Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

$(m_{1} + m_{2}) l_{1}^{2} θ_{1}^{}'' + m_{2} l_{1} l_{2} θ_{2}^{}'' + (m_{1} + m_{2}) l_{1} {gθ}_{1} = 0 m_{2} l_{2}^{2} θ_{2}^{}'' + m_{2} l_{1} l_{2} θ_{1}^{}'' + m_{2} l_{2} {gθ}_{2} = 0$
When $θ_{1}$ and $θ_{2}$ are small angles. Solve the system when
$m_{1} = 3 kg, m_{2} = 2 kg, l_{1} = l_{2} = 5 m, θ_{1} (0) = π / 6, θ_{2} {(0) = θ}_{1}^{} {' (0) = θ}_{2}^{}' (0) = 0$ .

The solutions for $θ_{1}$ and $θ_{2}$ are:

$θ_{1} (t) = \frac{π}{12} \cos 1.1 t + \frac{π}{12} \cos 2.31 t θ_{2} (t) = \frac{5}{12} \cos 1.1 t - \frac{5}{12} \cos 2.31 t$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Substituting the values of m1,m2,l1,l2

Substituting the given values for $m_{1}, m_{2}, l_{1}, l_{2}$ into the given system and taking $g = 9.8 m / s^{2}$ one will get:

$(3 + 2) 5^{2} θ_{1}^{} {'' + 2 \times 5 \times 5 θ}_{2}^{}'' + (3 + 2) 5 \times 9 . 8 θ_{1} = 0 2 \times 5^{2} θ_{2}^{} {'' + 2 \times 5 \times 5 θ}_{1}^{}'' + 2 \times 5 \times 9.8 θ_{2} = 0$

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

${5 θ}_{1}^{} {'' + 2 θ}_{2}^{}'' + 9.8 θ_{1} = 0 {5 θ}_{2}^{} {'' + 5 θ}_{1}^{}'' + 9.8 θ_{2} = 0$

One will rewrite this system in operator form:

$(5 D^{2} + 9.8) [θ_{1}] + 2 D^{2} [θ_{2}] = 0 5 D^{2} [θ_{1}] + (5 D^{2} + 9.8) [θ_{2}] = 0$

Finding θ1

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

$\begin{array}{rcl} {(5 D^{2} + 9.8)}^{2} [θ_{1}] + 2 D^{2} (5 D^{2} + 9.8) [θ_{2}] & = & 0 \\ - 10 D^{4} - 2 D^{2} (5 D^{2} + 9.8) [θ_{2}] & = & 0 \\ (25 D^{4} + 98 D^{2} + 96.04 - 10 D^{4}) [θ_{1}] & = & 0 \\ (15 D^{4} + 98 D^{2} + 96.04) [θ_{1}] & = & 0 \end{array}$

Now, one can find a general solution for $θ_{1}$ . The auxiliary equation is:

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

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

So, the general solution for $θ_{1}$ is $θ_{1} (t) = c_{1} \cos 1.1 t + c_{2} \sin 1.1 t + c_{3} \cos 2.31 t + c_{4} \sin 2.31 t$

Findingθ2

Now, one can find a solution for $θ_{2}$ from the first equation of the system

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

Integration

Integrating the previous equation twice one will get:

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

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

$5 D^{2} [θ_{1}] + 5 D^{2} [θ_{2}] + 9.8 [θ_{2}] = 0$

Finding the solution for θ1 and θ2

Since one does not have any constant nor a term multiplying t in $D^{2} [θ_{1}]$ and $D^{2} [θ_{2}]$ , one cannot have it in $θ_{1}$ , and therefore $A = B = 0$ , so the solution for $θ_{2}$ is $θ_{2} (t) = 1 . 57 c_{1} \cos 1.1 t + 1.57 c_{2} \sin 1.1 t - 1.57 c_{3} \cos 2.31 t - 1.57 \sin 2.31 t$

The initial conditions give us:

$θ_{1} (0) = c_{1} + c_{3} = \frac{π}{6}, θ_{2} (0) = 1 . 57 c_{1} - 1.57 c_{3} = 0 c_{1} = c_{3}, 2 c_{1} = \frac{π}{6}, c_{1} = c_{3} = \frac{π}{12} θ_{1}^{}' (0) = 1 . 1 c_{2} + 2.31 c_{4} = 0, θ_{2}^{}' (0) = 1 . 73 c_{2} - 3.65 c_{4} = 0 c_{2} = 2.11 c_{4}, 4.62 c_{4} = 0, c_{2} = c_{4} = 0$

Finally, the solutions for $θ_{1}$ and $θ_{2}$ are

$\begin{array}{rcl} θ_{1} (t) = \frac{π}{12} \cos 1.1 t + \frac{π}{12} \cos 2.31 t \\ θ_{2} (t) = 1 . 57 \frac{π}{12} \cos 1.1 t - 1.57 \frac{π}{12} \cos 2.31 t \\ θ_{1} (t) = \frac{π}{12} \cos 1.1 t + \frac{π}{12} \cos 2.31 t \\ θ_{2} (t) = \frac{5}{12} \cos 1.1 t - \frac{5}{12} \cos 2.31 t \end{array}$

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