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1987 Paper 1 Q13
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: 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: \[ x(t) = \begin{cases} \frac{v_0}{\mu} \sin \mu t & 0 \leq t \leq \frac{\pi}{\mu} \\ -\frac{v_0}{\kappa} \l 1-e^{-\kappa(t-\frac{\pi}{\mu})}\r & t > \frac{\pi}{\mu}\end{cases}\]
