1. Consider a particle of mass $m$ moving vertically in a fluid with quadratic drag force $Dv^2$, where $v$ is its velocity and $D > 0$ is a constant. The particle is also acted on by gravity, with acceleration due to gravity $g$.
   • Consider dropping the particle from rest through the fluid, so that its velocity is $v=\dot z≤0$, with $z$ measured upwards. Show that the equation of motion may be written as$$m \dot{v}=-m g+D v^{2}$$Show that this may be integrated to$$t=-\int_0^v\frac{\mathrm{d} u}{g-\frac{D u^2}m}$$By evaluating the integral, hence show that the solution is$$v(t)=-\sqrt{\frac{mg}D} \tanh \left(\sqrt{\frac{Dg}m} t\right)$$What is the terminal velocity?
     Solution.
     By Newton's second law,$$m \dot{v}=-m g+D v^{2}⇒\mathrm dt=-{\mathrm dv\over g-\frac{Dv^2}m}$$Integrating, we have$$t=-\int_0^v{\mathrm du\over g-\frac{Du^2}m}=\sqrt{\frac m{Dg}}\tanh^{-1}\left(-v\sqrt{\frac{D}{mg}}\right)$$Therefore$$v(t)=-\sqrt{\frac{mg}D} \tanh \left(\sqrt{\frac{Dg}m} t\right)$$Take $t→∞$, we get the terminal velocity $v(t)=-\sqrt{\frac{mg}D}$.
   • Now consider projecting the particle upwards through the fluid, starting at $z = 0$ with speed $u$. Show that the equation of motion may be written as $$\frac{\mathrm{d}\left(v^{2}\right)}{\mathrm{d} z}=2 \dot{v}=-2 g-\frac{2 D v^{2}}{m}$$Regarding this as an equation for $v^2(z)$, by integrating it show that the maximum height reached is$$z_{\max}=\frac m{2D} \log \left(1+\frac{D u^{2}}{m g}\right)$$What happens as $D→0$?
     Solution.
     $\frac{\mathrm d(v^2)}{\mathrm dz}=v\frac{\mathrm dv}{\mathrm dz}=\frac{\mathrm dz}{\mathrm dt}\frac{\mathrm dv}{\mathrm dz}=\frac{\mathrm dv}{\mathrm dt}=\dot v$. The second equality follows from Newton's second law $m \dot{v}=-m g-D v^2$[motion is upward so drag force is downward so $-Dv^2$].
     Solving as a first-order linear differential equation,$$v^2(z)=-\frac{mg}D+C e^{-\frac{2Dz}m}$$Since $v^2(0)=u^2$,$$v^2(z)=-\frac{mg}D+\left(u^2+\frac{mg}D\right)e^{-\frac{2Dz}m}⇒z=\frac m{2D}\log\left(u^2+\frac{mg}D\over v^2+\frac{mg}D\right)$$Because $v^2≥0$, we have$$z_{\max}=\frac m{2D}\log\left(u^2+\frac{mg}D\over \frac{mg}D\right)=\frac m{2D} \log \left(1+\frac{D u^{2}}{m g}\right)$$As $D→0$, using $\log(1+x)\sim x$,$$z_{\max}=\frac{u^2}{2g}$$
2. A particle of mass $m$ moves along the $x$ axis with one end attached to a spring of spring constant $k > 0$, and is subjected to an additional force $F_0\cos Ωt$.
   
   • Show that the equation of motion is $\ddot{x}+\omega^2x=A \cos \Omega t$ where $x = 0$ corresponds to the unstretched position of the spring, $ω=\sqrt{k/m}$, and $A=F_0/m$.
     Solution.
     By Newton's second law,$$m\ddot x=-kx+F_0\cosΩt⇒\ddot{x}+\omega^2x=A \cos \Omega t$$
   • Suppose that $x =\dot x = 0$ at time $t = 0$. Verify that if $Ω\neω$ then $$x(t)=\frac{A}{\omega^{2}-\Omega^{2}}(\cos \Omega t-\cos \omega t)$$satisfies the equation of motion and initial conditions, while when $Ω = ω$ then$$x(t)=\frac{A}{2 \omega} t \sin \omega t$$does. What is the qualitative difference between the two solutions?
     Solution.
     When $Ω≠ω$, solving as a second-order linear differential equation,$$x(t)={A\overω^2-Ω^2}\cosΩt+C_1\sinωt+C_2\cosωt$$Since $x(0)=\dot x(0)=0$,$$x(t)=\frac{A}{\omega^{2}-\Omega^{2}}(\cos \Omega t-\cos \omega t)$$When $Ω = ω$, solving as a second-order linear differential equation,$$x(t)={A\over2ω}t\sinωt+C_1\sinωt+C_2\cosωt$$Since $x(0)=\dot x(0)=0$,$$x(t)={A\over2ω}t\sinωt$$When $Ω≠ω$, $x(t)$ is bounded; when $Ω=ω$, $x(t)$ is unbounded.
     
3. Consider a particle of charge $q$ moving in a constant electromagnetic field. Without loss of generality we take the magnetic field $\mathbf B = (0, 0, B) ≠ 0$ to point along the $z$ axis, while
   the electric field $\mathbf E = (E_1, E_2, E_3)$ is constant, but arbitrary.
   
