Referring to the coupled mass-spring system discussed in Example , suppose an external force$\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{37}\mathbf{cos}\mathbf{3}\mathbf{t}$ is applied to the second object of mass $\mathbf{1}\mathbf{}\mathbf{kg}$. The displacement functions $\mathbf{x}\left(\mathbf{t}\right)$and $\mathbf{y}\left(\mathbf{t}\right)$now satisfy the system

$$\begin{gathered} \mathbf{\left(}\mathbf{16}\mathbf{\right)}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,} \\ \mathbf{\left(}\mathbf{17}\mathbf{\right)}\mathbf{(}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{(}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{37}\mathbf{cos}\mathbf{3}\mathbf{t} \end{gathered}$$

(a) Show that $\mathbf{x}\left(\mathbf{t}\right)$satisfies the equation $\mathbf{\left(}\mathbf{18}\mathbf{\right)}{\mathbf{x}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{37}\mathbf{cos}\mathbf{3}\mathbf{t}$

(b) Find a general solution $\mathbf{x}\left(\mathbf{t}\right)$ to the equation (18). [Hint: Use undetermined coefficients with ${\mathbf{x}}_{\mathbf{p}}\mathbf{=}\mathbf{Acos}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{3}\mathbf{t}$.]

(c) Substitute$\mathbf{x}\left(\mathbf{t}\right)$ back into (16) to obtain a formula for $\mathbf{y}\left(\mathbf{t}\right)$.

(d) If both masses are displaced2mto the right of their equilibrium positions and then released, find the displacement functions $\mathbf{x}\left(\mathbf{t}\right)$ and $\mathbf{y}\left(\mathbf{t}\right)$.

Let's rewrite the given system in operator form:

$$\begin{gathered} \left(2{\mathrm{D}}^2+6\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right]=0 \\ -2\left[\mathrm{x}\right]+\left({\mathrm{D}}^2+2\right)\left[\mathrm{y}\right]=37\cos 3\mathrm{t} \end{gathered}$$

One wants to obtain the equation for x so we will eliminate y from the system. To do so one will multiply the first equation by $\left({\mathrm{D}}^2+2\right)$ and the second by 2 and then add them together:

$\begin{array}{rcl}& & \left({\mathrm{D}}^2+6\right)\left({\mathrm{D}}^2+2\right)\left[\mathrm{x}\right]-2\left({\mathrm{D}}^2+2\right)\left[\mathrm{y}\right]=0\\ & & -4\left[\mathrm{x}\right]+2\left({\mathrm{D}}^2+2\right)\left[\mathrm{y}\right]=2\times 37\cos 3\mathrm{t}\\ & & \left(\left(2{\mathrm{D}}^2+6\right)\left({\mathrm{D}}^2+2\right)-4\right)\left[\mathrm{x}\right]=2\times 37\cos 3\mathrm{t}\end{array}$

$$\begin{gathered} \left(2{\mathrm{D}}^4+4{\mathrm{D}}^2+6{\mathrm{D}}^2+12-4\right)\left[\mathrm{x}\right]=2\times 37\cos 3\mathrm{t} \\ 2\left({\mathrm{D}}^4+5{\mathrm{D}}^2+4\right)\left[\mathrm{x}\right]=2\times 37\cos 3\mathrm{t} \\ \left({\mathrm{D}}^4+5{\mathrm{D}}^2+4\right)\left[\mathrm{x}\right]=37\cos 3\mathrm{t} \end{gathered}$$

Rewriting this back in standard form we have that $\mathrm{x}\left(\mathrm{t}\right)$satisfies

${\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)+5\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)=37\cos 3\mathrm{t}$.

To find a general solution, one will first find a homogeneous solution for $\mathrm{x}\left(\mathrm{t}\right)$.

