SOLUTION:
The motion in each normal mode is de-coupled from each other mode. Each motion is a simple harmonic motion in terms of normal coordinates, and reduces to second order differential equation of the form\begin{equation} \ddot{\eta }_{i}+\omega _{i}^{2}\eta _{i}=0 \tag{1} \end{equation} Where i ranges over the number of modes. The number of modes is equal to the number of independent degrees of freedoms in the system. Each mode oscillates at frequency \omega _{i}. Since this is a simple harmonic motion, its energy is given by\begin{equation} E_{i}=\frac{1}{2}m_{i}\dot{\eta }_{i}^{2}+\frac{1}{2}k_{i}\eta _{i}^{2} \tag{2} \end{equation} Where k_{i} is the effective stiffness of the mode and \omega _{i}^{2}=\frac{k_{i}}{m_{i}}. Therefore k_{i}=m_{i}\omega _{i}^{2}.
To show that E is conserved, we need to show that \frac{\partial E}{\partial t}=0. Hence from (2) \frac{\partial E_{i}}{\partial t}=m_{i}\dot{\eta }_{i}\ddot{\eta }_{i}+\left ( m_{i}\omega _{i}^{2}\right ) \eta _{i}\dot{\eta }_{i} But from (1) we see that \ddot{\eta }_{i}=-\omega _{i}^{2}\eta _{i}. Substituting into the above gives\begin{align*} \frac{\partial E_{i}}{\partial t} & =m_{i}\dot{\eta }_{i}\left ( -\omega _{i}^{2}\eta _{i}\right ) +\left ( m_{i}\omega _{i}^{2}\right ) \eta _{i}\dot{\eta }_{i}\\ & =0 \end{align*}
Therefore energy in each mode is constant.
SOLUTION:
Kinetic energy is T=\frac{1}{2}M\dot{z}^{2}+\frac{1}{2}I_{1}\dot{\theta }_{1}^{2}+\frac{1}{2}I_{2}\dot{\theta }_{2}^{2} Where I_{1} is moment of inertia of plate around axis y, and I_{2} is moment of inertia of plate around axis x. These are (from tables) :\begin{align*} I_{1} & =\frac{1}{12}MW^{2}\\ I_{2} & =\frac{1}{12}ML^{2} \end{align*}
The potential energy is\begin{align*} U & =4\left ( \frac{1}{2}Kz^{2}\right ) +4\left ( \frac{1}{2}K\left ( \frac{W}{2}\theta _{1}\right ) ^{2}\right ) +4\left ( \frac{1}{2}K\left ( \frac{L}{2}\theta _{2}\right ) ^{2}\right ) \\ & =2Kz^{2}+2K\left ( \frac{W}{2}\theta _{1}\right ) ^{2}+2K\left ( \frac{L}{2}\theta _{2}\right ) ^{2}\\ & =2Kz^{2}+\frac{1}{2}KW^{2}\theta _{1}^{2}+\frac{1}{2}KL^{2}\theta _{2}^{2} \end{align*}
Where small angle approximation is used in the above. Hence the Lagrangian is\begin{align*} L & =T-U\\ & =\frac{1}{2}M\dot{z}^{2}+\frac{1}{2}I_{1}\dot{\theta }_{1}^{2}+\frac{1}{2}I_{2}\dot{\theta }_{2}^{2}-2Kz^{2}-\frac{1}{2}KW^{2}\theta _{1}^{2}-\frac{1}{2}KL^{2}\theta _{2}^{2} \end{align*}
Equation of motion for z\begin{align*} \frac{\partial L}{\partial z} & =-4Kz\\ \frac{\partial L}{\partial \dot{z}} & =M\dot{z} \end{align*}
Hence M\ddot{z}+4Kz=0 Equation of motion for \theta _{1}\begin{align*} \frac{\partial L}{\partial \theta _{1}} & =-KW^{2}\theta _{1}\\ \frac{\partial L}{\partial \dot{\theta }_{1}} & =I_{1}\dot{\theta }_{1} \end{align*}
Hence I_{1}\ddot{\theta }_{1}+KW^{2}\theta _{1}=0 Similarly, we find I_{2}\ddot{\theta }_{2}+KL^{2}\theta _{2}=0 Therefore \begin{align*} \left [ M\right ] \mathbf{\ddot{q}}+\left [ K\right ] \mathbf{q} & =0\\\begin{pmatrix} M & 0 & 0\\ 0 & I_{1} & 0\\ 0 & 0 & I_{1}\end{pmatrix}\begin{pmatrix} \ddot{z}\\ \ddot{\theta }_{1}\\ \ddot{\theta }_{2}\end{pmatrix} +\begin{pmatrix} 4K & 0 & 0\\ 0 & KW^{2} & 0\\ 0 & 0 & KL^{2}\end{pmatrix}\begin{pmatrix} z\\ \theta _{1}\\ \theta _{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Which