Example 1 The string of length \(l\) and fixed on both ends is plucked at its midpoint and then released. Determine the subsequent motion. The string is shown in Figure 1.
Figure 1 - Initial displacement of string fixed at both ends and plucked at its midpoint.
Solution:The differential equation that describes the sting can be written as: \begin{eqnarray} c^2 \frac{\partial^2 w}{\partial x^2} &=& \frac{\partial^2 w}{\partial x^2}\nonumber\\ \end{eqnarray}
The general solution \(w_n(x,t)\) can be written as: \begin{eqnarray} w_n(x,t) &=& \sum_{n=1}^\infty w_n(x,t) = \sum_{n=1}^\infty \sin \frac{n\pi x}{l}\left[C_n\cos\frac{nc\pi t}{l} + D_n \sin\frac{nc\pi t}{l}\right]\nonumber \end{eqnarray} where: \begin{eqnarray} \sum_{n=1}^\infty C_n \sin\frac{n \pi x}{l} &=& w_0(x)\nonumber\\ \sum_{n=1}^\infty \frac{nc\pi}{l}D_n \sin \frac{n\pi x}{l} &=& \dot{w}_0(x).\nonumber \end{eqnarray} Since there is no initial velocity \(\dot{w}_0(x) = 0\) the coefficient D_n is also equal to zero. The solution in reduced form can be written as: \begin{eqnarray} w(x,t) &=& \sum_{n=!}^\infty C_n\sin \frac{n \pi x}{l} \cos \frac{nc\pi t}{l}\nonumber \end{eqnarray} where: \begin{eqnarray} C_n &=& \frac{2}{l}\int_0^l w_0(x) \sin \frac{n \pi x}{l}dx.\nonumber \end{eqnarray} The initial deflection \(w_0(x)\) is written as: \begin{eqnarray} w_0(x) &=& \begin{cases}\frac{2hx}{l}\quad 0 \leq x \leq \frac{l}{2}\\ \frac{2h(l-x)}{l}\quad \frac{l}{2}\leq x \leq l\end{cases}\nonumber\\ \end{eqnarray} By substituting previous expression into the expression for \(C_n\) the following is obtained: \begin{eqnarray} C_n &=& \frac{2}{l}\left(\int_0^{l/2} \frac{2h x}{l} \sin \frac{n\pi x}{l}dx+ \int_{l/2}^l \frac{2h}{l}(l-x) \sin \frac{n\pi x}{l}dx\right)\nonumber\\ &=& \begin{cases}\frac{8h}{\pi^2 n^2}\sin\frac{n\pi}{2} \quad n = 1,3,5,...\\ 0 \quad n = 2,4,6,...\end{cases} \end{eqnarray} Using the relation \(\sin\frac{n\pi}{2} = (-1)^\frac{n-1}{2}\quad n=1,3,5,...\) the desired solution can be written as: \begin{eqnarray} w(x,t) &=& \frac{8h}{\pi^2}\left(\sin\frac{\pi x}{l}\cos\frac{\pi ct}{l} - \frac{1}{9}\sin\frac{3\pi x}{l}\cos\frac{3\pi c t }{l}+\cdots\right).\nonumber \end{eqnarray} Example 2 Find the natural frequencies and free-vibration solution of a bar fixed at one end and free at the other.
