Q33E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

 

Step-by-Step Solution

Answer

For all answers follow all steps.

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

Step 4: Sketch for &lambda;<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 \(\left( {\frac{{{\bf{3 - }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right){\bf{and}}\left( {\frac{{{\bf{3 + }}\sqrt {\bf{5}} }}{{\bf{2}}}{\bf{,0}}} \right)\)