 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q5.3-30E

Expert-verified Found in: Page 1 ### 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 # Spring Pendulum. Let a mass be attached to one end of a spring with spring constant k and the other end attached to the ceiling. Let ${{\mathbf{l}}}_{{\mathbf{o}}}$ be the natural length of the spring, and let l(t) be its length at time t. If ${\mathbf{\theta }}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}$ is the angle between the pendulum and the vertical, then the motion of the spring pendulum is governed by the system $\mathbf{l}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{gcos\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{k}}{\mathbf{m}}\mathbf{\left(}\mathbf{l}\mathbf{-}{{\mathbf{l}}}_{{\mathbf{o}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\phantom{\rule{0ex}{0ex}}{{\mathbf{l}}}^{{\mathbf{2}}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{gl}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{sin\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$Assume g = 1, k = m = 1, and ${{\mathbf{l}}}_{{\mathbf{o}}}$= 4. When the system is at rest, ${\mathbf{l}}{\mathbf{=}}{{\mathbf{l}}}_{{\mathbf{o}}}{\mathbf{+}}\frac{\mathbf{mg}}{\mathbf{k}}{\mathbf{=}}{\mathbf{5}}$.a. Describe the motion of the pendulum when $\mathbf{l}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.b. When the pendulum is both stretched and given an angular displacement, the motion of the pendulum is more complicated. Using the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution, sketch the graphs of the length l and the angular displacement u on the interval [0,10] if $\mathbf{l}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

a. The pendulum would just move up and down in a periodic way.

b. See the table.

See the step by step solution

## Transform the equation

Here, the equation is:

$\mathrm{l}\text{'}\text{'}\left(\mathrm{t}\right)-\mathrm{l}\left(\mathrm{t}\right)\mathrm{\theta }\text{'}\left(\mathrm{t}\right)-\mathrm{gcos\theta }\left(\mathrm{t}\right)+\frac{\mathrm{k}}{\mathrm{m}}\left(\mathrm{l}-{\mathrm{l}}_{\mathrm{o}}\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{l}}^{2}\left(\mathrm{t}\right)\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)+2\mathrm{l}\left(\mathrm{t}\right)\mathrm{l}\text{'}\left(\mathrm{t}\right)\mathrm{\theta }\text{'}\left(\mathrm{t}\right)+\mathrm{gl}\left(\mathrm{t}\right)\mathrm{sin\theta }\left(\mathrm{t}\right)=0$

The system can be written as;

${\mathrm{x}}_{1}=\mathrm{l}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}=\mathrm{l}\text{'}\phantom{\rule{0ex}{0ex}}=\mathrm{x}{\text{'}}_{1}\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}=\mathrm{\theta }\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{4}=\mathrm{\theta }\text{'}\phantom{\rule{0ex}{0ex}}=\mathrm{x}{\text{'}}_{3}$

The transform equation is;

$\mathrm{x}{\text{'}}_{1}={\mathrm{x}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{2}=\mathrm{l}\text{'}\text{'}\phantom{\rule{0ex}{0ex}}={\mathrm{x}}_{1}{\mathrm{x}}_{4}+\mathrm{cos}\left({\mathrm{x}}_{3}\right)-{\mathrm{x}}_{1}+4\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{3}={\mathrm{x}}_{4}\phantom{\rule{0ex}{0ex}}\mathrm{x}{\text{'}}_{4}=\frac{-2{\mathrm{x}}_{2}{\mathrm{x}}_{4}-\mathrm{sin}\left({\mathrm{x}}_{3}\right)}{{\mathrm{x}}_{1}}$

The initial conditions are;

${\mathrm{x}}_{1}\left(0\right)=\mathrm{l}\left(0\right)=5.5\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{2}\left(0\right)=\mathrm{l}\text{'}\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}\left(0\right)=\mathrm{\theta }\left(0\right)=0.5\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{4}\left(0\right)=\mathrm{\theta }\text{'}\left(0\right)=0$

## Apply Runge-Kutte method

The values are;

 T L $\theta$ 0 5.5 0.5 0.1 5.49 0.499 0.5 5.41 0.488 1 5.13 0.454 2 4.105 0.289 2.5 3.47 0.131 3 2.89 -0.105 4 2.11 -0.835 5 2.13 -1.511 6 3.344 -1.633 9.9 7.6868 -0.696

## Graphs

Graph for l(t) Graph for $\mathrm{\theta }\left(\mathrm{t}\right)$. This is the required result.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades. 