Q5.6-8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A double pendulum swinging in a vertical plane under the influence of gravity (see Figure 5.35) satisfies the system

 

$$\begin{gathered} \left({\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\right){\mathbf{l}}_{\mathbf{1}}^{\mathbf{2}}{\mathbf{\theta }}_{\mathbf{1}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}{\mathbf{\theta }}_{\mathbf{2}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left({\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\right){\mathbf{l}}_{\mathbf{1}}{\mathbf{g\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{0} \\ {\mathbf{m}}_{\mathbf{2}}{\mathbf{l}}_{\mathbf{2}}^{\mathbf{2}}{\mathbf{\theta }}_{\mathbf{2}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{l}}_{\mathbf{1}}{\mathbf{l}}_{\mathbf{2}}{\mathbf{\theta }}_{\mathbf{1}}^{}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{l}}_{\mathbf{2}}{\mathbf{g\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$ 

When  ${\mathit{\theta }}_{\mathbf{1}}$ and ${\mathit{\theta }}_{\mathbf{2}}$ are small angles. Solve the system when 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{kg}\mathbf{,}{\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{kg}\mathbf{,}{\mathbf{l}}_{\mathbf{1}}\mathbf{=}{\mathbf{l}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{m}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\pi }\mathbf{/}\mathbf{6}\mathbf{,}{\mathbf{\theta }}_{\mathbf{2}}{\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\theta }}_{\mathbf{1}}^{}{\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\theta }}_{\mathbf{2}}^{}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

The solutions for ${\mathrm{\theta }}_1$  and  ${\mathrm{\theta }}_2$ are:

$$\begin{gathered} {\mathrm{\theta }}_1\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\cos 1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\cos 2.31\mathrm{t} \\ {\mathrm{\theta }}_2\left(\mathrm{t}\right)=\frac{5}{12}\cos 1.1\mathrm{t}-\frac{5}{12}\cos 2.31\mathrm{t} \end{gathered}$$

Substituting the values of m 1 , m 2 , l 1 , l 2

Substituting the given values for ${\mathrm{m}}_1,{\mathrm{m}}_2,{\mathrm{l}}_1,{\mathrm{l}}_2$  into the given system and taking $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$ one will get:

 

$$\begin{gathered} (3+2)5^2{\mathrm{\theta }}_1^{}{\text{'}\text{'}+2\times 5\times 5\mathrm{\theta }}_2^{}\text{'}\text{'}+(3+2)5\times 9.8{\mathrm{\theta }}_1=0 \\ 2\times 5^2{\mathrm{\theta }}_2^{}{\text{'}\text{'}+2\times 5\times 5\mathrm{\theta }}_1^{}\text{'}\text{'}+2\times 5\times 9.8{\mathrm{\theta }}_2=0 \end{gathered}$$

 

Dividing the first equation by  25 and the second by 10 one will get:

 

$$\begin{gathered} {5\mathrm{\theta }}_1^{}{\text{'}\text{'}+2\mathrm{\theta }}_2^{}\text{'}\text{'}+9.8{\mathrm{\theta }}_1=0 \\ {5\mathrm{\theta }}_2^{}{\text{'}\text{'}+5\mathrm{\theta }}_1^{}\text{'}\text{'}+9.8{\mathrm{\theta }}_2=0 \end{gathered}$$ 

 

One will rewrite this system in operator form:

 

$$\begin{gathered} \left(5{\mathrm{D}}^2+9.8\right)\left[{\mathrm{\theta }}_1\right]+2{\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]=0 \\ 5{\mathrm{D}}^2\left[{\mathrm{\theta }}_1\right]+\left(5{\mathrm{D}}^2+9.8\right)\left[{\mathrm{\theta }}_2\right]=0 \end{gathered}$$

Finding θ 1

One will eliminate ${\mathrm{\theta }}_2$  from the system. To do so one will multiply the first equation by $\left({\mathrm{D}}^2+9.8\right)$  and the second by  $-2D^2$ and then add those two equations together.

 

$\begin{array}{rcl}{\left(5{\mathrm{D}}^2+9.8\right)}^2\left[{\mathrm{\theta }}_1\right]+2{\mathrm{D}}^2\left(5{\mathrm{D}}^2+9.8\right)\left[{\mathrm{\theta }}_2\right]& =& 0\\ -10{\mathrm{D}}^4-2{\mathrm{D}}^2\left(5{\mathrm{D}}^2+9.8\right)\left[{\mathrm{\theta }}_2\right]& =& 0\\ \left(25{\mathrm{D}}^4+98{\mathrm{D}}^2+96.04-10{\mathrm{D}}^4\right)\left[{\mathrm{\theta }}_1\right]& =& 0\\ \left(15{\mathrm{D}}^4+98{\mathrm{D}}^2+96.04\right)\left[{\mathrm{\theta }}_1\right]& =& 0\end{array}$ 

