Home Work THREE

• 4.1.5 Since
  $$\cos[x] + \sin[x]=\sqrt{2}\sin(\frac{\pi}{4}+x),$$
  this problem has all the properties of
  $$\dot\theta=\sin{\theta}$$
  after change of variables $\theta=x+\frac{\pi}{4}$.
• 4.1.9 A vector field of on a line is determined as a unique correspondence between $t$ and $x(t)$ with each $t$ giving corresponding $x(t)$. The periodic motion is a motion when two values of $t$ give the same value of $x$, i.e. when $x(t+T)=x(t)$ with a period $T$. This can not occur on a line.

  A vector field on a circle interprets $\theta(t)=\theta_0$ and $\theta(t+T)=\theta_0+2\pi$ as one point. Therefore if the particle “moves” with nonzero $\dot\theta$ it will necessarily return to the same point. This fundamental difference would mage proofs of 2.6.2 and 2.7.2 invalid.

• 4.2.2 This command in Mathematica will plot a graph for you:

  Plot[ Sin[8 t] + Sin[9 t], {t, -20, 20}]
  

  Now use the fact that

  $$\sin(x)+\sin(y) = 2 \cos\frac{x-y}{2}\sin\frac{x+y}{2}$$
  to write
  $$\sin(8 t)+\sin(9t)=2\cos\frac{t}{2}\sin\frac{17 t}{2}.$$
  Since $17/2>1/2$ then $\sin\frac{17 t}{2}$ is a fast oscillation with the modulated amplitude of $2\cos\frac{t}{2}$. The period of amplitude modulation is therefore $4\pi$. Please use the following to see it

   
  Plot[{2 Cos[t/2], Sin[8 t] + Sin[9 t]}, {t, -20, 20}]
  

• 4.3.4

  There are two candidates for the fixed points here, one is $\theta=0$ and the second one is $\theta=\pi$. Stability and behaviour of these two points depends on $\mu$.

  For $\mu<-1$ there are two fixed points, one is at $\theta=0=2\pi$ and another at $\theta=Pi$. The $\theta=0$ is stable, while $\theta=\pi$ is unstable.

  For $\mu=-1$ there is only one unstable fixed point at $\theta=\pi$. The formerly stable fixed point at $\theta=0=2\pi$ is now replaced with the singularity.

  For $-1<\mu<1$ both fixed points become unstable. How this is possible, since two unstable fixed points on a circle would imply change of direction of motion. The answer is that there are two singularities between the fixed points. Singularities are located at the two solutions of

  $$\cos(\theta)=-\mu.$$

  For $\mu=1$ there is a singularity at $\theta=\pi$ and unstable fixed point at $\theta=0=2\pi$.

  For $\mu>1$ singularities dissapear with $\theta=\pi$ being stable and $\theta=0=2\pi$ being unstable.

• 4.5.1 Equation for the phase difference $\psi(t)$ will be the same as equation for the phase difference in the book with the $\sin$ function being replaced by the $f(\theta)$ function:
  $$\dot \psi(t) = \Omega-\omega-A f(\psi(t)).$$
  The range of entrainment is when $\dot\psi(t)=0$ is possible, i.e. when
  $$\omega-\frac{ A \pi}{2}<\Omega<\omega+\frac{A \pi}{2}.$$
  The difference with the textbook is due to the fact that $f(x)$ is between $\pm\frac{\pi}{2}$, which is different from sine function, which is bounded between $\pm 1$.

  The Phase Difference is equal to $\psi^*$ that is given by

  $$\psi^* = \frac{\Omega-\omega}{A},$$
  or
  $$\psi^* = \pi+\frac{\Omega-\omega}{A}.$$

  The equation for the drift time $T_{\rm drift}$ is

  $$T_{\rm drift}=\frac{\psi^*}{\Omega}.$$

  We can calculate the period also:
  

  $$T = \int d t = \int \frac{d t}{d \psi} d \psi = \int\limits_{-\frac{\pi}{2}}^{\frac{3 \pi}{2}}\frac{d \psi}{\Omega-\omega- A f(\psi)} =$$

  $$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{d \psi}{\Omega-... ...frac{2}{A}\ln\frac{\Omega-\omega-A\frac{\pi}{2}}{\Omega=\omega+A\frac{\pi}{2}}.$$