Standing Waves on a String

• The wave equation and boundary conditions in a string of length $L=1~m$ with both fixed ends are as follows:

  $$\frac{d^2y(x)}{dx^2}+k^2y(x)=0~~~~\&~~~~y(0)=y(1)=0$$

  
  

  Firstly, transform this quadratic equation into a system of linear (first degree) equations:

  
  

  $$y \rightarrow y_1$$

  $$\frac{dy}{dx} \rightarrow y_2$$

  
  
  
  

  with these values ( $y_1~ and~y_2$), the system of equations to be solved: (Now, we have a set of equations.)

  	$$\frac{dy_1}{dr}$$	$$=y_2$$
  	$$\frac{dy_2}{dr}$$	$$=-k^2y_1$$

  
  

  and the boundary conditions are:

  
  

  $$y(0) =$$ 0

  $$y(1) =$$ 0

  
  
• Here, the trial-and-error approach differs from the previously discussed boundary value problem.
• Different estimates for $y_2(0)$ values do not make it zero at the other boundary.

• Instead, an estimated value for the $k$ eigenvalue is taken and a solution search is initiated.
• Searching continues by increasing the value of $k$ until the boundary condition (here $y(1)=0$) at the other end is satisfied.
• For example, the solution at the other boundary is $y_{1k}(1)$ for a given value of $k$.
• Accordingly, the next step is find the root of the following equation:

  $$F(k)=y_{1k}(1)-y(1)=0$$

  When we encountered an eigenvalue $k$, then $F(k)$ will change sign as indicating the root.
• (Example py-file: The program to find the 5 smallest of the $k$ eigenvalues in a string: standingwawes.py)
• The program can find the eigenvalues $k=n\pi/L=\pi, 2\pi, 3\pi,\ldots$ on a string of length L=1 m.
• However, the error margin is to be increased by increasing eigenvalues (see $k_n/\pi$ values).

Figure 5.13: Solution for the Eigenvalue Problem for the ODE: $\frac {dy_2}{dx}=-k^2y_1$.