## 2 Theory
The boundary conditions at the joint between the beam and the elastic support can be obtained because the sum of the forces at the joint must be zero. These conditions are at \(\overline{x}=0\):  
\[M_{\overline{x}=0}=EI(x)w^{\prime\prime}-K_{\Phi 1}w^{\prime}=0, \tag{7}\]  
and  
\[V_{\overline{x}=0}=-(EI(x)w^{\prime\prime})^{\prime}+F_{0}w^{\prime}+\rho I(x )\ddot{w}^{\prime}-K_{W1}w-m_{pi}\ddot{w}=0, \tag{8}\]  
where \(M_{\overline{x}=0}\) and \(V_{x=0}\) are the bending moment and shear force at \(\overline{x}=0\), respectively, at \(\overline{x}=0\). At \(\overline{x}=1\):  
\[\mathcal{U}=\frac{1}{2}\int_{0}^{L}EI(x)\big{(}w^{\prime\prime}\big{)}^{2}dx+ \frac{1}{2}\int_{0}^{L}F_{0}(w\prime)^{2}dx+\frac{1}{2}K_{W1}(w_{x=0})^{2}+ \frac{1}{2}K_{W2}(w_{x=L})^{2}+\frac{1}{2}K_{\Phi 1}(w\prime_{x=0})^{2}+ \frac{1}{2}K_{\Phi 2}(w\prime_{x=L})^{2}, \tag{1}\]0  
and  
\[\mathcal{U}=\frac{1}{2}\int_{0}^{L}EI(x)\big{(}w^{\prime\prime}\big{)}^{2}dx+ \frac{1}{2}\int_{0}^{L}F_{0}(w\prime)^{2}dx+\frac{1}{2}K_{W1}(w_{x=0})^{2}+ \frac{1}{2}K_{W2}(w_{x=L})^{2}+\frac{1}{2}K_{\Phi 1}(w\prime_{x=0})^{2}+ \frac{1}{2}K_{\Phi 2}(w\prime_{x=L})^{2}, \tag{1}\]1  
where \(M_{\overline{x}=1}\) and \(V_{\overline{x}=1}\) are the bending moment and shear force at \(\overline{x}=1\), respectively, at \(\overline{x}=1\). Eventually, the bending moments and shear forces at both ends become zero. When considering the free-free end conditions, Equations (7)-(10) can be adopted for the bending moments and shear forces at both ends. This differs from previous studies [30;31] and simplifies the process. Assuming the harmonic oscillation has an angular frequency (\(\omega\)),  
\[\mathcal{U}=\frac{1}{2}\int_{0}^{L}EI(x)\big{(}w^{\prime\prime}\big{)}^{2}dx+ \frac{1}{2}\int_{0}^{L}F_{0}(w\prime)^{2}dx+\frac{1}{2}K_{W1}(w_{x=0})^{2}+ \frac{1}{2}K_{W2}(w_{x=L})^{2}+\frac{1}{2}K_{\Phi 1}(w\prime_{x=0})^{2}+ \frac{1}{2}K_{\Phi 2}(w\prime_{x=L})^{2}, \tag{1}\]2  
where \(W(x)\) is the amplitude of \(w(x,t)\). By substituting Equations (3) and (11) into the differential equation, the nondimensional form of Equation (4) can be rewritten as follows:  
\[ \begin{array}{ll}(1-\zeta)^{n+2}W^{\prime\prime\prime\prime}&-2(n+2)(1-\zeta )^{n+1}W^{\prime\prime}\\ &+\Big{(}(n+1)(n+2)(1-\zeta)^{n}+D_{2}(1-\zeta)^{n+2}-D_{3}\Big{)}W^{\prime \prime}\\ &-(n+2)D_{2}(1-\zeta)^{n+1}W^{\prime}-D_{1}(1-\zeta)^{n}W=0,\end{array}  \tag{12}\]  
For \(n=1\), the nondimensional differential equation of a single tapered beam is given by  
\[ \begin{array}{c}\big{(}\zeta^{3}-3\zeta^{2}+3\zeta-1\big{)}W^{\prime\prime \prime\prime}+6\big{(}\zeta^{2}-2\zeta+1\big{)}W^{\prime\prime\prime}+\big{\{} \overline{D}_{1}\zeta^{3}+\overline{D}_{2}\zeta^{2}+\overline{D}_{3}\zeta+ \overline{D}_{4}\big{\}}W^{\prime\prime}\\ +\overline{D}_{5}\big{(}\zeta^{2}-2\zeta+1\big{)}W^{\prime}+\overline{D}_{6}( \zeta-1)W=0,\end{array}  \tag{14}\]  
where \(\overline{D}_{1}=D_{2}\), \(\overline{D}_{2}=-3D_{2}\), \(\overline{D}_{3}=3(D_{2}+2)\), \(\overline{D}_{4}=-6-D_{2}+D_{3}\), \(\overline{D}_{5}=3D_{2}\), \(\overline{D}_{6}=-D_{1}\). The solution of Equation (14) can be assumed by  
\[W(\zeta,k)=\sum_{i=0}^{\infty}a_{i+1}(k)\zeta^{k+i}. \tag{15}\]  
By substituting Equation (15) into Equation (14), the indicial equation is determined as  
\[k(k-1)(k-2)(k-3)a_{1}=0, \tag{16}\]  
and the recurrence relationship can be obtained as  
\[ \begin{array}{c}a_{i+5}(k)=\frac{3(k+i+2)}{(k+i+4)}a_{i+4}(k)+\frac{ \overline{D}_{4}-3(k+i)(k+i+3)}{(k+i+4)(k+i+3)}a_{i+3}(k)\\ +\frac{(k+i)\big{\{}(k+i-1)(k+i+4)+\overline{D}_{3}\big{\}}+\overline{D}_{5}}{( k+i+4)(k+i+3)(k+i+2)}a_{i+2}(k)\\ +\frac{(k+i)\big{\{}\overline{D}_{2}(k+i-1)-2\overline{D}_{5}\big{\}}- \overline{D}_{6}}{(k+i+4)(k+i+3)(k+i+2)(k+i+1)}a_{i+1}(k)\\ +\frac{(k+i-1)\big{\{}\overline{D}_{1}(k+i-2)+\overline{D}_{5}\big{\}}+ \overline{D}_{6}}{(k+i+4)(k+i+3)(k+i+2)(k+i+1)}a_{i}(k).\end{array}  \tag{17}\]  
The Frobenius coefficients are as follows:  
\[