Solution: The differential equations for free-vibration unifrom bar can be written as: \begin{eqnarray} c^2\frac{\partial^2 u}{\partial x^2} u(x,t) &=& \frac{\partial^2 u}{\partial t^2}(x,t)\nonumber \end{eqnarray} The solution of previous equation can be written as: \begin{eqnarray} u(x,t) &=& \left(A\cos \frac{\omega x}{c} + B \sin \frac{\omega x}{c}\right)(C\cos \omega t + D\sin \omega t).\nonumber \end{eqnarray} In this example the bar is fixed at \(x=0\) and free at \(x = l\). The boundary conditions are: \begin{eqnarray} u(0,t) &=& 0, \quad t\geq 0\nonumber\\ \frac{\partial u}{\partial x}(l,t) &=& 0, \quad t \geq 0.\nonumber \end{eqnarray} Applying the boudnary condition to the solution the follwing si obtained: \begin{eqnarray} u(0,t) &=& 0\Rightarrow \left(A\cos \frac{\omega \cdot 0}{c} + B \sin \frac{\omega \cdot 0}{c}\right)(C\cos \omega t + D\sin \omega t) = 0\nonumber \\ A &=& 0\nonumber\\ \frac{\partial u}{\partial x} &=& \left(\frac{B \omega \cos\left(\frac{\omega x}{c}\right)}{c} - \frac{A \omega \sin\left(\frac{\omega x}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t)\nonumber\\ \frac{\partial u}{\partial x} &=& \left(\frac{B \omega \cos\left(\frac{\omega x}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t)\nonumber\\ &&\left(\frac{ B \omega \cos\left(\frac{\omega l}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t) = 0\nonumber\\ &&\left(\frac{ B \omega \cos\left(\frac{\omega l}{c}\right)}{c}\right) = 0\nonumber\\ &&B\frac{\omega}{c}\cos\frac{\omega l}{c} = 0\nonumber \end{eqnarray} The eigenvalues or natural frequencies are given by: \begin{eqnarray} \frac{\omega_n l }{c} &=& (2n+1)\frac{\pi}{2}\quad n = 0,1,2,...\nonumber\\ \omega_n &=& \frac{(2n+1)\pi c}{2l} \quad n = 0,1,2,...\nonumber \end{eqnarray} The free-vibration solution can be written as: \begin{eqnarray} u(x,t) &=& \sum_{n=0}^\infty u_n(x,t)\nonumber\\ &=& \sum_{n=0}^\infty \sin\frac{(2n+1)\pi x}{2l} \left[C_n \cos\frac{(2n+1)\pi c t}{2l} + D_n \sin \frac{(2n+1)\pi c t}{2l}\right]\nonumber\\ \end{eqnarray} Example 3 Determine the natural freuqncies of a bar with one ened fixed and a mass attached at the other end as shown in Figure 2
Figure 2 - Bar fixed at one end with body of mass \(m\) attached to it.
Solution: The differenatial equation that describes free vibration of uniform bar can be written as: \begin{eqnarray} c^2\frac{\partial^2 u}{\partial x^2} (x,t) &=& \frac{\partial^2 x}{\partial t^2}(x,t).\nonumber \end{eqnarray} Solution of the previous differential equation is: \begin{eqnarray} u(x,t) &=& \left(A\cos \frac{\omega x}{c} + B \sin \frac{\omega x}{c}\right)(C\cos \omega t + D\sin \omega t) \end{eqnarray} The boundary condition at the fixed end (\(x=0\)) is \(u(0,t)\) = 0. By substituting boundary condition into solution (previous equation) the following is obtained: \begin{eqnarray} u(x,t) &=& A(C\cos \omega t + D\sin \omega t).\nonumber \end{eqnarray} At the end \(x=l\), the tensile force in the bar must be equal to the inertia force of the vibrating mass \(M\), so: \begin{eqnarray} AE\frac{\partial^2 u}{\partial x^2}(x,t) &=& -M\frac{\partial^2 u}{\partial t^2}(l,t)\nonumber \end{eqnarray} Substituting u(x,t) into previous differential equation the following expression is obtained: \begin{eqnarray} AE\frac{\omega}{c}\cos\frac{\omega l}{c}(C\cos\omega t + D \sin \omega t) &=& M \omega^2 \sin\frac{qomega l}{c} (C\cos \omega t + D\sin \omega t)\nonumber \end{eqnarray} The previous expression can be simplified and written as: \begin{eqnarray} \frac{AE\omega}{c}\cos\frac{\omega l}{c} &=& M\omega^2 \sin \frac{\omega l}{c}\nonumber \end{eqnarray} The previous expression will be divided with \(\cos\frac{\omega l}{c}\), and new variables will be introduced \(\alpha = \frac{\omega l}{c}, \beta = \frac{AEl}{c^2M} = \frac{m}{M}\). The previous expression can now be written as: \begin{eqnarray} \alpha \tan \alpha &=&\beta \nonumber \end{eqnarray} The equation \(\alpha \sin \alpha = \beta \) is frequency equation and the solution will give the natural freuqncies of the system. If the mass of the bare is negligible when compared to the mass attached to the bar the following relations are valid: \begin{eqnarray} c &=& \sqrt{\frac{E}{\rho}} = \sqrt(\frac{EAl}{m}) = \infty\nonumber\\ \alpha &=& \frac{\omega l}{c} = 0 \nonumber\\ \tan \frac{\omega l}{c} &\simeq& \frac{\omega l}{c}\nonumber \end{eqnarray} The natural freuqency equation is reduced to: \begin{eqnarray} \left(\frac{\omega l}{c}\right)^2 &=& \beta\Big/\sqrt{}\nonumber\\ \omega_1 &=& \frac{c}{l}\sqrt{\beta} = \frac{c}{l}\sqrt{\frac{\rho A l}{M}}\nonumber \end{eqnarray} Example 4Derive the natural freuqnecy equation and normal modes function of a unifrom beam fixed at \(x = 0\) and simply supported at \(x = t\).