leads to \begin{align*} \det \begin{pmatrix} 4K-M\omega ^{2} & 0 & 0\\ 0 & KW^{2}-I_{1}\omega ^{2} & 0\\ 0 & 0 & KL^{2}-I_{2}\omega ^{2}\end{pmatrix} & =0\\ 4K^{3}L^{2}W^{2}-MK^{2}L^{2}\omega ^{2}W^{2}-4I_{1}K^{2}L^{2}\omega ^{2}-4I_{2}K^{2}\omega ^{2}W^{2}+MI_{1}KL^{2}\omega ^{4}+MI_{2}K\omega ^{4}W^{2}+4I_{1}I_{2}K\omega ^{4}-MI_{1}I_{2}\omega ^{6} & =0\\ \left ( KL^{2}-\omega ^{2}I_{2}\right ) \left ( KW^{2}-\omega ^{2}I_{1}\right ) \left ( M\omega ^{2}-4K\right ) & =0 \end{align*}
Therefore \begin{align*} \omega _{1} & =\sqrt{\frac{KL^{2}}{I_{2}}}\\ \omega _{2} & =\sqrt{\frac{KW^{2}}{I_{1}}}\\ \omega _{3} & =\sqrt{\frac{4K}{M}} \end{align*}
Using I_{1}=\frac{1}{12}MW^{2},I_{2}=\frac{1}{12}ML^{2}, the above become\begin{align*} \omega _{1} & =\sqrt{12\frac{KL^{2}}{ML^{2}}}=2\sqrt{3\frac{K}{M}}\\ \omega _{2} & =\sqrt{12\frac{KW^{2}}{MW^{2}}}=2\sqrt{3\frac{K}{M}}\\ \omega _{3} & =\sqrt{\frac{4K}{M}}=2\sqrt{\frac{K}{M}} \end{align*}
Hence \frac{\omega _{1}}{\omega _{2}}=\frac{1}{1},\frac{\omega _{1}}{\omega _{3}}=\sqrt{3},\frac{\omega _{2}}{\omega _{3}}=\sqrt{3}. Therefore \omega _{1}:\omega _{2}:\omega _{3}=1:1:\sqrt{3} Or \omega _{1}:\omega _{2}:\omega _{3}=\frac{1}{\sqrt{3}}:\frac{1}{\sqrt{3}}:1
SOLUTION:
Kinetic energy is\begin{align*} T & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \left ( \dot{x}+l\dot{\theta }\cos \theta \right ) ^{2}+\left ( l\dot{\theta }\sin \theta \right ) ^{2}\right ) \\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+l^{2}\dot{\theta }^{2}\cos ^{2}\theta +2\dot{x}l\dot{\theta }\cos \theta +l^{2}\dot{\theta }^{2}\sin ^{2}\theta \right ) \\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+2\dot{x}l\dot{\theta }\cos \theta +l^{2}\dot{\theta }^{2}\right ) \end{align*}
And potential energy is U=-mgl\cos \theta Hence the Lagrangian \begin{align*} L & =T-U\\ & =\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m\left ( \dot{x}^{2}+2\dot{x}l\dot{\theta }\cos \theta +l^{2}\dot{\theta }^{2}\right ) +mgl\cos \theta \end{align*}
Now we find equations of motions. For \theta \begin{align*} \frac{\partial L}{\partial \theta } & =-m\dot{x}l\dot{\theta }\sin \theta -mgl\sin \theta \\ \frac{\partial L}{\partial \dot{\theta }} & =\frac{1}{2}m\left ( 2\dot{x}l\cos \theta +2l^{2}\dot{\theta }\right ) \\ & =m\left ( \dot{x}l\cos \theta +l^{2}\dot{\theta }\right ) \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }} & =m\left ( \ddot{x}l\cos \theta -\dot{x}l\dot{\theta }\sin \theta +l^{2}\ddot{\theta }\right ) \end{align*}
Hence
\begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =0\nonumber \\ m\left ( \ddot{x}l\cos \theta -\dot{x}l\dot{\theta }\sin \theta +l^{2}\ddot{\theta }\right ) +m\dot{x}l\dot{\theta }\sin \theta +mgl\sin \theta & =0\nonumber \\ m\ddot{x}l\cos \theta +ml^{2}\ddot{\theta }+mgl\sin \theta & =0 \tag{1} \end{align}
Now we find equation of motion for x\begin{align*} \frac{\partial L}{\partial x} & =0\\ \frac{\partial L}{\partial \dot{x}} & =M\dot{x}+m\left ( \dot{x}+l\dot{\theta }\cos \theta \right ) \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & =M\ddot{x}+m\left ( \ddot{x}+l\ddot{\theta }\cos \theta -l\dot{\theta }^{2}\sin \theta \right ) \end{align*}
