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Chapter 13: Problem 4

Draw the phase portrait of the damped linear oscillator, whose displacement \(x(t)\) satisfies \(\ddot{x}+\mu \dot{x}+\omega_{0}^{2} x=0\), in the phase plane \((x, y)\), where \(y=\dot{x} .\) Distinguish the cases (a) under- (or light) damping \(0<\mu<2 \omega_{0}\), (b) over-damping \(\quad \mu>2 \omega_{0}\) (c) critical damping \(\quad \mu=2 \omega_{0}\)

Short Answer

Expert verified
For the damped linear oscillator described by the equation \(\ddot{x}+\mu \dot{x}+\omega_{0}^{2} x=0\), we have analyzed the motion for different damping coefficients and distinguished three cases: 1. Under-damping (0 < \(\mu\) < \(2\omega_{0}\)): The phase portrait shows spirals converging to the origin, representing oscillatory behavior with decreasing amplitude. 2. Over-damping (\(\mu\) > \(2\omega_{0}\)): The phase portrait shows two families of curves converging to the origin, corresponding to different quadrants where the trajectories start, representing non-oscillatory, decaying motion. 3. Critical damping (\(\mu\) = \(2\omega_{0}\)): The phase portrait displays unique trajectories for each initial condition, all converging to the origin, representing the fastest possible decay without oscillation.

Step by step solution

01

Find the general solution of the given equation for each case

(a) Under-damping (0 < \(\mu\) < \(2\omega_{0}\)): In this case, the roots of the characteristic equation are complex: \(\lambda_{1,2} = \frac{-\mu \pm \sqrt{\mu^2 - 4\omega_0^2}}{2}\) The general solution is: \(x(t) = e^{-\frac{\mu}{2}t}(A \cos(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t)+B \sin(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t))\) (b) Over-damping (\(\mu\) > \(2\omega_{0}\)): In this case, the roots of the characteristic equation are real and distinct: \(\lambda_{1,2} = \frac{-\mu \pm \sqrt{\mu^2 - 4\omega_0^2}}{2}\) The general solution is: \(x(t) = C e^{\lambda_1 t} + D e^{\lambda_2 t}\) (c) Critical damping (\(\mu\) = \(2\omega_{0}\)): In this case, the roots of the characteristic equation are real and equal: \(\lambda_{1,2} = \frac{-\mu}{2}\) The general solution is: \(x(t) = (E + Ft)e^{-\frac{\mu}{2}t}\)
02

Find the velocity function for each case

(a) Under-damping: \(y(t) = \dot{x}(t) = e^{-\frac{\mu}{2}t}(-\frac{\mu}{2}(A \cos(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t)+B \sin(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t)) - \sqrt{\omega_0^2 - \frac{\mu^2}{4}}(- A \sin(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t) + B \cos(\sqrt{\omega_0^2 - \frac{\mu^2}{4}}t)))\) (b) Over-damping: \(y(t) = \dot{x}(t) = C \lambda_1 e^{\lambda_1 t} + D \lambda_2 e^{\lambda_2 t}\) (c) Critical damping: \(y(t) = \dot{x}(t) = -\frac{\mu}{2}(E + Ft)e^{-\frac{\mu}{2}t} + Fe^{-\frac{\mu}{2}t}\)
03

Draw phase portraits for each case

The phase portraits represent various possible trajectories for the given system depending on the initial conditions. (a) Under-damping: In this case, the phase portrait of the motion has the shape of spirals converging to the origin. (b) Over-damping: In this case, the phase portrait shows two families of curves converging to the origin, one corresponding to trajectories that start in the first/second quadrant and the other corresponding to trajectories starting in the third/fourth quadrant. (c) Critical damping: In this case, the phase portrait shows a unique trajectory for each initial condition, converging to the origin.

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Most popular questions from this chapter

For the pendulum equation \(\ddot{\theta}+(g / l) \sin \theta=0\) find the equations of the trajectories in the phase plane and sketch the phase portrait of the system. Show that on the separatrices $$ \theta(t)=(2 n+1) \pi \pm 4 \arctan [\exp (\omega t+\alpha)] $$ where \(n\) is an integer, \(\alpha=\) constant and \(\omega=\sqrt{g / l}\). [Hint: Use the substitution \(\left.u=\tan \frac{1}{4}\\{\theta-(2 n+1) \pi\\} .\right]\)

