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Found in: Page 272

### 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

# 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.Setting$${\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}$$, convert the second-order equation to an equivalent first-order system.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.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.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$${\bf{\lambda = 3}}$$.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.

See the 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.

Now,

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

## Step 3: Find critical points.

Now,

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

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$$\left( {\frac{{{\bf{\lambda \pm }}\sqrt {{{\bf{\lambda }}^{\bf{2}}}{\bf{ - 4}}} }}{{\bf{2}}}{\bf{,0}}} \right)$$.

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 $$\left( {\frac{{{\bf{3 - }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right){\bf{and}}\left( {\frac{{{\bf{3 + }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right)$$