Problem source

A particle of mass $m$ moves along the $x$-axis. At time $t=0$ it passes through $x=0$ with velocity $v_{0} > 0$. The particle is acted on by a force $\mathrm{F}(x)$, directed along the $x$-axis and measured in the direction of positive $x$, which is given by
\[
\mathrm{F}(x)=\begin{cases}
-m\mu^{2}x & \qquad(x\geqslant0),\\
-m\kappa\dfrac{\mathrm{d}x}{\mathrm{d}t} & \qquad(x <  0),
\end{cases}
\]
where $\mu$ and $\kappa$ are positive constants. Obtain the particle's subsequent position as a function of time, and give a rough sketch of the $x$-$t$ graph. 

Solution source

Using Newton's second law in the form, $\F(x) = m \ddot{x}$.

Our two different differential equations can be solved as follows:

When $x \geq 0$
$-\mu^2x = \ddot{x} \Rightarrow x = A\sin \mu t + B \cos \mu t$ when $x \geq 0$.

And when $x < 0$
$-\kappa \dot{x} = \ddot{x} \Rightarrow \dot{x} = Ce^{-\kappa t} \Rightarrow x = De^{-\kappa t} + E$ when $x < 0$

Following the trajectory of the particle:
At $t = 0, x = 0, \dot{x} = v_0 > 0$, so $x = \frac{v_0}{\mu} \sin \mu t$ until $t = \frac{\pi}{\mu}$.

When $t = \frac{\pi}{\mu}$ the particle will head into the negative $x$-axis with velocity $-v_0$. At which point our initial conditions for our differential equations give us that $De^{-\frac{\pi\kappa}{\mu}} + E = 0, -\kappa De^{-\frac{\pi\kappa}{\mu}} = -v_0 \Rightarrow  De^{-\frac{\pi\kappa}{\mu}}  = \frac{v_0}{\kappa}, E = -\frac{v_0}{\kappa}$. 

To summarise:

\[ x(t) = \begin{cases} \frac{v_0}{\mu} \sin \mu t & 0 \leq t \leq \frac{\pi}{\mu}  \\
-\frac{v_0}{\kappa} \l  1-e^{-\kappa(t-\frac{\pi}{\mu})}\r & t > \frac{\pi}{\mu}\end{cases}\]


\begin{center}
\begin{tikzpicture}[scale=2]

    \draw[->] (-0.5, 0) -- (3, 0);
    \draw[->] (0, -1) -- (0, 1);
    \node at (0,1) [above] {$x$};
    \node at (3,0) [right] {$t$};
    \draw[domain = 0:1, samples=180, variable = \x]  plot ({\x},{sin(180*\x}); 
    \draw[domain = 1:3, samples=180, variable = \x]  plot ({\x},{-(1-exp(-2*(\x-1)))}); 

\end{tikzpicture}
\end{center}