Now, one can find a general solution for ${\theta }_1$. The auxiliary equation is:

 

$15{\mathrm{r}}^4+98{\mathrm{r}}^2+96.04=0$ , and its roots are:

 

$$\begin{gathered} {\mathrm{r}}^2=\frac{-98\pm \sqrt{9604-5762.4}}{30}=\frac{-98\pm 61.98}{30} \\ {\mathrm{r}}_{1,2}^2=-1.2, {\mathrm{r}}_{3,4}^2=-5.33 \\ {\mathrm{r}}_{1,2}=\pm \sqrt{1.2}\mathrm{i}, {\mathrm{r}}_{3,4}=\pm \sqrt{5.33}\mathrm{i} \\ {\mathrm{r}}_{1,2}=\pm 1.1\mathrm{i}, {\mathrm{r}}_{3,4}=\pm 2.31\mathrm{i} \end{gathered}$$ 

 

So, the general solution for ${\theta }_1$  is ${\mathrm{\theta }}_1\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 1.1\mathrm{t}+{\mathrm{c}}_2\sin 1.1\mathrm{t}+{\mathrm{c}}_3\cos 2.31\mathrm{t}+{\mathrm{c}}_4\sin 2.31\mathrm{t}$

Finding θ 2

Now, one can find a solution for ${\mathrm{\theta }}_2$ from the first equation of the system

$\begin{array}{rcl}2{\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]& =& -\left(5{\mathrm{D}}^2+9.8\right)\left[{\mathrm{\theta }}_1\right]\\ \mathrm{D}\left[{\mathrm{\theta }}_1\right]& =& -1.1{\mathrm{c}}_1\sin 1.1\mathrm{t}+1.1{\mathrm{c}}_2\cos 1.1\mathrm{t}-2.31{\mathrm{c}}_3\sin 2.31\mathrm{t}+2.31{\mathrm{c}}_4\cos 2.31\mathrm{t}\\ {\mathrm{D}}^2\left[{\mathrm{\theta }}_1\right]& =& -1.2{\mathrm{c}}_1\cos 1.1\mathrm{t}-1.2{\mathrm{c}}_2\sin 1.1\mathrm{t}-5.33{\mathrm{c}}_3\cos 2.31\mathrm{t}-5.33{\mathrm{c}}_4\sin 2.31\mathrm{t}2\\ {\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]& =& -5{\mathrm{D}}^2\left[{\mathrm{\theta }}_1\right]-9.8{\mathrm{\theta }}_1\\ & =& 6{\mathrm{c}}_1\cos 1.1\mathrm{t}+6{\mathrm{c}}_2\sin 1.1\mathrm{t}+26.65{\mathrm{c}}_3\cos 2.31\mathrm{t}+26.65{\mathrm{c}}_4\sin 2.31\mathrm{t}-9.8{\mathrm{c}}_1\cos 1.1\mathrm{t}\\ & & -9.8{\mathrm{c}}_2\sin 1.1\mathrm{t}-9.8{\mathrm{c}}_3\cos 2.31\mathrm{t}-9.8{\mathrm{c}}_4\sin 2.31\mathrm{t}\\ & =& -3.8{\mathrm{c}}_1\cos 1.1\mathrm{t}-3.8{\mathrm{c}}_2\sin 1.1\mathrm{t}+16.85{\mathrm{c}}_3\cos 2.31\mathrm{t}+16.85{\mathrm{c}}_4\sin 2.31\mathrm{t}\\ {\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]& =& -1.9{\mathrm{c}}_1\cos 1.1\mathrm{t}-1.9{\mathrm{c}}_2\sin 1.1\mathrm{t}+8.42{\mathrm{c}}_3\cos 2.31\mathrm{t}+8.42{\mathrm{c}}_4\sin 2\end{array}$

Integration

Integrating the previous equation twice one will get:

