By inspection\[ \left [ k\right ] =\begin{bmatrix} k_{1}+k_{2} & -k_{2}\\ -k_{2} & k_{2}+k_{3}\end{bmatrix} =\begin{bmatrix} 27 & -18\\ -18 & 36 \end{bmatrix} \] And \[ \left [ m\right ] =\begin{bmatrix} m_{1} & 0\\ 0 & m_{2}\end{bmatrix} =\begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix} \] The system is\begin{equation} \begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}\begin{Bmatrix} \ddot{x}_{1}\\ \ddot{x}_{2}\end{Bmatrix} +\begin{bmatrix} 27 & -18\\ -18 & 36 \end{bmatrix}\begin{Bmatrix} x_{1}\\ x_{2}\end{Bmatrix} =\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \tag{1} \end{equation} The above is solved using modal analysis in order to decouple the system. The first step is to determine the eigenvalues.\begin{align*} \left [ A\right ] & =\left [ m\right ] ^{-1}\left [ k\right ] \\ & =\frac{1}{2}\begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 27 & -18\\ -18 & 36 \end{bmatrix} \\ & =\begin{bmatrix} 1 & 0\\ 0 & \frac{1}{2}\end{bmatrix}\begin{bmatrix} 27 & -18\\ -18 & 36 \end{bmatrix} \\ & =\begin{bmatrix} 27 & -18\\ -9 & 18 \end{bmatrix} \end{align*}
To find the eigenvalues of \(\left [ A\right ] \) we solve \(\left \vert A-\lambda I\right \vert =0\) or\begin{align*} \begin{vmatrix} 27-\lambda & -18\\ -9 & 18-\lambda \end{vmatrix} & =0\\ \lambda ^{2}-45\lambda +324 & =0 \end{align*}
Hence\begin{align*} \lambda _{1} & =9\\ \lambda _{2} & =36 \end{align*}
Which implies\begin{align*} \omega _{n\left ( 1\right ) } & =3\text{ rad/s}\\ \omega _{n\left ( 2\right ) } & =9\text{ rad/s} \end{align*}
Now we find the eigenvectors \(u_{i}\) or the shape vectors. For \(\lambda _{1}=9\)\begin{align*} \left [ A\right ] \begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} & =\lambda _{1}\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} \\\begin{bmatrix} 27 & -18\\ -9 & 18 \end{bmatrix}\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} & =9\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} \\\begin{bmatrix} 27u_{1}-18u_{2}\\ -9u_{1}+18u_{2}\end{bmatrix} & =\begin{Bmatrix} 9u_{1}\\ 9u_{2}\end{Bmatrix} \end{align*}
Using first equation only gives\[ 27u_{1}-18u_{2}=9u_{1}\] We always normalized to \(u_{1}=1\), hence the above gives\begin{align*} 27-18u_{2} & =9\\ u_{2} & =1 \end{align*}
Therefore the first eigenvector is\[ \vec{u}_{1}=\begin{Bmatrix} 1\\ 1 \end{Bmatrix} \] To find the second eigenvector. For \(\lambda _{2}=36\)\begin{align*} \left [ A\right ] \begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} & =\lambda _{2}\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} \\\begin{bmatrix} 27 & -18\\ -9 & 18 \end{bmatrix}\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} & =36\begin{Bmatrix} u_{1}\\ u_{2}\end{Bmatrix} \\\begin{bmatrix} 27u_{1}-18u_{2}\\ -9u_{1}+18u_{2}\end{bmatrix} & =\begin{Bmatrix} 36u_{1}\\ 35u_{2}\end{Bmatrix} \end{align*}
Using first equation only gives\[ 27u_{1}-18u_{2}=36u_{1}\] We always normalized to \(u_{1}=1\), hence the above gives\begin{align*} 27-18u_{2} & =36\\ u_{2} & =-\frac{1}{2} \end{align*}
Therefore the second eigenvector is\[ \vec{u}_{2}=\begin{Bmatrix} 1\\ -\frac{1}{2}\end{Bmatrix} \] Hence the modal matrix is\[ \left [ u\right ] =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} \] Using the modal matrix, we can now decouple the original system given above in (1) which is\begin{equation} \left [ m\right ] \begin{Bmatrix} \ddot{x}_{1}\\ \ddot{x}_{2}\end{Bmatrix} +\left [ k\right ] \begin{Bmatrix} x_{1}\\ x_{2}\end{Bmatrix} =\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \tag{2} \end{equation} Let \(\begin{Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end{Bmatrix} =\left [ u\right ] \begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} \), then