## 2 Methods  
The system under consideration is a finite length rod with changes in its area with respect to the longitudinal position. This problem is analytically modeled using the rod longitudinal differential equation. The model uses the following assumptions: (1) the excitation is at a fixed frequency; (2) the displacement field is one-dimensional in the longitudinal direction; (3) the (varying) height and width of the rod are small compared to its length; (4) the rod has uniform stress distribution; (5) the area can be represented as a strictly positive continuous analytical function with respect to position on the interval of the rod and has a continuous first derivative; (6) the change of area is symmetric about the centroid of the rod; and (7) the particle motion is linear. The rod longitudinal motion equation with non-uniform area is [1]  
\[EA(x)\frac{\partial^{2}u(x,t)}{\partial x^{2}}+E\frac{\partial A(x)}{\partial x }\frac{\partial u(x,t)}{\partial x}-\rho A(x)\frac{\partial^{2}u(x,t)}{ \partial t^{2}}=f(x,t), \tag{1}\]  
where \(u(x,t)\) is longitudinal displacement, \(x\) is spatial position, \(t\) is time, \(E\) is [PERSON]'s modulus, \(A(x)\) is the cross-sectional area, \(\rho\) is density and \(f(x,t)\) is the applied force per unit length in the longitudinal direction. Note the second term in Equation (1) that is present because of the spatially varying area. This term is the longitudinal stress in the rod multiplied by the change of area with respect to the spatial variable \(x\). The forcing function and the displacement are modeled as harmonic in time at fixed frequency, i.e., \(u(x,t)=U(x)\exp(-\mathrm{i}\omega t)\) and \(f(x,t)=F(x)\exp(-\mathrm{i}\omega t)\), which transforms Equation (1) into  
\[EA(x)\frac{\mathrm{d}^{2}U(x)}{\mathrm{d}x^{2}}+E\frac{\mathrm{d}A(x)}{ \mathrm{d}x}\frac{\mathrm{d}U(x)}{\mathrm{d}x}+\rho\omega^{2}A(x)U(x)=F(x). \tag{2}\]  
This is a second order ordinary linear differential equation with non-constant coefficients. To facilitate a solution to this problem, it is necessary to specify boundary conditions. For this method, three possible combinations of boundary conditions are possible: (1) fixed-fixed, (2) fixed-free or (3) free-free. The eigenfunctions (or modes) for this problem with the rod having a uniform cross-sectional area are the basis for this solution method, and for the fixed-fixed boundary condition these are  
\[\phi_{n}(x)=\sin\Bigl{(}\frac{n\pi x}{L}\Bigr{)}\qquad n=1,\;2,\;3\;\cdots, \tag{3}\]  
Figure 1: Fixed-free rod with a linear change of area. for the fixed-free boundary condition they are  
\[\phi_{n}(x)=\sin\left[\frac{(2n-1)\pi x}{2L}\right]\qquad n=1,\;2,\;3\;\cdots, \tag{4}\]  
and for the free-free boundary condition they are  
\[\phi_{n}(x)=\cos\Bigl{(}\frac{n\pi x}{L}\Bigr{)}\qquad n=0,\;1,\;2,\;3\;\cdots. \tag{5}\]  
The solution to the longitudinal displacement is now written as a series of indexed coefficients multiplied by the eigenfunctions from the uniform area differential equation. This equation, for the fixed-free boundary condition, is  
\[U(x)=\sum\limits_{n=1}^{n=\infty}U_{n}\sin\left[\frac{(2n-1)\pi x}{2L}\right], \tag{6}\]  
where \(U_{n}\) is the unknown wave propagation coefficient whose solution is sought. The fixed-free boundary condition is analyzed in this section as it corresponds to the example problems presented later in the paper. Equation (6) is now inserted into Equation (2), and this results in  
\[ \begin{array}{l}A(x)\sum\limits_{n=1}^{n=\infty}\left\{\rho\omega^{2}-E \Bigl{[}\frac{(2n-1)\pi}{2L}\Bigr{]}^{2}\right\}U_{n}\sin\Bigl{[}\frac{(2n-1) \pi x}{2L}\Bigr{]}+\\ E\frac{\mathrm{d}A(x)}{\mathrm{d}x}\sum\limits_{n=1}^{n=\infty}\left[\frac{(2 n-1)\pi}{2L}\right]U_{n}\cos\Bigl{[}\frac{(2n-1)\pi x}{2L}\Bigr{]}=F(x).\end{array}  \tag{7}\]  
Equation (7) is now multiplied by a single \(p\)-indexed eigenfunction, which yields  
\[