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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 272
Fundamentals Of Differential Equations And Boundary Value Problems

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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Short Answer

A Problem of Current Interest. The motion of an ironbar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{x}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\) for \({\bf{ - }}{{\bf{x}}_{\bf{o}}}{\bf{ < x < \lambda }}\)where the constants \({{\bf{x}}_{\bf{o}}}\) and \({\bf{\lambda }}\) are, respectively, the distances from the bar to the wall and to the wire when thebar is at equilibrium (rest) with the current off.

  1. Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
  2. Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by\({\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \), where C is a constant.
  3. Show that if \({\bf{\lambda < 2}}\) there are no critical points in the xy-phase plane, whereas if \({\bf{\lambda > 2}}\) there are two critical points. For the latter case, determine these critical points.
  4. Physically, the case \({\bf{\lambda < 2}}\)corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\).
  5. From your phase plane diagrams in part (d), describe the possible motions of the bar when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\), under various initial conditions.

For all answers follow all steps.

See the step by step solution

Step by Step Solution

Step 1: The ordering system

Let\({\bf{v = x'}}\,\,{\bf{then}}\,\,{\bf{v' = x''}}\).

The system is:

\(\begin{array}{c}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = v}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\end{array}\)

Step 2: Solve phase plane equation.


\(\begin{array}{c}\frac{{{\bf{dv}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{{\bf{ - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}}}{{\bf{v}}}\\\int {{\bf{vdv}}} {\bf{ = }}\int {{\bf{ - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}{\bf{dx}}} \\\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = - }}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - ln(\lambda - x) + C}}\\{{\bf{v}}^{\bf{2}}}{\bf{ = - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x) + C}}\\{\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \end{array}\)

Step 3: Find critical points.


\(\begin{array}{c}{\bf{ - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}{\bf{ = 0}}\\{\bf{ - x(\lambda - x) + 1 = 0}}\\{{\bf{x}}^{\bf{2}}}{\bf{ - \lambda x + 1 = 0}}\\{\bf{x = }}\frac{{{\bf{\lambda \pm }}\sqrt {{{\bf{\lambda }}^{\bf{2}}}{\bf{ - 4}}} }}{{\bf{2}}}\end{array}\)

Now it is clear that \({\bf{\lambda < 2}}\)the result is negative so no real solutions. If \({\bf{\lambda > 2}}\) real solutions exist the critical points are\(\left( {\frac{{{\bf{\lambda \pm }}\sqrt {{{\bf{\lambda }}^{\bf{2}}}{\bf{ - 4}}} }}{{\bf{2}}}{\bf{,0}}} \right)\).

Step 4: Sketch for λ<2

For λ=3

Step 5: Motion of the bar

When \({\bf{\lambda = 1}}\) the bar will always be attracted to the magnet.

When\({\bf{\lambda = }}3\), the critical values are \(\left( {\frac{{{\bf{3 - }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right){\bf{and}}\left( {\frac{{{\bf{3 + }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right)\)

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