Q5.3-24E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length l(t)of the wire varies with time in some predetermined fashion. If

U(t) is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem ${\mathbf{l}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{l}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{gl}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{o}}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\theta }}_{\mathbf{1}}$ where g is the acceleration due to gravity. Assume that $\mathbf{l}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{l}}_{\mathbf{o}}\mathbf{+}{\mathbf{l}}_{\mathbf{1}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{-}\varphi \mathbf{)}$ where ${\mathbf{l}}_1$ is much smaller than ${\mathbf{l}}_{\mathbf{o}}$. (This might be a model for a person on a swing, where the pumping action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take g = 1. Using the Runge– Kutta algorithm with h = 0.1, study the motion of the pendulum when ${\mathbf{\theta }}_{\mathbf{o}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{l}}_{\mathbf{o}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{l}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{\omega }\mathbf{=}\mathbf{1}\mathbf{,}\mathit{\varphi }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}$. In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle ${\mathbf{\theta }}_{\mathbf{o}}$?

Step-by-Step Solution

Answer

The result can get by the Runge-Kutta method, and yes, the pendulum ever attains an angle greater in absolute value than the initial angle ${\theta }_o$.

Transform the equation

Here, the equation ${\mathrm{l}}^2\left(\mathrm{t}\right)\mathrm{q}\text{'}\text{'}\left(\mathrm{t}\right)+2\mathrm{l}\left(\mathrm{t}\right)\mathrm{l}\text{'}\left(\mathrm{t}\right)\mathrm{q}\text{'}\left(\mathrm{t}\right)+\mathrm{gl}\left(\mathrm{t}\right)\sin \left(\mathrm{q}\left(\mathrm{t}\right)\right)=0$

 

The system can be written as:

$$\begin{gathered} x_1=\theta \left(t\right) \\ {x}_2=\theta \text{'} \\ =x{\text{'}}_1 \end{gathered}$$

The transform equation is:

 $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}{\mathbf{x}}_{\mathbf{2}} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{)}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{sinx}}_{\mathbf{1}}}{\mathbf{1}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{cos}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{)}} \end{gathered}$$

 

The initial conditions are:

$$\begin{gathered} x_1\left(0\right)=\theta \left(0\right) \\ =0.05 \\ {x}_2\left(0\right)=\theta \text{'}\left(0\right) \\ =0 \end{gathered}$$

Apply the Runge-Kutta method

Apply Matlab to find the results. And some results are:

t	$\theta$
0	0.05
0.1	0.049
0.5	0.044
1	0.028
1.5	0.0039
2	-0.023
3	-0.060
5	0.01505
6	0.4816
7	0.03861
8	-0.0079
9	-0.05499
10	-0.04677

From the table, we can see that the pendulum does indeed attain an angle greater in the absolute value than the initial angle of 0.050.05. For instance, at the time $t=9.4$ the absolute value of the angle is $\left|\theta \left(9.4\right)\right|=|-0.061082|=0.061082>0.05$

Therefore, the pendulum attains an angle greater in absolute value than the initial angle ${\theta }_0$.