Optimizing Resonance Frequency of Vibrating Carbon Nanotubes Using Mathematica

  • Thread starter Excom
  • Start date
In summary, the equation describes the resonance frequency of a beam under axial tension and has two variables, x and y. The equation cannot be solved in terms of elementary functions, but it is possible to approximate the solution using a contour plot and FindRoot in Mathematica.
  • #1
Excom
61
0
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?
 
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  • #2
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 [itex]f(x)= xe^x[/itex]) but I wouldn't like to try!
 
  • #3
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:
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:
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
how the result of this equation done
f= 1/t= 1.44/(RA+ 2RB)*C
 

1. How do I know which method to use when solving an equation?

There are several methods for solving equations, such as substitution, elimination, and graphing. The best method to use depends on the type of equation you are dealing with. For linear equations, substitution or elimination may be more efficient, while for quadratic equations, factoring or the quadratic formula may be better options. It is important to understand the characteristics of each type of equation to determine the most appropriate method.

2. What are the basic steps for solving an equation?

The basic steps for solving an equation include isolating the variable on one side of the equation, using inverse operations to cancel out any constants or coefficients, and simplifying the equation until the variable is left alone. It is important to perform the same operations on both sides of the equation to maintain its equality.

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4. What should I do if there are variables on both sides of the equation?

If there are variables on both sides of the equation, the first step is to move all the variables to one side of the equation and all the constants to the other side. This can be done by using inverse operations to cancel out any constants or coefficients. Once the variables are on one side, follow the basic steps for solving an equation to isolate the variable and find its value.

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