Hence\begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} & =0\nonumber \\ M\ddot{x}+m\left ( \ddot{x}+l\ddot{\theta }\cos \theta -l\dot{\theta }^{2}\sin \theta \right ) & =0\nonumber \\ \ddot{x}\left ( M+m\right ) +ml\ddot{\theta }\cos \theta -ml\dot{\theta }^{2}\sin \theta & =0 \tag{2} \end{align}
Now we can write them in matrix form \left [ M\right ] \mathbf{\ddot{q}}+\left [ K\right ] \mathbf{q}=0, from (1) and (2) we obtain, after using small angle approximation \cos \theta \approx 1,\sin \theta \approx \theta and also \dot{\theta }^{2}\approx 0\begin{pmatrix} M+m & ml\\ ml & ml^{2}\end{pmatrix}\begin{pmatrix} \ddot{x}\\ \ddot{\theta }\end{pmatrix} +\begin{pmatrix} 0 & 0\\ 0 & mgl \end{pmatrix}\begin{pmatrix} x\\ \theta \end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} Now assuming solution is \mathbf{q}\left ( t\right ) =\mathbf{a}e^{i\omega t}, then the above can be rewritten as\begin{equation} \begin{pmatrix} -\omega ^{2}\left ( M+m\right ) & -\omega ^{2}ml\\ -\omega ^{2}ml & mgl-ml^{2}\omega ^{2}\end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} \tag{1} \end{equation} These have non-trivial solution when\begin{align*} \det \begin{pmatrix} -\omega ^{2}\left ( M+m\right ) & -\omega ^{2}ml\\ -\omega ^{2}ml & mgl-ml^{2}\omega ^{2}\end{pmatrix} & =0\\ Ml^{2}m\omega ^{4}-glm^{2}\omega ^{2}-Mglm\omega ^{2} & =0\\ \omega ^{2}\left ( Ml^{2}m\omega ^{2}-glm^{2}-Mglm\right ) & =0 \end{align*}
Hence \omega =0 is one eigenvalue and \omega =\sqrt{\frac{g}{l}\frac{m+M}{M}} is another.\begin{align*} & \fbox{$\omega _1=0$}\\ & \fbox{$\omega _2=\sqrt{\frac{g}{l}\frac{\left ( M+m\right ) }{M}}$} \end{align*}
Now that we found \omega _{i} we go back to (1) to find corresponding eigenvectors. For \omega _{1}, (1) becomes\begin{pmatrix} 0 & 0\\ 0 & mgl \end{pmatrix}\begin{pmatrix} a_{11}\\ a_{21}\end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} Hence from the second equation above 0a_{11}+mgla_{21}=0 So a_{11} can be any value, and a_{21}=0. So the following is a valid first eigenvector \mathbf{a}_{1}=\begin{pmatrix} a_{11}\\ 0 \end{pmatrix} For \omega _{2} (1) becomes\begin{pmatrix} -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \left ( M+m\right ) & -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) ml\\ -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) ml & mgl-ml^{2}\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \end{pmatrix}\begin{pmatrix} a_{12}\\ a_{22}\end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} From first equation we find\begin{align*} -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \left ( M+m\right ) a_{12}-\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) mla_{22} & =0\\ \left ( M+m\right ) a_{12}+mla_{22} & =0 \end{align*}
Hence a_{12}=-\frac{ml}{\left ( M+m\right ) }a_{22}. So the following is a valid second eigenvector \mathbf{a}_{2}=\begin{pmatrix} -\frac{ml}{\left ( M+m\right ) }a_{22}\\ a_{22}\end{pmatrix} Therefore\begin{align*} x & =a_{11}\eta _{1}+a_{12}\eta _{2}\\ \theta & =a_{21}\eta _{1}+a_{22}\eta _{2} \end{align*}
Where \eta _{i} are the normal coordinates. Using relation found earlier, then\begin{align} x & =a_{11}\eta _{1}\tag{2}\\ \theta & =-\frac{ml}{\left ( M+m\right ) }a_{22}\eta _{1}+a_{22}\eta _{2} \tag{3} \end{align}
