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Q5.6-8E

Expert-verified
Found in: Page 288

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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A double pendulum swinging in a vertical plane under the influence of gravity (see Figure 5.35) satisfies the system$\left({\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\right){{\mathbf{l}}}_{{\mathbf{1}}}^{{\mathbf{2}}}{{\mathbf{\theta }}}_{{\mathbf{1}}}^{}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{l}}}_{{\mathbf{1}}}{{\mathbf{l}}}_{{\mathbf{2}}}{{\mathbf{\theta }}}_{{\mathbf{2}}}^{}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}\left({\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\right){{\mathbf{l}}}_{{\mathbf{1}}}{{\mathbf{g\theta }}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{0}}\phantom{\rule{0ex}{0ex}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{l}}}_{{\mathbf{2}}}^{{\mathbf{2}}}{{\mathbf{\theta }}}_{{\mathbf{2}}}^{}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{l}}}_{{\mathbf{1}}}{{\mathbf{l}}}_{{\mathbf{2}}}{{\mathbf{\theta }}}_{{\mathbf{1}}}^{}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{{\mathbf{l}}}_{{\mathbf{2}}}{{\mathbf{g\theta }}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}$ When ${{\mathbit{\theta }}}_{{\mathbf{1}}}$ and ${{\mathbit{\theta }}}_{{\mathbf{2}}}$ are small angles. Solve the system when ${{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{kg}}{\mathbf{,}}{{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{kg}}{\mathbf{,}}{{\mathbf{l}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{l}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{m}}{\mathbf{,}}{{\mathbf{\theta }}}_{{\mathbf{1}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\pi }\mathbf{/}\mathbf{6}\mathbf{,}{{\mathbf{\theta }}}_{{\mathbf{2}}}{\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\theta }}_{{\mathbf{1}}}^{}{\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\theta }}_{{\mathbf{2}}}^{}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

The solutions for ${\mathrm{\theta }}_{1}$ and ${\mathrm{\theta }}_{2}$ are:

${\mathrm{\theta }}_{1}\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\mathrm{cos}1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\mathrm{cos}2.31\mathrm{t}\phantom{\rule{0ex}{0ex}}{\mathrm{\theta }}_{2}\left(\mathrm{t}\right)=\frac{5}{12}\mathrm{cos}1.1\mathrm{t}-\frac{5}{12}\mathrm{cos}2.31\mathrm{t}$

See the step by step solution

Substituting the values of m1,m2,l1,l2

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:

$\left(3+2\right){5}^{2}{\mathrm{\theta }}_{1}^{}{\text{'}\text{'}+2×5×5\mathrm{\theta }}_{2}^{}\text{'}\text{'}+\left(3+2\right)5×9.8{\mathrm{\theta }}_{1}=0\phantom{\rule{0ex}{0ex}}2×{5}^{2}{\mathrm{\theta }}_{2}^{}{\text{'}\text{'}+2×5×5\mathrm{\theta }}_{1}^{}\text{'}\text{'}+2×5×9.8{\mathrm{\theta }}_{2}=0$

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

${5\mathrm{\theta }}_{1}^{}{\text{'}\text{'}+2\mathrm{\theta }}_{2}^{}\text{'}\text{'}+9.8{\mathrm{\theta }}_{1}=0\phantom{\rule{0ex}{0ex}}{5\mathrm{\theta }}_{2}^{}{\text{'}\text{'}+5\mathrm{\theta }}_{1}^{}\text{'}\text{'}+9.8{\mathrm{\theta }}_{2}=0$

One will rewrite this system in operator form:

$\left(5{\mathrm{D}}^{2}+9.8\right)\left[{\mathrm{\theta }}_{1}\right]+2{\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{2}\right]=0\phantom{\rule{0ex}{0ex}}5{\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{1}\right]+\left(5{\mathrm{D}}^{2}+9.8\right)\left[{\mathrm{\theta }}_{2}\right]=0$

Finding  θ1

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 $-2{D}^{2}$ 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:

${\mathrm{r}}^{2}=\frac{-98±\sqrt{9604-5762.4}}{30}=\frac{-98±61.98}{30}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1,2}^{2}=-1.2,{\mathrm{r}}_{3,4}^{2}=-5.33\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1,2}=±\sqrt{1.2}\mathrm{i},{\mathrm{r}}_{3,4}=±\sqrt{5.33}\mathrm{i}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1,2}=±1.1\mathrm{i},{\mathrm{r}}_{3,4}=±2.31\mathrm{i}$

So, the general solution for ${\theta }_{1}$ is ${\mathrm{\theta }}_{1}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}+{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}$

Findingθ2

Now, one can find a solution for ${\mathrm{\theta }}_{2}$ from the first equation of the system

$\begin{array}{rcl}2{\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{2}\right]& =& -\left(5{\mathrm{D}}^{2}+9.8\right)\left[{\mathrm{\theta }}_{1}\right]\\ \mathrm{D}\left[{\mathrm{\theta }}_{1}\right]& =& -1.1{\mathrm{c}}_{1}\mathrm{sin}1.1\mathrm{t}+1.1{\mathrm{c}}_{2}\mathrm{cos}1.1\mathrm{t}-2.31{\mathrm{c}}_{3}\mathrm{sin}2.31\mathrm{t}+2.31{\mathrm{c}}_{4}\mathrm{cos}2.31\mathrm{t}\\ {\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{1}\right]& =& -1.2{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}-1.2{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}-5.33{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}-5.33{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}2\\ {\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{2}\right]& =& -5{\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{1}\right]-9.8{\mathrm{\theta }}_{1}\\ & =& 6{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}+6{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}+26.65{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}+26.65{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}-9.8{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}\\ & & -9.8{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}-9.8{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}-9.8{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}\\ & =& -3.8{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}-3.8{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}+16.85{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}+16.85{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}\\ {\mathrm{D}}^{2}\left[{\mathrm{\theta }}_{2}\right]& =& -1.9{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}-1.9{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}+8.42{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}+8.42{\mathrm{c}}_{4}\mathrm{sin}2\end{array}$

