User:EML4507.s13.team4ever.Tobin/R.6.1

Honor Pledge:
On my honor I have neither given nor received unauthorized aid in doing this problem.

Problem Description:
We are tasked with completing Pb.53.9 on p.53.19b by transforming the generalized eigenvalue problem into a standard eigenvalue problem.

Given
$$m_1=3, m_2=2$$ $$k_1=10,k_2=20,k_3=15$$

Solution
Solving the FBD of both $$m_1 $$ and $$ m_2 $$ gives us the equations of motion. $$m= \begin{bmatrix} m_1 & 0\\ 0 & m_2\\ \end{bmatrix}= \begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix} $$ $$k= \begin{bmatrix} k_1-k_2 & k_2\\ k_2 & -(k_2+k_3)\\ \end{bmatrix}=\begin{bmatrix} -10 & 20\\ 20 & -35\\ \end{bmatrix}$$ Equation of Motion in Matrix Form. $$\begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix}\begin{bmatrix} \ddot{x_1}\\ \ddot{x_2}\\ \end{bmatrix}+\begin{bmatrix} -10 & 20\\ 20 & -35\\ \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ Now transforming the generalized eigenvalue problem into the standard eigenvalue problem.

$$kx={\lambda}mx$$.............................................. (1)  The $$ m, m^{-1/2},$$ and $$ m^{1/2}$$ matrix will be needed so they are solved and provided below. $$m=\begin{bmatrix} 3 & 0\\ 0 & 2\\ \end{bmatrix},m^{-1/2}=\begin{bmatrix} 1/\sqrt{3} & 0\\ 0 & 1/\sqrt{2}\\ \end{bmatrix},m^{1/2}=\begin{bmatrix} \sqrt{3} & 0\\ 0 & \sqrt{2}\\ \end{bmatrix}$$ Derivation of the new value $$x^*$$.First premultiply by $$ m^{-1/2}$$ $$m^{-1/2}kx={\lambda}m^{-1/2}mx={\lambda}m^{1/2}x$$ Let, $$x^*=m^{1/2}x$$ Then, $$ x=m^{-1/2}x^*$$ Plugging Into Eq. 1 above. $$(m^{-1/2}km^{1/2})x^*=k^*x^*={\lambda}x^*=0$$ Where, $$k^*=m^{-1/2}km^{1/2}=\begin{bmatrix} 3.333 & 8.165\\ 8.165 & -17.5\\ \end{bmatrix}$$ Now, $$(k^*-{\lambda}I)x^*=0$$............................................ (2)  First, Solving for the eigenvalues by setting the determinant of $$(k^*-{\lambda}I)$$equal to zero. $$det(k^*-{\lambda}I)=det(\begin{bmatrix} -3.333-{\lambda} & 8.165\\ 8.165 & -17.5-{\lambda}\\ \end{bmatrix})=0$$ $$(-3.33-{\lambda})(-17.5-{\lambda})=0$$ $${\lambda}^2 +20.83{\lambda}-8.392 = 0$$ Solving for $$ {\lambda}_1 $$ and $${\lambda}_2$$ using the quadratic. $${\lambda}_1=0.395$$ and $${\lambda}_2=-21.225$$ Plugging these values in one at a time into Eq.2 above will yield two respective eigenvectors. Solving for $$ x^*_1$$, $$\begin{bmatrix} 3.333-0.935 & 8.165\\ 8.165 & -17.50.395\\ \end{bmatrix}\begin{bmatrix} x^*_{11}\\ x^*_{12}\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ $$-3.725x^*_{11}+8.165x^*_{12}=0$$ $$8.165x^*_{11}-17.895x^*_{12}=0$$ So, $$x^*_1=\begin{bmatrix} 2.19\\ 1\\ \end{bmatrix}$$ Now solving for $$x^*_2$$, $$\begin{bmatrix} 3.333+21.225 & 8.165\\ 8.165 & -17.5+21.225\\ \end{bmatrix}\begin{bmatrix} x^*_{21}\\ x^*_{22}\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$ $$17.895x^*_{21}+8.165x^*_{22}=0$$ $$8.165x^*_{21}+3.725x^*_{22}=0$$ So, $$x^*_2=\begin{bmatrix} -0.456\\ 1\\ \end{bmatrix}$$

Final Solution
Using the transformation $$x=M^{-1/2}x^*$$, it is possible to obtain the eigenvectors $$x_1$$ and $$ x_2$$. $$ x_1=M^{-1/2}x^*_1=\begin{bmatrix} 1/\sqrt{3}& 0\\ 0& 1/\sqrt{2}\\ \end{bmatrix}\begin{bmatrix} 2.19\\ 1\\ \end{bmatrix}=\begin{bmatrix} 1.264\\ 0.707\\ \end{bmatrix}$$ $$ x_2=M^{-1/2}x^*_2=\begin{bmatrix} 1/\sqrt{3}& 0\\ 0& 1/\sqrt{2}\\ \end{bmatrix}\begin{bmatrix} -0.456\\ 1\\ \end{bmatrix}=\begin{bmatrix} -0.263\\ 0.707\\ \end{bmatrix}$$ So, $$x_1=\begin{bmatrix} 1.264\\ 0.707\\ \end{bmatrix}$$ $$x_2=\begin{bmatrix} -0.263\\ 0.707\\ \end{bmatrix}$$