the above becomes\[ \left [ m\right ] \left [ u\right ] \begin{Bmatrix} \ddot{q}_{1}\left ( t\right ) \\ \ddot{q}_{2}\left ( t\right ) \end{Bmatrix} +\left [ k\right ] \left [ u\right ] \begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} =\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \] Premultiplying both sides by \(\left [ u\right ] ^{T}\) gives\begin{equation} \left [ u\right ] ^{T}\left [ m\right ] \left [ u\right ] \begin{Bmatrix} \ddot{q}_{1}\left ( t\right ) \\ \ddot{q}_{2}\left ( t\right ) \end{Bmatrix} +\left [ u\right ] ^{T}\left [ k\right ] \left [ u\right ] \begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} =\left [ u\right ] ^{T}\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \tag{4} \end{equation} But \[ \left [ u\right ] ^{T}\left [ m\right ] \left [ u\right ] =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} ^{T}\begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} =\begin{bmatrix} 3 & 0\\ 0 & \frac{3}{2}\end{bmatrix} \] And \[ \left [ u\right ] ^{T}\left [ k\right ] \left [ u\right ] =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} ^{T}\begin{bmatrix} 27 & -18\\ -18 & 36 \end{bmatrix}\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} =\begin{bmatrix} 27 & 0\\ 0 & 54 \end{bmatrix} \] Then (4) becomes \[\begin{bmatrix} 3 & 0\\ 0 & \frac{3}{2}\end{bmatrix}\begin{Bmatrix} \ddot{q}_{1}\left ( t\right ) \\ \ddot{q}_{2}\left ( t\right ) \end{Bmatrix} +\begin{bmatrix} 27 & 0\\ 0 & 54 \end{bmatrix}\begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} =\begin{Bmatrix} 3\sin 4t\\ 3\sin 4t \end{Bmatrix} \] Hence we obtain 2 ODEs\begin{align*} 3\ddot{q}_{1}\left ( t\right ) +27q_{1}\left ( t\right ) & =3\sin 4t\\ \frac{3}{2}\ddot{q}_{2}\left ( t\right ) +54q_{2}\left ( t\right ) & =3\sin 4t \end{align*}
Or\begin{align} \ddot{q}_{1}\left ( t\right ) +9q_{1}\left ( t\right ) & =\sin 4t\tag{5}\\ \ddot{q}_{2}\left ( t\right ) +36q_{2}\left ( t\right ) & =2\sin 4t\tag{6} \end{align}
Note There is a short cut to obtain the above (5,6) equations directly as follows. Starting with (2), we just write\begin{align*} \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\begin{Bmatrix} \ddot{q}_{1}\left ( t\right ) \\ \ddot{q}_{2}\left ( t\right ) \end{Bmatrix} +\begin{bmatrix} \omega _{n\left ( 1\right ) }^{2} & 0\\ 0 & \omega _{n\left ( 2\right ) }^{2}\end{bmatrix}\begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} & =\left [ u\right ] ^{-1}\left [ m\right ] ^{-1}\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \\\begin{Bmatrix} \ddot{q}_{1}\left ( t\right ) +9q_{1}\left ( t\right ) \\ \ddot{q}_{2}\left ( t\right ) ++36q_{2}\left ( t\right ) \end{Bmatrix} & =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} ^{-1}\begin{bmatrix} 1 & 0\\ 0 & 2 \end{bmatrix} ^{-1}\begin{Bmatrix} 3\sin 4t\\ 0 \end{Bmatrix} \\ & =\begin{Bmatrix} \sin 4t\\ 2\sin 4t \end{Bmatrix} \end{align*}
Which is the same as (5,6). This short cut just needs finding \(\left [ u\right ] ^{-1}\left [ m\right ] ^{-1}\). Use this short cut for the exam.
Solving (5)
The homogeneous solution is \[ q_{1,h}\left ( t\right ) =A_{1}\cos 3t+B_{1}\sin 3t \] And to find the particular solution, we guess \(q_{1,p}=C\sin 4t\), hence \(\dot{q}_{1,p}=4C\cos 4t\) and \(\ddot{q}_{1,p}=-16C\sin 4t\). Plug-in in (5) gives\begin{align*} -16C\sin 4t+9\left ( C\sin 4t\right ) & =\sin 4t\\ -7C_{1}\sin 4t & =\sin 4t\\ C_{1} & =-\frac{1}{7} \end{align*}
Hence \(q_{1,p}=-\frac{1}{7}\sin 4t\) and the complete solution is\[ q_{1}\left ( t\right ) =A_{1}\cos 3t+B_{1}\sin 3t-\frac{1}{7}\sin 4t \] Now we do the same to solve (6)
