Question

Answer the questions posed in Exercise 54 for the system.

$\left|\begin{array}{c}\frac{d\theta }{dt}=\omega \\ \frac{d\omega }{dt}=-p\theta -q\omega \end{array}\right|$

where $\mathit{p}$and $\mathit{q}$are positive and ${\mathit{q}}^{\mathbf{2}}\mathbf{>}\mathbf{4}\mathit{p}$.

Short answer

(a) The zero state is a stable equilibrium and the sketch of the system is in explanation.

(b) All trajectories converge to origin and it observe that for condition $\omega <0$door reaches $\theta <0$.

Step-by-step solution

(a) Obtain equation from the system.

Let the equations from the given determinant are as follows:

$$\begin{gathered} \frac{\mathbf{d}\mathbf{\theta }}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{\omega } \\ \frac{\mathbf{d}\mathbf{\omega }}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{p}\mathit{\theta }\mathbf{-}\mathit{q}\mathit{\omega } \end{gathered}$$

The corresponding matrix for the system is:

$A=\left[\begin{array}{cc}0& 1\\ -p& -q\end{array}\right]$

Calculate the Eigen values of given system and sketch of the graph.

Eigen values are given by the characteristic equation

$\begin{array}{rcl}\left|A-\lambda I\right|& =& 0\\ \left|\begin{array}{cc}-\lambda & 1\\ -p& -q-\lambda \end{array}\right|& =& 0\\ {\lambda }^2+q\lambda +p& =& 0\\ \lambda & =& \frac{-q\pm \sqrt{q^2-4q}}{2},q^2>4q\end{array}$

Thus, the Eigen values of are negative.

Hence, zero state is a stable equilibrium.

The phase portrait is as follows:

(b) Movement of door for different trajectory.

Let $E_{{\lambda }_1}$, $E_{{\lambda }_2}$are Eigen spaces of ${\lambda }_1$and role="math" localid="1665036517865" ${\lambda }_2$. Since both eigenvalues are negative.

Hence, the solution is:

$\overrightarrow{x}\left(t\right)=C_1{e}^{{\lambda }_{1}t}\overrightarrow{v_1}+C_2{e}^{{\lambda }_{2}t}\overrightarrow{v_2}$

Here, the values are $\overrightarrow{x}=\left[\begin{array}{c}\theta \\ \omega \end{array}\right]$, role="math" localid="1665035976107" $\overrightarrow{v_1}$corresponding to ${\lambda }_1=\frac{-q-\sqrt{q^2-4q}}{2}$and $\overrightarrow{v_2}$ corresponding to ${\lambda }_2=\frac{-q+\sqrt{q^2-4q}}{2}$.

As $t\to \infty ,\overrightarrow{x}\left(t\right)\to 0$solution converges to origin.

Also, $C_1$and$C_2$ are arbitrary constants.

Dominant terms is $C_2{e}^{{\lambda }_{2}t}\overrightarrow{v_2}$as ${\lambda }_2$is larger in magnitude, therefore distant trajectories are parallel to $E_{{\lambda }_2}$.

All trajectories converge to origin.Along the trajectory 1, the movement of the door slow down continuously until it reaches to the closed state as along this trajectory w decreases continuously.

Door slams if the trajectory 2 is followed that is initially if :

$\frac{\omega \left(0\right)}{\theta \left(0\right)}<{\lambda }_2=\frac{-q\sqrt{q^2+4q}}{2}$

From figure it observe that for condition $\omega <0$door reaches$\theta <0$ .

Hence, all trajectories converge to origin and it observe that for condition $\omega <0$door reaches$\theta <0$ .