Hence from (2) \eta _{1}=-\frac{x}{a_{11}} And now (3) can be written as \theta =-\frac{ml}{\left ( M+m\right ) }a_{22}\frac{x}{a_{11}}+a_{22}\eta _{2} Therefore \eta _{2}=\frac{\theta }{a_{22}}+\frac{1}{a_{11}}\frac{mlx}{\left ( M+m\right ) } To sketch the mode shapes. Looking at \mathbf{a}_{1}=\begin{pmatrix} a_{11}\\ 0 \end{pmatrix} and \mathbf{a}_{2}=\begin{pmatrix} -\frac{ml}{\left ( M+m\right ) }a_{22}\\ a_{22}\end{pmatrix} and normalizing we can write
\begin{pmatrix} 1\\ 0 \end{pmatrix} ,\begin{pmatrix} -\frac{ml}{\left ( M+m\right ) }\\ 1 \end{pmatrix} So in the first mode shape, the mass M moves with the pendulum fixed to it in the same orientation all the time. So the whole system just slides along x with \theta =0 all the time. In the second mode, x move by \frac{-ml}{\left ( M+m\right ) } factor to \theta motion. For example, for M\ll m\,, then mode 2 is \begin{pmatrix} -l\\ 1 \end{pmatrix} , hence antisymmetric mode. If M=m then we get \begin{pmatrix} -\frac{l}{2}\\ 1 \end{pmatrix} antisymmetric, but now the ratio changes. So the second mode shape is antisymmetric, but the ratio depends on the ratio of m to M.
This is extra and can be ignored if needed. I was not sure if we should use s=l\theta as the generalized coordinate instead of \theta in order to make all the coordinates of same units. So this is repeat of the above, but using s=l\theta transformation. Starting with equations of motion\begin{align*} \ddot{x}\left ( M+m\right ) +ml\ddot{\theta }\cos \theta -ml\dot{\theta }^{2}\sin \theta & =0\\ m\ddot{\theta }+m\ddot{x}\frac{\cos \theta }{l}+m\frac{g}{l}\sin \theta & =0 \end{align*}
Will now use s=l\theta transformation, and use s as the second degree of freedom, which is the small distance the pendulum mass swings by. This is so that both x and s has same units of length to make it easier to work with the shape functions. Hence the equations of motions become\begin{align*} \ddot{x}\left ( M+m\right ) +ml\frac{\ddot{s}}{l}\cos \left ( \frac{s}{l}\right ) -ml\frac{\dot{s}^{2}}{l^{2}}\sin \left ( \frac{s}{l}\right ) & =0\\ m\frac{\ddot{s}}{l}+m\ddot{x}\frac{\cos \left ( \frac{s}{l}\right ) }{l}+m\frac{g}{l}\sin \left ( \frac{s}{l}\right ) & =0 \end{align*}
We first apply small angle approximation, which implies \cos \frac{s}{l}\rightarrow 1,\sin \left ( \frac{s}{l}\right ) \rightarrow \frac{s}{l} and also \frac{\dot{s}^{2}}{l^{2}}\rightarrow 0, therefore the equations of motions becomes\begin{align*} \ddot{x}\left ( M+m\right ) +m\ddot{s} & =0\\ m\frac{\ddot{s}}{l}+m\ddot{x}\frac{1}{l}+m\frac{g}{l}\frac{s}{l} & =0 \end{align*}
And now we write the matrix form\begin{pmatrix} M+m & m\\ m & m \end{pmatrix}\begin{pmatrix} \ddot{x}\\ \ddot{s}\end{pmatrix} +\begin{pmatrix} 0 & 0\\ 0 & m\frac{g}{l}\end{pmatrix}\begin{pmatrix} x\\ s \end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} Now assuming solution is \mathbf{q}\left ( t\right ) =\mathbf{a}e^{i\omega t}, then the above can be rewritten as\begin{equation} \begin{pmatrix} -\omega ^{2}\left ( M+m\right ) & -\omega ^{2}m\\ -\omega ^{2}m & m\frac{g}{l}-m\omega ^{2}\end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} \tag{1} \end{equation} These have non-trivial solution when\begin{align*} \det \begin{pmatrix} -\omega ^{2}\left ( M+m\right ) & -\omega ^{2}m\\ -\omega ^{2}m & m\frac{g}{l}-m\omega ^{2}\end{pmatrix} & =0\\ -\frac{1}{l}\left ( gm^{2}\omega ^{2}-Mlm\omega ^{4}+Mgm\omega ^{2}\right ) & =0\\ \omega ^{2}\left ( \frac{gm^{2}}{l}-Mm\omega ^{2}+M\frac{g}{l}m\right ) & =0\\ \omega ^{2}\left ( M\omega ^{2}-\left ( \frac{g}{l}\left ( m+M\right ) \right ) \right ) & =0 \end{align*}