$\begin{array}{rcl}\mathrm{D}\left[{\mathrm{\theta }}_2\right]& =& \int \left(-1.9{\mathrm{c}}_1\cos 1.1\mathrm{t}-1.9{\mathrm{c}}_2\sin 1.1\mathrm{t}+8.42{\mathrm{c}}_3\cos 2.31\mathrm{t}+8.42{\mathrm{c}}_4\sin 2.31\mathrm{t}\right)\mathrm{dt}\\ & =& -1.9{\mathrm{c}}_1\int \cos 1.1\mathrm{tdt}-1.9{\mathrm{c}}_2\int \sin 1.1\mathrm{tdt}+8.42{\mathrm{c}}_3\int \cos 2.31\mathrm{tdt}+8.42{\mathrm{c}}_4\int \sin 2.31\mathrm{tdt}\\ & =& -1.73{\mathrm{c}}_1\sin 1.1\mathrm{t}+1.73{\mathrm{c}}_2\cos 1.1\mathrm{t}+3.65{\mathrm{c}}_3\sin 2.31\mathrm{t}-3.65{\mathrm{c}}_4\cos 2.31\mathrm{t}+\mathrm{A}\\ {\mathrm{\theta }}_2& =& \int \left(-1.73{\mathrm{c}}_1\sin 1.1\mathrm{t}+1.73{\mathrm{c}}_2\cos 1.1\mathrm{t}+3.65{\mathrm{c}}_3\sin 2.31\mathrm{t}-3.65{\mathrm{c}}_4\cos 2.31\mathrm{t}+\mathrm{A}\right)\mathrm{dt}\\ & =& -1.73{\mathrm{c}}_1\int \sin 1.1\mathrm{tdt}+1.73{\mathrm{c}}_2\int \cos 1.1\mathrm{tdt}+3.65{\mathrm{c}}_3\int \sin 2.31\mathrm{tdt}-3.65{\mathrm{c}}_4\int \cos 2.31\mathrm{tdt}\\ & & +\mathrm{A}\int \mathrm{dt}\\ & =& 1.57{\mathrm{c}}_1\cos 1.1\mathrm{t}+1.57{\mathrm{c}}_2\sin 1.1\mathrm{t}-1.57{\mathrm{c}}_3\cos 2.31\mathrm{t}-1.57\sin 2.31\mathrm{t}+\mathrm{At}+\mathrm{B}\end{array}$

But this equation must satisfy the second equation of the system, which is:

 $5{\mathrm{D}}^2\left[{\mathrm{\theta }}_1\right]+5{\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]+9.8\left[{\mathrm{\theta }}_2\right]=0$

Finding the solution for θ 1 and θ 2

Since one does not have any constant nor a term multiplying  t in  ${\mathrm{D}}^2\left[{\mathrm{\theta }}_1\right]$ and ${\mathrm{D}}^2\left[{\mathrm{\theta }}_2\right]$, one cannot have it in ${\mathrm{\theta }}_1$, and therefore $\mathrm{A}=\mathrm{B}=0$, so the solution for   ${\theta }_2$is ${\mathrm{\theta }}_2\left(\mathrm{t}\right)=1.57{\mathrm{c}}_1\cos 1.1\mathrm{t}+1.57{\mathrm{c}}_2\sin 1.1\mathrm{t}-1.57{\mathrm{c}}_3\cos 2.31\mathrm{t}-1.57\sin 2.31\mathrm{t}$

 

The initial conditions give us:

 

$$\begin{gathered} {\mathrm{\theta }}_1\left(0\right)={\mathrm{c}}_1+{\mathrm{c}}_3=\frac{\mathrm{\pi }}{6}, {\mathrm{\theta }}_2\left(0\right)=1.57{\mathrm{c}}_1-1.57{\mathrm{c}}_3=0 \\ {\mathrm{c}}_1={\mathrm{c}}_3,2{\mathrm{c}}_1=\frac{\mathrm{\pi }}{6},{\mathrm{c}}_1={\mathrm{c}}_3=\frac{\mathrm{\pi }}{12} \\ {\mathrm{\theta }}_1^{}\text{'}\left(0\right)=1.1{\mathrm{c}}_2+2.31{\mathrm{c}}_4=0, {\mathrm{\theta }}_2^{}\text{'}\left(0\right)=1.73{\mathrm{c}}_2-3.65{\mathrm{c}}_4=0 \\ {\mathrm{c}}_2=2.11{\mathrm{c}}_4,4.62{\mathrm{c}}_4=0,{\mathrm{c}}_2={\mathrm{c}}_4=0 \end{gathered}$$ 

Finally, the solutions for ${\mathrm{\theta }}_1$  and ${\mathrm{\theta }}_2$  are

$\begin{array}{rcl}& & {\mathrm{\theta }}_1\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\cos 1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\cos 2.31\mathrm{t}\\ & & {\mathrm{\theta }}_2\left(\mathrm{t}\right)=1.57\frac{\mathrm{\pi }}{12}\cos 1.1\mathrm{t}-1.57\frac{\mathrm{\pi }}{12}\cos 2.31\mathrm{t}\\ & & {\mathrm{\theta }}_1\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\cos 1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\cos 2.31\mathrm{t}\\ & & {\mathrm{\theta }}_2\left(\mathrm{t}\right)=\frac{5}{12}\cos 1.1\mathrm{t}-\frac{5}{12}\cos 2.31\mathrm{t}\end{array}$