Solution: The boundary conditions are \begin{eqnarray} W(0) &=& 0,\quad\frac{dW}{dx}(0) = 0,\quad W(l) = 0\nonumber\\ EI\frac{d^2W}{dx^2}(l) &=& 0\quad \frac{d^2W}{dx^2}(x) = 0. \nonumber \end{eqnarray} By applying the first boundary condition \(W(0) = 0\) to the equation: \begin{eqnarray} W(x) &=& C_1\cos\beta x + C_2 \sin \beta x + C_3 \cosh\beta x + C_4 \sinh\beta x\nonumber\\ \end{eqnarray} the following is obtained: \begin{eqnarray} W(0) &=& C_1 + C_3 = 0\nonumber \end{eqnarray} By applying the second boundary condition \(\frac{dW}{dx}(0) = 0\) to solution W(x) the follwoing is obtained: \begin{eqnarray} \frac{dW}{dx}\Big|_{x=0} &=& \beta(-C_1\sin\beta x + C_2 \cos \beta x + C_3\sinh\beta x + C_4 \cosh\beta x)_{x=0} = 0\nonumber\\ \beta (C_2 + C_4) = 0\nonumber \end{eqnarray} The solution can now be written as: \begin{eqnarray} W(x) &=& C_1(\cos\beta x - \cosh\beta x) + C_2(\sin\beta x - \sinh\beta x)\nonumber \end{eqnarray} Applying the rest of the conditions the following expressions are obtained: \begin{eqnarray} C_1(\cos\beta l - \cosh\beta l) + C_2 (\sin\beta l-\sinh \beta l) &=& 0\nonumber\\ -C_1(\cos\beta l +\cosh\beta l) - C_2 (\sin\beta l + \sinh \beta l) &=& 0\nonumber \end{eqnarray} For a nontrivial solution of \(C_1\) and \(C_2\), the determinat of their coefficients must be zero i.e. \begin{eqnarray} \begin{vmatrix} (\cos\beta l - \cosh \beta l) & (\sin\beta l - \sinh\beta l)\\ -(\cos\beta l + \cosh \beta l) & -(\sin\beta l + \sinh \beta l) \end{vmatrix}&=& 0\nonumber \end{eqnarray} Solving the determinant the frequency equation is obtained: \begin{eqnarray} \cos\beta l \sin\beta l - \sin\beta l \cosh \beta l &=& 0\nonumber \\ \tan \beta l &=& \tanh \beta l\nonumber \end{eqnarray} The roots of the previous equation gives the natural freuqncy of vibration: \begin{eqnarray} \omega_n &=& (\beta_n l)^2 \sqrt{\frac{EI}{\rho A l^4}}\nonumber\\ \end{eqnarray} If the value \(C_2\) corresponding to \(\beta_n\) is denoted as \(C_{2n}\), it can be expressed in terms of \(C_{1n}\) as: \begin{eqnarray} C_{2n} &=& -C_{1n}\left(\frac{\cos \beta_n l - \cosh \beta_n l}{\sin\beta_n l - \sinh\beta_n l}\right).\nonumber \end{eqnarray} The \(W_n(x)\) can be wirtten as: \begin{eqnarray} W_n(x) &=& C_{1n}\left((cos\beta_n x - \cosh \beta_n x) - \left(\frac{\cos \beta_n l - \cosh \beta_n l}{\sin\beta_n l - \sinh\beta_n l}\right)(\sin\beta_n x- \sinh\beta_n x)\right)\nonumber\\ \end{eqnarray} The normal modes of vibration from: \begin{eqnarray} w_n(x,t) &=& W_n(x)(A_n\cos\omega_n t + B_n \sin \omega_n t)\nonumber \end{eqnarray}