Integration

Integrating the previous equation twice one will get:

$\begin{array}{rcl}\mathrm{D}\left[{\mathrm{\theta }}_{2}\right]& =& \int \left(-1.9{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}-1.9{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}+8.42{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}+8.42{\mathrm{c}}_{4}\mathrm{sin}2.31\mathrm{t}\right)\mathrm{dt}\\ & =& -1.9{\mathrm{c}}_{1}\int \mathrm{cos}1.1\mathrm{tdt}-1.9{\mathrm{c}}_{2}\int \mathrm{sin}1.1\mathrm{tdt}+8.42{\mathrm{c}}_{3}\int \mathrm{cos}2.31\mathrm{tdt}+8.42{\mathrm{c}}_{4}\int \mathrm{sin}2.31\mathrm{tdt}\\ & =& -1.73{\mathrm{c}}_{1}\mathrm{sin}1.1\mathrm{t}+1.73{\mathrm{c}}_{2}\mathrm{cos}1.1\mathrm{t}+3.65{\mathrm{c}}_{3}\mathrm{sin}2.31\mathrm{t}-3.65{\mathrm{c}}_{4}\mathrm{cos}2.31\mathrm{t}+\mathrm{A}\\ {\mathrm{\theta }}_{2}& =& \int \left(-1.73{\mathrm{c}}_{1}\mathrm{sin}1.1\mathrm{t}+1.73{\mathrm{c}}_{2}\mathrm{cos}1.1\mathrm{t}+3.65{\mathrm{c}}_{3}\mathrm{sin}2.31\mathrm{t}-3.65{\mathrm{c}}_{4}\mathrm{cos}2.31\mathrm{t}+\mathrm{A}\right)\mathrm{dt}\\ & =& -1.73{\mathrm{c}}_{1}\int \mathrm{sin}1.1\mathrm{tdt}+1.73{\mathrm{c}}_{2}\int \mathrm{cos}1.1\mathrm{tdt}+3.65{\mathrm{c}}_{3}\int \mathrm{sin}2.31\mathrm{tdt}-3.65{\mathrm{c}}_{4}\int \mathrm{cos}2.31\mathrm{tdt}\\ & & +\mathrm{A}\int \mathrm{dt}\\ & =& 1.57{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}+1.57{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}-1.57{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}-1.57\mathrm{sin}2.31\mathrm{t}+\mathrm{At}+\mathrm{B}\end{array}$

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 θ1 and  θ2

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 ${\mathrm{\theta }}_{2}\left(\mathrm{t}\right)=1.57{\mathrm{c}}_{1}\mathrm{cos}1.1\mathrm{t}+1.57{\mathrm{c}}_{2}\mathrm{sin}1.1\mathrm{t}-1.57{\mathrm{c}}_{3}\mathrm{cos}2.31\mathrm{t}-1.57\mathrm{sin}2.31\mathrm{t}$

The initial conditions give us:

${\mathrm{\theta }}_{1}\left(0\right)={\mathrm{c}}_{1}+{\mathrm{c}}_{3}=\frac{\mathrm{\pi }}{6},{\mathrm{\theta }}_{2}\left(0\right)=1.57{\mathrm{c}}_{1}-1.57{\mathrm{c}}_{3}=0\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{1}={\mathrm{c}}_{3},2{\mathrm{c}}_{1}=\frac{\mathrm{\pi }}{6},{\mathrm{c}}_{1}={\mathrm{c}}_{3}=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}{\mathrm{\theta }}_{1}^{}\text{'}\left(0\right)=1.1{\mathrm{c}}_{2}+2.31{\mathrm{c}}_{4}=0,{\mathrm{\theta }}_{2}^{}\text{'}\left(0\right)=1.73{\mathrm{c}}_{2}-3.65{\mathrm{c}}_{4}=0\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{2}=2.11{\mathrm{c}}_{4},4.62{\mathrm{c}}_{4}=0,{\mathrm{c}}_{2}={\mathrm{c}}_{4}=0$

Finally, the solutions for ${\mathrm{\theta }}_{1}$ and ${\mathrm{\theta }}_{2}$ are

$\begin{array}{rcl}& & {\mathrm{\theta }}_{1}\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\mathrm{cos}1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\mathrm{cos}2.31\mathrm{t}\\ & & {\mathrm{\theta }}_{2}\left(\mathrm{t}\right)=1.57\frac{\mathrm{\pi }}{12}\mathrm{cos}1.1\mathrm{t}-1.57\frac{\mathrm{\pi }}{12}\mathrm{cos}2.31\mathrm{t}\\ & & {\mathrm{\theta }}_{1}\left(\mathrm{t}\right)=\frac{\mathrm{\pi }}{12}\mathrm{cos}1.1\mathrm{t}+\frac{\mathrm{\pi }}{12}\mathrm{cos}2.31\mathrm{t}\\ & & {\mathrm{\theta }}_{2}\left(\mathrm{t}\right)=\frac{5}{12}\mathrm{cos}1.1\mathrm{t}-\frac{5}{12}\mathrm{cos}2.31\mathrm{t}\end{array}$

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