Sketch the phase portrait and describe the dynamical trajectories for a particle of mass \(m\) moving under the following forces (1) \(\quad F=m\left(\frac{3}{2} x^{2}-x\right)\) (2) \(\quad F=2 m x\left(x^{2}-1\right)\) (3) \(\quad F=-m \sin x\) (4) \(\quad F=-m x \dot{x} /(1+\dot{x})\) In the fourth case, what at happens as the trajectories approach the line \(v=-1\) in the phase plane?

According to Newton's second law, \(F = m \ddot{x}\). For each force, we will replace the force expressions to determine the equation of motion. (1) \(F = m\left(\frac{3}{2} x^{2}-x\right)\) \(\Rightarrow m \ddot{x} = m\left(\frac{3}{2} x^{2}-x\right)\) \(\Rightarrow \ddot{x} = \frac{3}{2} x^{2}-x\) (2) \(F = 2 m x\left(x^{2}-1\right)\) \(\Rightarrow m \ddot{x} = 2 m x\left(x^{2}-1\right)\) \(\Rightarrow \ddot{x} = 2 x\left(x^{2}-1\right)\) (3) \(F = -m \sin x\) \(\Rightarrow m \ddot{x} = -m \sin x\) \(\Rightarrow \ddot{x} = - \sin x\) (4) \(F = -m x \dot{x} /(1+\dot{x})\) \(\Rightarrow m \ddot{x} = -m x \dot{x} /(1+\dot{x})\) \(\Rightarrow \ddot{x} = - x \dot{x} /(1+\dot{x})\) Now, we have the equations of motion for each case.

To analyze the phase portraits and trajectories, we should look at possible fixed points (where \(\dot{x} = 0\) and \(\ddot{x} = 0\)) and the behavior of the trajectories as \(x\) changes. Each given force will be analyzed separately: (1) Equilibrium Points: \(x=0\) and \(x=2/3\). For \(x<0\), \(m\ddot x>0\) so motion is oscillatory, so \(x=2/3\) is a stable equilibrium point, and \(x=0\) is an unstable equilibrium point due to the negative slope. (2) Equilibrium Points: \(x=-1, x=0, x=1\). For \(x<-1\), \(m\ddot x>0\), for \(x \in(-1, 0)\), \(m\ddot x<0\), and for \(1>x>0\), \(m\ddot x>0\) and for \(x>1\), \(m\ddot x>0\). Thus, we have two stable equilibrium points at \(x=-1\) and \(x=1\) and one unstable equilibrium point at \(x=0\). (3) Equilibrium Points: multiple equilibria at \(x = n\pi\) where \(n\) is any integer. For \(x \in (2n\pi, (2n+1)\pi)\), we have \(m\ddot x<0\), and for \(x \in ((2n-1)\pi, 2n\pi)\), we have \(m\ddot x>0\). Therefore, odd multiples of \(\pi\) are stable equilibrium points while even multiples are unstable. (4) Equilibrium Points: The only equilibrium point is at \(x=0\). Note that for \(x > 0\), we have a positive motion, while for \(x < 0\), we have a negative motion. Behavior changes near the line \(v=-1\) in the phase plane which is the next step.

According to the given force, \(F=-m x \dot{x} /(1+\dot{x})\), when \(\dot{x}=-1\): \(F=-m x \dot{x} / (1 - 1)\) which leads to a division by zero situation. As the trajectories approach the line \(v=-1\) in the phase plane, we will have an infinite force, indicating that motion in the line \(v=-1\) is not physically possible. In practice, the motion will be prevented, and the trajectories will never reach the line \(v=-1\) in the phase plane.