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

Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 287), using Laplace transforms.

$x (t) = - \frac{8}{17} \cos t - \frac{9}{17} \cos \sqrt{\frac{20}{3} t} y (t) = - \frac{6}{17} \cos t + \frac{6}{17} \cos \sqrt{\frac{20}{3}} t$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Given Information

The differential equation of a coupled mass spring oscillator is

${\begin{matrix} x^{''} + 4 x - 4 y = 0, & x (0) = - 1, x^{'} (0) = 0 \\ 3 y^{''} - 6 x + 11 y = 0 & y (0) = 0, y^{'} (0) = 0 \end{matrix}$

Step 2: Determining the coupled mass–spring oscillator motion

The mechanism in Exercises 5.6 Problem 1 is

${\begin{matrix} x^{''} + 4 x - 4 y = 0, & x (0) = - 1, x^{'} (0) = 0 \\ 3 y^{''} - 6 x + 11 y = 0 & y (0) = 0, y^{'} (0) = 0 \end{matrix}$

Using the Laplace transform, we can obtain

${\begin{cases} L {x " + 4 x - 4 y} = 0 \\ L {3 y " - 6 x + 11 y} = 0 \end{cases}$

$\Rightarrow {\begin{cases} s^{2} X (s) - s x (0) - x^{'} (0) + 4 X (s) - 4 Y (s) = 0 \\ 3 [s^{2} Y (s) - s y (0) - y^{'} (0)] - 6 X (s) + 11 Y (s) = 0 \end{cases}$

$\Rightarrow {\begin{cases} s^{2} X (s) + s + 4 X (s) - 4 Y (s) = 0 \\ 3 s^{2} Y (s) - 6 X (s) + 11 Y (s) = 0 \end{cases}$

$\Rightarrow {\begin{cases} (s^{2} + 4) X (s) - 4 Y (s) = - s \\ (3 s^{2} + 11) Y (s) - 6 X (s) = 0 \end{cases}$

$\Rightarrow {\begin{cases} X (s) = \frac{4}{s^{2} + 1} Y (s) - \frac{s}{s^{2} + 1} \\ (3 s^{2} + 11) Y (s) - 6 X (s) = 0 \end{cases}$

We now have

$\begin{matrix} (3 s^{2} + 11) Y (s) - 6 (\frac{4}{s^{2} + 1} Y (s) - \frac{s}{s^{2} + 1}) = 0 \\ (3 s^{2} + 11) Y (s) - \frac{24}{s^{2} + 1} Y (s) + \frac{6 s}{s^{2} + 1} = 0 \\ (3 s^{2} + 11 - \frac{24}{s^{2} + 1}) Y (s) = - \frac{6 s}{s^{2} + 1} \\ \frac{(3 s^{2} + 11) (s^{2} + 1) - 24}{s^{2} + 1} Y (s) = - \frac{6 s}{s^{2} + 1} \\ (3 s^{4} + 23 s^{2} + 20) Y (s) = 6 s \\ Y (s) = - \frac{6 s}{3 s^{4} + 23 s^{2} + 20} \end{matrix}$

We can get partial fractions by using partial fractions.

$Y (s) = - \frac{6 s}{17 (s^{2} + 1)} + \frac{6 s}{17 (s^{2} + \frac{20}{3})}$

As a result, if we use the inverse Laplace, we get

$y (t) = - \frac{6}{17} \cos t + \frac{6}{17} \cos \sqrt{\frac{20}{3} t}$

We now have X(s):

$\begin{array}{rcl} (3 s^{2} + 11) \frac{(- 6 s)}{3 s^{4} + 23 s^{2} + 20} - 6 X (s) & = & 0 \\ X (s) & = & \frac{- s (3 s^{2} + 11)}{(s^{2} + 1) (s^{2} + \frac{20}{3})} \end{array}$

We can get partial fractions by using partial fractions.

$X (s) = - \frac{8 s}{17 (s^{2} + 1)} - \frac{9 s}{17 (s^{2} + \frac{20}{3})}$

As a result, if we use the inverse Laplace, we get

$x (t) = - \frac{8}{17} \cos t - \frac{9}{17} \cos \sqrt{\frac{20}{3}} t$

Step 3: Determining the Result

The solution is obtained as:

$x (t) = - \frac{8}{17} \cos t - \frac{9}{17} \cos \sqrt{\frac{20}{3} t}$

$y (t) = - \frac{6}{17} \cos t + \frac{6}{17} \cos \sqrt{\frac{20}{3}} t$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 287), using Laplace transforms.

Step by Step Solution

Step 1:Given Information

Step 2: Determining the coupled mass–spring oscillator motion

Step 3: Determining the Result

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 287), using Laplace transforms.

Step by Step Solution

Step 1:Given Information

Step 2: Determining the coupled mass–spring oscillator motion

Step 3: Determining the Result

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