Solution:The differential equation that describes the sting can be written as: \begin{eqnarray} c^2 \frac{\partial^2 w}{\partial x^2} &=& \frac{\partial^2 w}{\partial x^2}\nonumber\\ \end{eqnarray}
The general solution \(w_n(x,t)\) can be written as: \begin{eqnarray} w_n(x,t) &=& \sum_{n=1}^\infty w_n(x,t) = \sum_{n=1}^\infty \sin \frac{n\pi x}{l}\left[C_n\cos\frac{nc\pi t}{l} + D_n \sin\frac{nc\pi t}{l}\right]\nonumber \end{eqnarray} where: \begin{eqnarray} \sum_{n=1}^\infty C_n \sin\frac{n \pi x}{l} &=& w_0(x)\nonumber\\ \sum_{n=1}^\infty \frac{nc\pi}{l}D_n \sin \frac{n\pi x}{l} &=& \dot{w}_0(x).\nonumber \end{eqnarray} Since there is no initial velocity \(\dot{w}_0(x) = 0\) the coefficient D_n is also equal to zero. The solution in reduced form can be written as: \begin{eqnarray} w(x,t) &=& \sum_{n=!}^\infty C_n\sin \frac{n \pi x}{l} \cos \frac{nc\pi t}{l}\nonumber \end{eqnarray} where: \begin{eqnarray} C_n &=& \frac{2}{l}\int_0^l w_0(x) \sin \frac{n \pi x}{l}dx.\nonumber \end{eqnarray} The initial deflection \(w_0(x)\) is written as: \begin{eqnarray} w_0(x) &=& \begin{cases}\frac{2hx}{l}\quad 0 \leq x \leq \frac{l}{2}\\ \frac{2h(l-x)}{l}\quad \frac{l}{2}\leq x \leq l\end{cases}\nonumber\\ \end{eqnarray} By substituting previous expression into the expression for \(C_n\) the following is obtained: \begin{eqnarray} C_n &=& \frac{2}{l}\left(\int_0^{l/2} \frac{2h x}{l} \sin \frac{n\pi x}{l}dx+ \int_{l/2}^l \frac{2h}{l}(l-x) \sin \frac{n\pi x}{l}dx\right)\nonumber\\ &=& \begin{cases}\frac{8h}{\pi^2 n^2}\sin\frac{n\pi}{2} \quad n = 1,3,5,...\\ 0 \quad n = 2,4,6,...\end{cases} \end{eqnarray} Using the relation \(\sin\frac{n\pi}{2} = (-1)^\frac{n-1}{2}\quad n=1,3,5,...\) the desired solution can be written as: \begin{eqnarray} w(x,t) &=& \frac{8h}{\pi^2}\left(\sin\frac{\pi x}{l}\cos\frac{\pi ct}{l} - \frac{1}{9}\sin\frac{3\pi x}{l}\cos\frac{3\pi c t }{l}+\cdots\right).\nonumber \end{eqnarray} Example 2 Find the natural frequencies and free-vibration solution of a bar fixed at one end and free at the other.
