# How do I solve this equation?

1. Aug 16, 2010

### Excom

Hi

In the article: "Beam to String Transition of Vibrating Carbon Nanotubes Under Axial Tension" I have found an equation that describes the resonance frequency of a beam under axial tension.

However, I have some problem solving it.

The equation looks like this:

x + y*Sinh[y + Sqrt[y^2 + x^2]]^0.5*Sin[-y + Sqrt[y^2 + x^2]]^0.5 -
x*Cosh[y + Sqrt[y^2 + x^2]]^0.5*Cos[-y + Sqrt[y^2 + x^2]]^0.5 = 0

x is the dimensionless natural frequency and y is the dimensionless natural frequency parameter.

I want to find the x for different y.

I have tried to solve this equation i Mathematica but without much success.

Can anyone help me?

2. Aug 16, 2010

### HallsofIvy

Since you have x both "inside" and "outside" a transcendental function, I doubt the equation can be solved in terms of elementary functions. You might be able to write the hyperbolic functions in terms of exponentials and use Lamberts W function (defined as the inverse function to $f(x)= xe^x$) but I wouldn't like to try!

3. Aug 16, 2010

### jackmell

Oh, come on guys. Let's try. I mean, he said nothing about getting it exactly right. So first just make a contour plot of it in some desired range, say y in 0 to 5:

Code (Text):
cp1 = ContourPlot[
x + y*Sinh[y + Sqrt[y^2 + x^2]]^0.5*Sin[-y + Sqrt[y^2 + x^2]]^0.5 -
x*Cosh[y + Sqrt[y^2 + x^2]]^0.5*Cos[-y + Sqrt[y^2 + x^2]]^0.5 ==
0, {y, 0, 5}, {x, 0, 20}]

That's below, and suppose I want the value of x when y is one. Well, from the plot, it looks about 8 and 15 right? So let's use Find root to get it closer:

Code (Text):

In[97]:=
f[x_, y_] :=
x + y*Sinh[y + Sqrt[y^2 + x^2]]^0.5*
Sin[-y + Sqrt[y^2 + x^2]]^0.5 -
x*Cosh[y + Sqrt[y^2 + x^2]]^0.5*
Cos[-y + Sqrt[y^2 + x^2]]^0.5
myx = Re[x /. FindRoot[f[x, 1] == 0,
{x, 9}]]
N[f[myx, 1]]
myx = Re[x /. FindRoot[f[x, 1] == 0,
{x, 15}]]
N[f[myx, 1]]

Out[98]=
8.781804090459223

Out[99]=
1.8616219676914625*^-12

Out[100]=
15.099644858445615

Out[101]=
6.752998160663992*^-10

Now I think I could pull at least a B for that effort. :)

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4. Nov 4, 2010