Hence \omega =0 is one eigenvalue and \omega =\sqrt{\frac{g}{l}\frac{\left ( M+m\right ) }{M}} is another.\begin{align*} & \fbox{$\omega _1=0$}\\ & \fbox{$\omega _2=\sqrt{\frac{g}{l}\frac{\left ( M+m\right ) }{M}}$} \end{align*}
Now that we found \omega _{i} we go back to (1) to find corresponding eigenvectors. For \omega _{1}, (1) becomes\begin{align*} \begin{pmatrix} 0 & 0\\ 0 & m\frac{g}{l}\end{pmatrix}\begin{pmatrix} a_{11}\\ a_{21}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\ 0a_{11}+m\frac{g}{l}a_{21} & =0 \end{align*}
Hence from the second equation above 0a_{11}+m\frac{g}{l}a_{21}=0 So a_{11} can be any value, and a_{21}=0. So the following is a valid first eigenvector \mathbf{a}_{1}=\begin{pmatrix} a_{11}\\ 0 \end{pmatrix} For \omega _{2} (1) becomes\begin{pmatrix} -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \left ( M+m\right ) & -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) m\\ -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) m & m\frac{g}{l}-m\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \end{pmatrix}\begin{pmatrix} a_{12}\\ a_{22}\end{pmatrix} =\begin{pmatrix} 0\\ 0 \end{pmatrix} From first equation we find\begin{align*} -\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) \left ( M+m\right ) a_{12}-\left ( \frac{g}{l}\frac{\left ( M+m\right ) }{M}\right ) ma_{22} & =0\\ \left ( M+m\right ) a_{12}+ma_{22} & =0 \end{align*}
Hence a_{12}=-\frac{m}{\left ( M+m\right ) }a_{22}. So the following is a valid second eigenvector \mathbf{a}_{2}=\begin{pmatrix} -\frac{m}{\left ( M+m\right ) }a_{22}\\ a_{22}\end{pmatrix} Therefore\begin{align*} x & =a_{11}\eta _{1}+a_{12}\eta _{2}\\ \theta & =a_{12}\eta _{1}+a_{22}\eta _{2} \end{align*}
Where \eta _{i} are the normal coordinates. Using relation found earlier, then\begin{align} x & =a_{11}\eta _{1}\tag{2}\\ \theta & =-\frac{m}{\left ( M+m\right ) }a_{22}\eta _{1}+a_{22}\eta _{2}\tag{3} \end{align}
Hence from (2) \eta _{1}=-\frac{x}{a_{11}} And now (3) can be written as \theta =-\frac{m}{\left ( M+m\right ) }a_{22}\frac{x}{a_{11}}+a_{22}\eta _{2} Therefore \eta _{2}=\frac{\theta }{a_{22}}+\frac{mx}{\left ( M+m\right ) }\frac{1}{a_{11}} To sketch the mode shapes. Looking at \mathbf{a}_{1}=\begin{pmatrix} a_{11}\\ 0 \end{pmatrix} and \mathbf{a}_{2}=\begin{pmatrix} -\frac{m}{\left ( M+m\right ) }a_{22}\\ a_{22}\end{pmatrix} and normalizing we can write
\begin{pmatrix} 1\\ 0 \end{pmatrix} ,\begin{pmatrix} -\frac{m}{\left ( M+m\right ) }\\ 1 \end{pmatrix} So in the first mode shape, the mass M moves with the pendulum fixed to it in the same orientation all the time. So the whole system just slides along x with \theta =0 all the time. In the second mode, x move by \frac{-m}{\left ( M+m\right ) } factor to \theta motion. For example, for M\ll m\,, then mode 2 is \begin{pmatrix} -1\\ 1 \end{pmatrix} , hence antisymmetric mode. If M=m then we get \begin{pmatrix} -\frac{1}{2}\\ 1 \end{pmatrix} antisymmetric, but now the ratio changes. So the second mode shape is antisymmetric, but the ratio depends on the ratio of m to M.
