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Mobile App Chapter 5: Q12E (page 304)
Using a numerical scheme such as Runge–Kutta or a software package, calculate the Poincare map for equation (7) when b = 0.3, \({\bf{\gamma }}\)= 1.2, and F = 0.2. (Notice that the closer you start to the limiting point, the sooner the transient part will die out.) Compare your map with Figure 5.46(a) on page 300. Redo for F = 0.28.
Short Answer
By using the software, the Poincare maps are plotted and the value in the first part is that a solution is \(\frac{{2\pi }}{\gamma }\) periodic, and in the second part a solution is sub harmonic of the period\(\frac{{4\pi }}{\gamma }\).
Step by step solution
Sketch the Poincare map for\({\bf{F = 0}}{\bf{.2}}\).
Here the equation of (7) is given as:
\({\bf{x''(t) + bx' - x(t) + }}{{\bf{x}}^{\bf{3}}}{\bf{ = Fsin\gamma t}}\)
Put the given values when\(b{\rm{ }} = {\rm{ }}0.3,\,\gamma = {\rm{ }}1.2\), and \(F = 0.2\)
In this case, the limit is set at a single point and it doesn't depend on the initial value of the equation. So we can conclude that a solution is \(\frac{{2\pi }}{\gamma }\) periodic.
Sketch the graph when \({\bf{F = 0}}{\bf{.28}}\)
Here a solution is the sub harmonic of the period\(\frac{{4\pi }}{\gamma }\).
Therefore, the value in the first part is that a solution is \(\frac{{2\pi }}{\gamma }\) periodic, and in the second part a solution is sub harmonic of the period\(\frac{{4\pi }}{\gamma }\).
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