The auxiliary equation is:

${\mathrm{r}}^4+5{\mathrm{r}}^2+4=0$, and its roots are:

$$\begin{gathered} {\mathrm{r}}^4+5{\mathrm{r}}^2+4=\left({\mathrm{r}}^2+1\right)\left({\mathrm{r}}^2+4\right)=0 \\ {\mathrm{r}}^2=-1,{\mathrm{r}}^2=-4 \\ {\mathrm{r}}_{1,2}=\pm \mathrm{i},{\mathrm{r}}_{3,4}=\pm 2\mathrm{i} \end{gathered}$$

Therefore, the homogeneous solution for x is:

${\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos 2\mathrm{t}+{\mathrm{c}}_4\sin 2\mathrm{t}$

One will find a particular solution by using the method of undetermined coefficients and assume that a particular solution has a form of ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}3\mathrm{t}+\mathrm{Bsin}3\mathrm{t}$.

One needs the second and the fourth derivative of${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$:

$$\begin{gathered} \mathrm{x}\text{'}=-3\mathrm{Asin}3\mathrm{t}+3\mathrm{Bcos}3\mathrm{t},\mathrm{x}\text{'}\text{'}=-9\mathrm{Acos}3\mathrm{t}-9\mathrm{Bsin}3\mathrm{t} \\ \mathrm{x}\text{'}\text{'}\text{'}=27\mathrm{Asin}3\mathrm{t}-27\mathrm{Bcos}3\mathrm{t},{\mathrm{x}}^{\left(4\right)}=81\mathrm{Acos}3\mathrm{t}+81\mathrm{Bsin}3\mathrm{t} \end{gathered}$$

Now, one has that,

$$\begin{gathered} {\mathrm{x}}_{\mathrm{p}}^{\left(4\right)}{\left(\mathrm{t}\right)+5\mathrm{x}}_{\mathrm{p}}^{}\text{'}\text{'}\left(\mathrm{t}\right)+4{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=81\mathrm{Acos}3\mathrm{t}+81\mathrm{Bsin}3\mathrm{t}+5(-9\mathrm{Acos}3\mathrm{t}-9\mathrm{Bsin}3\mathrm{t})+4(\mathrm{Acos}3\mathrm{t}+\mathrm{Bsin}3\mathrm{t}) \\ =40\mathrm{Acos}3\mathrm{t}+40\mathrm{Bsin}3\mathrm{t}=37\cos 3\mathrm{t} \\ \Rightarrow 40\mathrm{A}=37,40\mathrm{B}=0 \\ \iff \mathrm{A}=\frac{37}{40},\mathrm{B}=0 \end{gathered}$$

So, the particular solution for x is ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=37/40\cos 3\mathrm{t}$ and the general solution for x is $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{h}}+{\mathrm{x}}_{\mathrm{p}}={\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos 2\mathrm{t}+{\mathrm{c}}_4\sin 2\mathrm{t}+\frac{37}{40}\cos 3\mathrm{t}$.

The first equation of the given system gives us that $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)$

So, one need to find the second derivative of x and substitute it into the previous equation.

$\begin{array}{rcl}& & \mathrm{x}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{sint}+{\mathrm{c}}_2\mathrm{cost}-2{\mathrm{c}}_3\sin 2\mathrm{t}+2{\mathrm{c}}_4\cos 2\mathrm{t}-3\times \frac{37}{40}\sin 3\mathrm{t}\backslash hfill\\ & & \mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{cost}-{\mathrm{c}}_2\mathrm{sint}-4{\mathrm{c}}_3\cos 2\mathrm{t}-4{\mathrm{c}}_4\sin 2\mathrm{t}-9\times \frac{37}{40}\cos 3\mathrm{t}\end{array}$