SOLUTION:
Kinetic energy T=\frac{1}{2}m\dot{x}_{1}^{2}+\frac{1}{2}M\dot{x}_{2}^{2}+\frac{1}{2}m\dot{x}_{3}^{2} Potential energy U=\frac{1}{2}k\left ( x_{2}-x_{1}\right ) ^{2}+\frac{1}{2}k\left ( x_{3}-x_{2}\right ) ^{2} Hence the Lagrangian \begin{align*} L & =T-U\\ & =\frac{1}{2}m\dot{x}_{1}^{2}+\frac{1}{2}M\dot{x}_{2}^{2}+\frac{1}{2}m\dot{x}_{3}^{2}-\frac{1}{2}k\left ( x_{2}-x_{1}\right ) ^{2}-\frac{1}{2}k\left ( x_{3}-x_{2}\right ) ^{2} \end{align*}
EQM for x_{1}\begin{align*} \frac{\partial L}{\partial x_{1}} & =k\left ( x_{2}-x_{1}\right ) \\ \frac{\partial L}{\partial \dot{x}_{1}} & =m\dot{x}_{1}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{1}} & =m\ddot{x}_{1} \end{align*}
Therefore\begin{align} m\ddot{x}_{1}-k\left ( x_{2}-x_{1}\right ) & =0\nonumber \\ m\ddot{x}_{1}+kx_{1}-kx_{2} & =0\tag{1} \end{align}
EQM for x_{2}\begin{align*} \frac{\partial L}{\partial x_{2}} & =-k\left ( x_{2}-x_{1}\right ) +k\left ( x_{3}-x_{2}\right ) \\ \frac{\partial L}{\partial \dot{x}_{2}} & =M\dot{x}_{2}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{2}} & =M\ddot{x}_{2} \end{align*}
Therefore\begin{align} M\ddot{x}_{2}+k\left ( x_{2}-x_{1}\right ) -k\left ( x_{3}-x_{2}\right ) & =0\nonumber \\ M\ddot{x}_{2}+kx_{2}-kx_{1}-kx_{3}+kx_{2} & =0\nonumber \\ M\ddot{x}_{2}+2kx_{2}-kx_{1}-kx_{3} & =0\tag{2} \end{align}
EQM for x_{3}\begin{align*} \frac{\partial L}{\partial x_{3}} & =-k\left ( x_{3}-x_{2}\right ) \\ \frac{\partial L}{\partial \dot{x}_{3}} & =m\dot{x}_{3}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{3}} & =m\ddot{x}_{3} \end{align*}
Therefore\begin{align} m\ddot{x}_{3}+k\left ( x_{3}-x_{2}\right ) & =0\nonumber \\ m\ddot{x}_{3}+kx_{3}-kx_{2} & =0\tag{3} \end{align}
Now we can write equations (1,2,3) in matrix form \left [ M\right ] \mathbf{\ddot{q}}+\left [ K\right ] \mathbf{q}=0 to obtain\begin{pmatrix} m & 0 & 0\\ 0 & M & 0\\ 0 & 0 & m \end{pmatrix}\begin{pmatrix} \ddot{x}_{1}\\ \ddot{x}_{2}\\ \ddot{x}_{3}\end{pmatrix} +\begin{pmatrix} k & -k & 0\\ -k & 2k & -k\\ 0 & -k & k \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} Now assuming solution is \mathbf{q}\left ( t\right ) =\mathbf{a}e^{i\omega t}, then the above can be rewritten as\begin{equation} \begin{pmatrix} k-m\omega ^{2} & -k & 0\\ -k & 2k-M\omega ^{2} & -k\\ 0 & -k & k-m\omega ^{2}\end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \tag{4} \end{equation} These have non-trivial solution when\begin{align*} \det \begin{pmatrix} k-m\omega ^{2} & -k & 0\\ -k & 2k-M\omega ^{2} & -k\\ 0 & -k & k-m\omega ^{2}\end{pmatrix} & =0\\ \omega ^{2}\left ( k-m\omega ^{2}\right ) \left ( -Mm\omega ^{2}+Mk+2km\right ) & =0 \end{align*}
