Loading web-font TeX/Main/Regular

3.2 Solving slide 390 example

PDF (letter size)

PDF (legal size)

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 $x_{1}\left ( t\right ) ,x_{2}\left ( t\right )$