Substituting this into $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)$

One has,

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)=-{\mathrm{c}}_1\mathrm{cost}-{\mathrm{c}}_2\mathrm{sint}-4{\mathrm{c}}_3\cos 2\mathrm{t}-4{\mathrm{c}}_4\sin 2\mathrm{t}-9\times \frac{37}{40}\cos 3\mathrm{t} \\ +3\left({\mathrm{c}}_1\mathrm{cost}+{\mathrm{c}}_2\mathrm{sint}+{\mathrm{c}}_3\cos 2\mathrm{t}+{\mathrm{c}}_4\sin 2\mathrm{t}+\frac{37}{40}\cos 3\mathrm{t}\right) \\ \mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{c}}_1\mathrm{cost}+2{\mathrm{c}}_2\mathrm{sint}-{\mathrm{c}}_3\cos 2\mathrm{t}-{\mathrm{c}}_4\sin 2\mathrm{t}-3\times \frac{37}{20}\cos 3\mathrm{t} \end{gathered}$$

Both masses are displaced 2m to the right of their equilibrium position, so $\mathrm{x}\left(0\right)=\mathrm{y}\left(0\right)=2$, and since they are released, one has that$\mathrm{x}\text{'}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0$.

First, one will find the first derivative of y:

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-2{\mathrm{c}}_2\mathrm{sint}+2{\mathrm{c}}_2\mathrm{cost}+2{\mathrm{c}}_3\sin 2\mathrm{t}-2{\mathrm{c}}_4\cos 2\mathrm{t}-9\times \frac{37}{40}\sin 3\mathrm{t}$

So, from this initial condition,one has that

$\begin{array}{rcl}& & \mathrm{x}\left(0\right)={\mathrm{c}}_1\times 1+{\mathrm{c}}_2\times 0+{\mathrm{c}}_3\times 1+{\mathrm{c}}_4\times 0+\frac{37}{40}\times 1\\ & =& {\mathrm{c}}_1+{\mathrm{c}}_3+\frac{37}{40}\\ & =& 2\\ & & \mathrm{y}\left(0\right)=2{\mathrm{c}}_1\times 1+2{\mathrm{c}}_2\times 0-{\mathrm{c}}_3\times 1-{\mathrm{c}}_4\times 0-3\times \frac{37}{20}\times 1\\ & =& 2{\mathrm{c}}_1-{\mathrm{c}}_3-6\times \frac{37}{40}\\ & =& 2\end{array}$

Adding the previous two equations together one will get

$\begin{array}{rcl}3{\mathrm{c}}_1-5\times \frac{37}{40}& =& 4\\ {\mathrm{c}}_1& =& \frac{23}{8}\end{array}$

Substituting this into the second equation one will have${\mathrm{c}}_2=2{\mathrm{c}}_1-6\times \frac{37}{40}-2\iff {\mathrm{c}}_3=-\frac{9}{5}$

The third and the fourth initial condition give us

$$\begin{gathered} \mathrm{x}\text{'}\left(0\right)=-{\mathrm{c}}_1\times 0+{\mathrm{c}}_2\times 1-2{\mathrm{c}}_3\times 0+2{\mathrm{c}}_4\times 1-3\times \frac{37}{40}\times 0 \\ ={\mathrm{c}}_2+2{\mathrm{c}}_4 \\ =0 \\ \mathrm{y}\text{'}\left(0\right)=-2{\mathrm{c}}_1\times 0+2{\mathrm{c}}_2\times 1+2{\mathrm{c}}_3\times 0-2{\mathrm{c}}_4\times 1-9\times \frac{37}{40}\times 0 \\ =2{\mathrm{c}}_2-2{\mathrm{c}}_4 \\ =0 \end{gathered}$$

Adding those equations together one gets that $3{\mathrm{c}}_2=0\Rightarrow {\mathrm{c}}_2=0$

Substituting this into the first equation one has that $2{\mathrm{c}}_4=0\Rightarrow {\mathrm{c}}_4=0$

Finally, one has that the displacement functions x and y is:

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=\frac{23}{8}\mathrm{cost}-\frac{9}{5}\cos 2\mathrm{t}+\frac{37}{40}\cos 3\mathrm{t} \\ \mathrm{y}\left(\mathrm{t}\right)=\frac{23}{4}\mathrm{cost}+\frac{9}{5}\cos 2\mathrm{t}-3\times \frac{37}{40}\cos 3\mathrm{t} \end{gathered}$$