For the Galton board of Fig. \(13.25\) we may arrange things so that each piece of lead shot has an equal chance of rebounding just to the left or to the right at each direct encounter with a scattering pin at each level. Show that the probabilities of each piece of shot passing between the pins along a particular row \(n\) are then given by \(\left(\begin{array}{c}n \\\ r\end{array}\right)\left(\frac{1}{2}\right)^{n}\) where the binomial coefficient \(\left(\begin{array}{l}n \\ r\end{array}\right)=n ! /[(n-r) ! r !]\) and \(r=0,1, \ldots, n\). Use the result \(\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)=\left(\begin{array}{c}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) to generate the probability distribution for row \(n=\) 16. (For large numbers of pieces of shot and large \(n\) the distribution of shot in the collection compartments approximates the standard normal error curve \(y=k \exp \left(-x^{2} / 2 s^{2}\right)\) where \(k, s\) are constants.)

Consider the Lorenz system \((13.33)\). (a) Show that the origin \(P_{1}(0,0,0)\) is a critical point and that its stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+\beta)\left[\lambda^{2}+(\sigma+1) \lambda+\sigma(1-\rho)\right]=0 $$ Hence show that \(P_{1}\) is asymptotically stable only when \(0<\rho<1\) (b) Show that there are two further critical points $$ P_{2}, P_{3} \equiv[\pm \sqrt{\beta(\rho-1)}, \pm \sqrt{\beta(\rho-1)},(\rho-1)] $$ when \(\rho>1\), and that their stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ \lambda^{3}+\lambda^{2}(\sigma+\beta+1)+\lambda \beta(\sigma+\rho)+2 \sigma \beta(\rho-1)=0 $$ (c) Show that when \(\rho=1\) the roots of the cubic in (b) are \(0,-\beta,-(1+\sigma)\) and that in order for the roots to have the form \(-\mu, \pm \mathrm{i} \nu\) (with \(\mu, \nu\) real) we must have $$ \rho=\rho_{\text {crit }}=\frac{\sigma(\sigma+\beta+3)}{(\sigma-\beta-1)}>0 $$ (d) By considering how the roots of the cubic in (b) change continuously with \(\rho\) (with \(\left(\sigma, \beta\right.\) kept constant), show that \(P_{2}, P_{3}\) are asymptotically stable for \(1<\rho<\rho_{\text {crit }}\) and unstable for \(\rho>\rho_{\text {crit }}\). (e) Show that if \(\bar{z}=z-\rho-\sigma\), then $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(x^{2}+y^{2}+\bar{z}^{2}\right)=-\sigma x^{2}-y^{2}-\beta\left[\bar{z}+\frac{1}{2}(\rho+\sigma)\right]^{2}+\frac{1}{4} \beta(\rho+\sigma)^{2} $$ so that \(\left(x^{2}+y^{2}+\bar{z}^{2}\right)^{1 / 2}\) decreases for all states outside any sphere which contains a particular ellipsoid (implying the existence of an attractor).

For the Rikitake dynamo system (13.35): (a) Show that there are two real critical points at \(\left(\pm k, \pm 1 / k, \mu k^{2}\right)\) in the \(\left(X_{1}, X_{2}, Y\right)\) phase space, where \(k\) is given by \(A=\mu\left(k^{2}-1 / k^{2}\right)\). (b) Show that the stability of these critical points is determined by eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+2 \mu)\left[\lambda^{2}+\left(k^{2}+\frac{1}{k^{2}}\right)\right]=0 $$ so that the points are not asymptotically stable in this approximation. (For the full system they are actually unstable.) (c) Show that the divergence of the phase-space flow velocity is negative, so that the flow causes volume to contract. (d) Given that \(\bar{Y}=\sqrt{2}(Y-A / 2)\), show that $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}\right)=-\mu\left(X_{1}^{2}+X_{2}^{2}\right)+\sqrt{2} \bar{Y} $$ and use this result to determine in which region of the space the trajectories all have a positive inward component towards \(X_{1}=\) \(X_{2}=\bar{Y}=0\) on surfaces \(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}=\) constant

The relativistic equivalent of the simple harmonic oscillator equation for a spring with constant \(k\) and a rest mass \(m_{0}\) attached is $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m_{0} y}{\sqrt{1-y^{2} / c^{2}}}\right)+k x=0 \quad \text { with } \quad \dot{x}=y $$ where \(c\) is the speed of light. Show that the phase trajectories are given by $$ m_{0} c^{2} / \sqrt{1-y^{2} / c^{2}}+\frac{1}{2} k x^{2}=\text { constant } $$ and sketch the phase portrait for this system.
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