Solution: The differential equations for free-vibration unifrom bar can be written as: \begin{eqnarray} c^2\frac{\partial^2 u}{\partial x^2} u(x,t) &=& \frac{\partial^2 u}{\partial t^2}(x,t)\nonumber \end{eqnarray} The solution of previous equation can be written as: \begin{eqnarray} u(x,t) &=& \left(A\cos \frac{\omega x}{c} + B \sin \frac{\omega x}{c}\right)(C\cos \omega t + D\sin \omega t).\nonumber \end{eqnarray} In this example the bar is fixed at \(x=0\) and free at \(x = l\). The boundary conditions are: \begin{eqnarray} u(0,t) &=& 0, \quad t\geq 0\nonumber\\ \frac{\partial u}{\partial x}(l,t) &=& 0, \quad t \geq 0.\nonumber \end{eqnarray} Applying the boudnary condition to the solution the follwing si obtained: \begin{eqnarray} u(0,t) &=& 0\Rightarrow \left(A\cos \frac{\omega \cdot 0}{c} + B \sin \frac{\omega \cdot 0}{c}\right)(C\cos \omega t + D\sin \omega t) = 0\nonumber \\ A &=& 0\nonumber\\ \frac{\partial u}{\partial x} &=& \left(\frac{B \omega \cos\left(\frac{\omega x}{c}\right)}{c} - \frac{A \omega \sin\left(\frac{\omega x}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t)\nonumber\\ \frac{\partial u}{\partial x} &=& \left(\frac{B \omega \cos\left(\frac{\omega x}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t)\nonumber\\ &&\left(\frac{ B \omega \cos\left(\frac{\omega l}{c}\right)}{c}\right)(C\cos \omega t + D\sin \omega t) = 0\nonumber\\ &&\left(\frac{ B \omega \cos\left(\frac{\omega l}{c}\right)}{c}\right) = 0\nonumber\\ &&B\frac{\omega}{c}\cos\frac{\omega l}{c} = 0\nonumber \end{eqnarray} The eigenvalues or natural frequencies are given by: \begin{eqnarray} \frac{\omega_n l }{c} &=& (2n+1)\frac{\pi}{2}\quad n = 0,1,2,...\nonumber\\ \omega_n &=& \frac{(2n+1)\pi c}{2l} \quad n = 0,1,2,...\nonumber \end{eqnarray} The free-vibration solution can be written as: \begin{eqnarray} u(x,t) &=& \sum_{n=0}^\infty u_n(x,t)\nonumber\\ &=& \sum_{n=0}^\infty \sin\frac{(2n+1)\pi x}{2l} \left[C_n \cos\frac{(2n+1)\pi c t}{2l} + D_n \sin \frac{(2n+1)\pi c t}{2l}\right]\nonumber\\ \end{eqnarray} Example 3 Determine the natural freuqncies of a bar with one ened fixed and a mass attached at the other end as shown in Figure 2
Solution: The differenatial equation that describes free vibration of uniform bar can be written as: \begin{eqnarray} c^2\frac{\partial^2 u}{\partial x^2} (x,t) &=& \frac{\partial^2 x}{\partial t^2}(x,t).\nonumber \end{eqnarray} Solution of the previous differential equation is: \begin{eqnarray} u(x,t) &=& \left(A\cos \frac{\omega x}{c} + B \sin \frac{\omega x}{c}\right)(C\cos \omega t + D\sin \omega t) \end{eqnarray} The boundary condition at the fixed end (\(x=0\)) is \(u(0,t)\) = 0. By substituting boundary condition into solution (previous equation) the following is obtained: \begin{eqnarray} u(x,t) &=& A(C\cos \omega t + D\sin \omega t).\nonumber \end{eqnarray} At the end \(x=l\), the tensile force in the bar must be equal to the inertia force of the vibrating mass \(M\), so: \begin{eqnarray} AE\frac{\partial^2 u}{\partial x^2}(x,t) &=& -M\frac{\partial^2 u}{\partial t^2}(l,t)\nonumber \end{eqnarray} Substituting u(x,t) into previous differential equation the following expression is obtained: \begin{eqnarray} AE\frac{\omega}{c}\cos\frac{\omega l}{c}(C\cos\omega t + D \sin \omega t) &=& M \omega^2 \sin\frac{qomega l}{c} (C\cos \omega t + D\sin \omega t)\nonumber \end{eqnarray} The previous expression can be simplified and written as: \begin{eqnarray} \frac{AE\omega}{c}\cos\frac{\omega l}{c} &=& M\omega^2 \sin \frac{\omega l}{c}\nonumber \end{eqnarray} The previous expression will be divided with \(\cos\frac{\omega l}{c}\), and new variables will be introduced \(\alpha = \frac{\omega l}{c}, \beta = \frac{AEl}{c^2M} = \frac{m}{M}\). The previous expression can now be written as: \begin{eqnarray} \alpha \tan \alpha &=&\beta \nonumber \end{eqnarray} The equation \(\alpha \sin \alpha = \beta \) is frequency equation and the solution will give the natural freuqncies of the system. If the mass of the bare is negligible when compared to the mass attached to the bar the following relations are valid: \begin{eqnarray} c &=& \sqrt{\frac{E}{\rho}} = \sqrt(\frac{EAl}{m}) = \infty\nonumber\\ \alpha &=& \frac{\omega l}{c} = 0 \nonumber\\ \tan \frac{\omega l}{c} &\simeq& \frac{\omega l}{c}\nonumber \end{eqnarray} The natural freuqency equation is reduced to: \begin{eqnarray} \left(\frac{\omega l}{c}\right)^2 &=& \beta\Big/\sqrt{}\nonumber\\ \omega_1 &=& \frac{c}{l}\sqrt{\beta} = \frac{c}{l}\sqrt{\frac{\rho A l}{M}}\nonumber \end{eqnarray} Example 4Derive the natural freuqnecy equation and normal modes function of a unifrom beam fixed at \(x = 0\) and simply supported at \(x = t\).