Hence we have 3 normal frequencies. One of them is zero.\begin{align*} \omega _{1} & =0\\ \omega _{2} & =\sqrt{\frac{k}{m}}\\ \omega _{3} & =\sqrt{k\frac{M+2m}{Mm}} \end{align*}
For each normal frequency, there is a corresponding eigen shape vector. Now we find these eigen shapes. For \omega _{1}, and from (4)\begin{pmatrix} k & -k & 0\\ -k & 2k & -k\\ 0 & -k & k \end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} Hence \begin{align*} ka_{1}-ka_{2}+0a_{3} & =0\\ -ka_{1}+2ka_{2}-ka_{3} & =0\\ 0a_{1}-ka_{2}+ka_{3} & =0 \end{align*}
Or\begin{align*} a_{1}-a_{2} & =0\\ -a_{1}+2a_{2}-a_{3} & =0\\ -a_{2}+a_{3} & =0 \end{align*}
Hence a_{1}=a_{2} and a_{2}=a_{3}. So \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} is first eigenvector. Now we find the second one for \omega _{2}. From (4) and using \omega =\sqrt{\frac{k}{m}}\begin{align*} \begin{pmatrix} k-m\frac{k}{m} & -k & 0\\ -k & 2k-M\frac{k}{m} & -k\\ 0 & -k & k-m\frac{k}{m}\end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 0 & -k & 0\\ -k & 2k-M\frac{k}{m} & -k\\ 0 & -k & 0 \end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Hence \begin{align*} -ka_{2} & =0\\ -ka_{1}+\left ( 2k-M\frac{k}{m}\right ) a_{2}-ka_{3} & =0\\ -ka_{2} & =0 \end{align*}
Or\begin{align*} a_{2} & =0\\ -a_{1}+a_{2}\left ( 2-\frac{M}{m}\right ) -a_{3} & =0\\ a_{2} & =0 \end{align*}
hence a_{2}=0 and a_{1}=-a_{3}. So \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix} is second eigenvector. Now we find the third one for \omega _{3}. From (4) and using \omega =\sqrt{k\frac{M+2m}{Mm}}\begin{align*} \begin{pmatrix} k-m\left ( k\frac{M+2m}{Mm}\right ) & -k & 0\\ -k & 2k-M\left ( k\frac{M+2m}{Mm}\right ) & -k\\ 0 & -k & k-m\left ( k\frac{M+2m}{Mm}\right ) \end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} k-k\frac{M+2m}{M} & -k & 0\\ -k & 2k-k\frac{M+2m}{m} & -k\\ 0 & -k & k-k\frac{M+2m}{M}\end{pmatrix}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Hence \begin{align*} k\left ( 1-\frac{M+2m}{M}\right ) a_{1}-ka_{2} & =0\\ -ka_{1}+k\left ( 2-\frac{M+2m}{m}\right ) a_{2}-ka_{3} & =0\\ -ka_{2}+k\left ( 1-\frac{M+2m}{M}\right ) a_{3} & =0 \end{align*}
Or\begin{align*} \left ( 1-\frac{M+2m}{M}\right ) a_{1}-a_{2} & =0\\ -a_{1}+\left ( 2-\frac{M+2m}{m}\right ) a_{2}-a_{3} & =0\\ -a_{2}+\left ( 1-\frac{M+2m}{M}\right ) a_{3} & =0 \end{align*}
Solution is: a_{1}=a_{3},a_{2}=-\frac{2}{M}ma_{3} So \begin{pmatrix} 1\\ -\frac{2m}{M}\\ 1 \end{pmatrix} is third eigevector. To sketch the mode shapes, will use the following diagram