   • Assuming the particle has mass $m$, but ignoring gravity, show that Newton’s second law implies the coupled ODEs\begin{align*} m \ddot{x} &=q E_{1}+q B \dot{y} \\ m \ddot{y} &=q E_{2}-q B \dot{x} \\ m \ddot{z} &=q E_{3} \end{align*}for the position $\mathbf r = (x, y, z)$ of the particle.
     Proof. By Newton's second law,$$m\begin{pmatrix}\ddot x\\\ddot y\\\ddot z\end{pmatrix}=q\begin{pmatrix}E_1\\E_2\\E_3\end{pmatrix}+q\begin{pmatrix}\dot x\\\dot y\\\dot z\end{pmatrix}∧\begin{pmatrix}0\\0\\B\end{pmatrix}$$
   • Verify that\begin{align*} x(t) &=x_{0}+\frac{E_{2}}{B} t+R \cos (\omega t+\phi) \\ y(t) &=y_{0}-\frac{E_{1}}{B} t-R \sin (\omega t+\phi) \\ z(t) &=z_{0}+u t+\frac{q}{2 m} E_{3} t^{2} \end{align*}solves the equations of motion in part (a), where $ω = qB/m$ is the cyclotron frequency, $(x_0,y_0,z_0)$ is a constant vector, and $u,R$ and $φ$ are also constants.
     [Optional: For a more challenging version of this question, rather than verifying the solution, instead derive it, hence showing it is the general solution.]
     Solution.
     Let $w=x+iy$, from the first two equations, $m\ddot w=q(E_1+iE_2)-iqB\dot w$, solving as first-order linear differential equation for $\dot w$,$$\dot w=\frac{E_2-iE_1}B+Ce^{-\frac{iqBt}m}$$Separating real and imaginary parts,\begin{align*}\dot x&=\frac{E_2}B+C\cos\frac{qBt}m\\\dot y&=-\frac{E_1}B-C\sin\frac{qBt}m\end{align*}Integrating, we get\begin{align*} x(t) &=x_{0}+\frac{E_{2}}{B} t+R \cos (\omega t+\phi) \\ y(t) &=y_{0}-\frac{E_{1}}{B} t-R \sin (\omega t+\phi)\end{align*}Integrating $m\ddot{z} =qE_3$ twice, we have $z=C_1+C_2t+\frac{q}{2m}E_3t^2$, therefore$$ z(t)=z_{0}+u t+\frac q{2m} E_3t^2$$[More general approach: Eliminate $y$ by differentiating wrt $t$:
     $m\dddot x=qB\ddot y=\frac{qB}m\left(qE_2-qB\dot x\right)$
     $⇒\dddot x+\frac{q^2B^2}{m}\dot x=\frac{qB}mqE_2$ (second order equation for $x$)
     solution is $x(t)=x_{0}+\frac{E_{2}}{B} t+R \cos (\omega t+\phi)$]
4. Consider a particle of mass $m$ moving in a plane with position vector $\mathbf r = (x, y)$, subject to a force $\mathbf F = -k\mathbf r$, where $k > 0$ is constant.
   
   • Show that the general solution to the equation of motion is $\mathbf{r}(t)=\mathbf{A} \sin \omega t+\mathbf{B} \cos \omega t$ where $\mathbf A$ and $\mathbf B$ are constant vectors, and $ω =\sqrt{k/m}$. (You might find it  helpful to write out the vector equation of motion in terms of its components.)
   • Show that the solution in part (a) may be rewritten as $\mathbf{r}(t)=\mathbf{a} \sin (\omega t+\phi)+\mathbf{b} \cos (\omega t+\phi)$
     where now $\bf a$ and $\bf b$ are constant orthogonal vectors, and $φ$ is a constant phase.
   • Hence show that the trajectory of the particle traces out an ellipse, with centre the origin.
     
   Solution.
   • $m\ddot x=-kx⇒x=A_1\sinωt+B_1\cosωt$ and $m\ddot y=-ky⇒y=A_2\sinωt+B_2\cosωt$, therefore $\mathbf{r}(t)=\binom{A_1}{A_2}\sin \omega t+\binom{B_1}{B_2}\cos \omega t$.
   • $\mathbf{r}(t)=\mathbf{a}\sin(\omega t+\phi)+\mathbf{b}\cos(\omega t+\phi)$
     $\hphantom{\mathbf{r}(t)}=\mathbf{a}(\sin\phi\cos\omega t+\cos\phi\sin\omega t)+\mathbf{b}(\cos\phi\cos\omega t-\sin\phi\sin\omega t)$
     $\hphantom{\mathbf{r}(t)}=(\mathbf a\cos\phi-\mathbf b\sin\phi)\sin\omega t+(\mathbf{a}\sin\phi+\mathbf{b}\cos\phi)\cos\omega t$
     Solving\begin{align*}\mathbf A&=\mathbf a\cos\phi-\mathbf b\sin\phi\\\mathbf B&=\mathbf b\cos\phi+\mathbf a\sin\phi\end{align*}we have\begin{align*}\mathbf a&=\mathbf A\cos\phi+\mathbf B\sin\phi\\\mathbf b&=\mathbf B\cos\phi-\mathbf A\sin\phi\end{align*}$\mathbf a·\mathbf b=0$⇔$\cot2ϕ=\frac{{\bf A}^2-{\bf B}^2}{2{\bf AB}}$, therefore we can always find ϕ such that $\mathbf a·\mathbf b=0$.
   • Let $\mathbf r=\frac{\mathbf a}{|\mathbf a|}r_1+\frac{\mathbf b}{|\mathbf b|}r_2$, where$$\frac{r_1^2}{|\mathbf a|^2}+\frac{r_2^2}{|\mathbf b|^2}=\sin^2(\omega t+\phi)+\cos^2(\omega t+\phi)=1$$therefore the trajectory of the particle traces out an ellipse, with centre the origin, $|\mathbf a|$ and $|\mathbf b|$ the semiaxes.