The homogeneous solution is \[ q_{2,h}\left ( t\right ) =A_{2}\cos 6t+B_{2}\sin 6t \] And to find the particular solution, we guess \(q_{2,p}=C\sin 4t\), hence \(\dot{q}_{2,p}=4C\cos 4t\) and \(\ddot{q}_{2,p}=-16C\sin 4t\). Plug-in in (6) gives\begin{align*} -16C\sin 4t+36\left ( C\sin 4t\right ) & =2\sin 4t\\ 20C_{1}\sin 4t & =2\sin 4t\\ C_{1} & =\frac{1}{10} \end{align*}
Hence \(q_{2,p}=\frac{1}{10}\sin 4t\) and the complete solution is\[ q_{2}\left ( t\right ) =A_{2}\cos 6t+B_{2}\sin 6t+\frac{1}{10}\sin 4t \] Therefore the solution in principle coordinates is\begin{align} q_{1}\left ( t\right ) & =A_{1}\cos 3t+B_{1}\sin 3t-\frac{1}{7}\sin 4t\tag{5A}\\ q_{2}\left ( t\right ) & =A_{2}\cos 6t+B_{2}\sin 6t+\frac{1}{10}\sin 4t \tag{6A} \end{align}
Since \(\left \{ x\right \} =\left [ u\right ] \left \{ q\right \} \), then \(\left \{ q\right \} =\left [ u\right ] ^{-1}\left \{ x\right \} \). Therefore\begin{align*} \left \{ q\left ( 0\right ) \right \} & =\left [ u\right ] ^{-1}\left \{ x\left ( 0\right ) \right \} \\\begin{Bmatrix} q_{1}\left ( 0\right ) \\ q_{2}\left ( 0\right ) \end{Bmatrix} & =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix} ^{-1}\begin{Bmatrix} x_{1}\left ( 0\right ) \\ x_{2}\left ( 0\right ) \end{Bmatrix} \\ & =\begin{bmatrix} \frac{1}{3} & \frac{2}{3}\\ \frac{2}{3} & -\frac{2}{3}\end{bmatrix}\begin{Bmatrix} 3\\ 0 \end{Bmatrix} \\ & =\begin{Bmatrix} 1\\ 2 \end{Bmatrix} \end{align*}
And\begin{align*} \left \{ \dot{q}\left ( 0\right ) \right \} & =\left [ u\right ] ^{-1}\left \{ \dot{x}\left ( 0\right ) \right \} \\\begin{Bmatrix} \dot{q}_{1}\left ( 0\right ) \\ \dot{q}_{2}\left ( 0\right ) \end{Bmatrix} & =\begin{bmatrix} \frac{1}{3} & \frac{2}{3}\\ \frac{2}{3} & -\frac{2}{3}\end{bmatrix}\begin{Bmatrix} 0\\ 9 \end{Bmatrix} \\ & =\begin{Bmatrix} 6\\ -6 \end{Bmatrix} \end{align*}
Applying first initial conditions to (5A,6A) gives\begin{align*} 1 & =A_{1}\\ 2 & =A_{2} \end{align*}
Hence (5A,6A) becomes\begin{align} q_{1}\left ( t\right ) & =\cos 3t+B_{1}\sin 3t-\frac{1}{7}\sin 4t\tag{5B}\\ q_{2}\left ( t\right ) & =2\cos 6t+B_{2}\sin 6t+\frac{1}{10}\sin 4t \tag{6B} \end{align}
Taking derivatives\begin{align*} \dot{q}_{1}\left ( t\right ) & =-3\sin 3t+3B_{1}\cos 3t-\frac{4}{7}\cos 4t\\ \dot{q}_{2}\left ( t\right ) & =-12\sin 6t+6B_{2}\cos 6t+\frac{4}{10}\cos 4t \end{align*}
Applying the second initial conditions to the above gives\begin{align*} 6 & =3B_{1}-\frac{4}{7}\\ -6 & =6B_{2}+\frac{4}{10} \end{align*}
Solving gives \(B_{1}=\frac{46}{21},B_{2}=-\frac{16}{15}\). Hence (5B,6B) become\begin{align} q_{1}\left ( t\right ) & =\cos 3t+\frac{46}{21}\sin 3t-\frac{1}{7}\sin 4t\tag{5C}\\ q_{2}\left ( t\right ) & =2\cos 6t-\frac{16}{15}\sin 6t+\frac{1}{10}\sin 4t \tag{6C} \end{align}
The above is the solution in principle coordinates. Now we transform it back to normal coordinates. Since \(\left \{ x\right \} =\left [ u\right ] \left \{ q\right \} \), then \begin{align} \begin{Bmatrix} x_{1}\left ( t\right ) \\ x_{2}\left ( t\right ) \end{Bmatrix} & =\begin{bmatrix} u_{11} & u_{12}\\ u_{21} & u_{22}\end{bmatrix}\begin{Bmatrix} q_{1}\left ( t\right ) \\ q_{2}\left ( t\right ) \end{Bmatrix} \nonumber \\ & =\begin{bmatrix} 1 & 1\\ 1 & -\frac{1}{2}\end{bmatrix}\begin{Bmatrix} \cos \left ( 3t\right ) +\left ( \frac{46}{21}\right ) \sin \left ( 3t\right ) -\left ( \frac{1}{7}\right ) \sin \left ( 4t\right ) \\ 2\cos \left ( 6t\right ) -\left ( \frac{16}{15}\right ) \sin \left ( 6t\right ) +\frac{1}{10}\sin 4t \end{Bmatrix} \nonumber \\ & =\begin{Bmatrix} \cos 3t+2\cos 6t+\frac{46}{21}\sin 3t-\frac{3}{70}\sin 4t-\frac{16}{15}\sin 6t\\ \cos 3t-\cos 6t+\frac{46}{21}\sin 3t-\frac{27}{140}\sin 4t+\frac{8}{15}\sin 6t \end{Bmatrix} \nonumber \end{align}
The above is the final solution. Here is a plot of \(x_{1}\left ( t\right ) ,x_{2}\left ( t\right ) \)