Solution: The boundary conditions are \begin{eqnarray} W(0) &=& 0,\quad\frac{dW}{dx}(0) = 0,\quad W(l) = 0\nonumber\\ EI\frac{d^2W}{dx^2}(l) &=& 0\quad \frac{d^2W}{dx^2}(x) = 0. \nonumber \end{eqnarray} By applying the first boundary condition \(W(0) = 0\) to the equation: \begin{eqnarray} W(x) &=& C_1\cos\beta x + C_2 \sin \beta x + C_3 \cosh\beta x + C_4 \sinh\beta x\nonumber\\ \end{eqnarray} the following is obtained: \begin{eqnarray} W(0) &=& C_1 + C_3 = 0\nonumber \end{eqnarray} By applying the second boundary condition \(\frac{dW}{dx}(0) = 0\) to solution W(x) the follwoing is obtained: \begin{eqnarray} \frac{dW}{dx}\Big|_{x=0} &=& \beta(-C_1\sin\beta x + C_2 \cos \beta x + C_3\sinh\beta x + C_4 \cosh\beta x)_{x=0} = 0\nonumber\\ \beta (C_2 + C_4) = 0\nonumber \end{eqnarray} The solution can now be written as: \begin{eqnarray} W(x) &=& C_1(\cos\beta x - \cosh\beta x) + C_2(\sin\beta x - \sinh\beta x)\nonumber \end{eqnarray} Applying the rest of the conditions the following expressions are obtained: \begin{eqnarray} C_1(\cos\beta l - \cosh\beta l) + C_2 (\sin\beta l-\sinh \beta l) &=& 0\nonumber\\ -C_1(\cos\beta l +\cosh\beta l) - C_2 (\sin\beta l + \sinh \beta l) &=& 0\nonumber \end{eqnarray} For a nontrivial solution of \(C_1\) and \(C_2\), the determinat of their coefficients must be zero i.e. \begin{eqnarray} \begin{vmatrix} (\cos\beta l - \cosh \beta l) & (\sin\beta l - \sinh\beta l)\\ -(\cos\beta l + \cosh \beta l) & -(\sin\beta l + \sinh \beta l) \end{vmatrix}&=& 0\nonumber \end{eqnarray} Solving the determinant the frequency equation is obtained: \begin{eqnarray} \cos\beta l \sin\beta l - \sin\beta l \cosh \beta l &=& 0\nonumber \\ \tan \beta l &=& \tanh \beta l\nonumber \end{eqnarray} The roots of the previous equation gives the natural freuqncy of vibration: \begin{eqnarray} \omega_n &=& (\beta_n l)^2 \sqrt{\frac{EI}{\rho A l^4}}\nonumber\\ \end{eqnarray} If the value \(C_2\) corresponding to \(\beta_n\) is denoted as \(C_{2n}\), it can be expressed in terms of \(C_{1n}\) as: \begin{eqnarray} C_{2n} &=& -C_{1n}\left(\frac{\cos \beta_n l - \cosh \beta_n l}{\sin\beta_n l - \sinh\beta_n l}\right).\nonumber \end{eqnarray} The \(W_n(x)\) can be wirtten as: \begin{eqnarray} W_n(x) &=& C_{1n}\left((cos\beta_n x - \cosh \beta_n x) - \left(\frac{\cos \beta_n l - \cosh \beta_n l}{\sin\beta_n l - \sinh\beta_n l}\right)(\sin\beta_n x- \sinh\beta_n x)\right)\nonumber\\ \end{eqnarray} The normal modes of vibration from: \begin{eqnarray} w_n(x,t) &=& W_n(x)(A_n\cos\omega_n t + B_n \sin \omega_n t)\nonumber \end{eqnarray}