/2) + 100)) - 2*L^2*(cos(L*((g^4 + 10000)^(1/2) - 100)^(1/2))*cosh(L*((g^4 + 10000)^(1/2) + 100)^(1/2)) - 1)*((g^4 + 10000)^(1/2) - 100)^(1/2)*((g^4 + 10000)^(1/2) + 100)^(1/2) == 0

It's the characteristic equation of a 4x4 matrix. The infinite sollutions of this equation represent the dimensionless eigenfrequencies g(L) of an oscillating, clamped clamped beam with axial load.

My aim is to find the first 4 (lowest) eigenfrequencies depending on the length of the beam.

I've tried everything I can think of I am a beginner and I find it really difficult.

fsolve requires a guess, and I can't predict the periodicity of this function

I tried to get an analytical sollution with

s=solve(eqn,g,'return conditions',true)

Please someone enlighten me.

Thanks in advance.

[Image Analyst]

Well for one thing, char is a keyword, so you don't want to use that as the name of your variable.

[Walter Roberson]

define lowest in this context.

You use g^4 everywhere .you could replace it with nonnegative G for searching but every solution in G would also imply -G and i*G and -i*G so if there are infinite solutions then the largest of them also implies the one with greatest negative value . Therefore if there are infinite solutions then the four smallest are four negative infinities.

[Walter Roberson]

When L is an integer, there appears to be L+1 positive roots for g^4 (counting 0). The largest root can be on the order of 14000. The slopes are numerically relatively steep, making it difficult to find numeric solutions for given L.

[Jan]

@solo gouda: The nature of this forum is to share questions and solutions. After an answer was given by a member of the community, it is impolite to remove essential parts of the question, because the thread becomes meaningless afterwards.

Answer 1

[David Goodmanson]

Hi solo,

In situations like this it never hurts to plot out the function to see the zero crossings. I don't think people do that enough. Plotting gives you an idea of the regions to specify for the fzero or fsolve functions.

The amplitude of the function y (your char) rapidly increases with g. Since you are only trying to find the zero crossings, it helps to almost-normalize the function by dividing by cosh(bl) which is rapidly increasing in the same manner. That works since cosh is never zero for real argument.

L = 5;
g = 0:.001:14;
al=(((10^4+g.^4).^0.5 -10^2).^0.5)*L;
bl=(((10^4+g.^4).^0.5 +10^2).^0.5)*L;
y = 2*al.*bl.*(1-cos(al).*cosh(bl))+(bl.^2-al.^2).*sin(al).*sinh(bl);
y = y./cosh(bl);
figure(1)
plot(g,y)

I believe that there are an unlimited number of roots for most positive values of L, probably all of them. For L=5 there are 17 zero crossings for g<14, not counting the one at g=0. As you would expect, as L decreases the frequencies of the roots increase and vice versa.

[solo gouda]

you are a true hero of the internet. i tried to do complex things and didnt think of the 1-1 of math. Plots. thank you for pointing out the obvious. have a nice one