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
