M3.3 - Further dynamics

For motion along one signed axis, first write the resultant force with its direction, then choose the acceleration form that matches the variable in the force law. A varying force produces a differential equation rather than a constant-acceleration problem.

Define the coordinate origin and positive direction before inserting a force. Include every force component in the resultant. Separate variables or integrate with limits, use the supplied motion condition to determine the constant, and accept only positions for which $v^2\ge0$ and the force model is defined.

A $2$ kg particle moves in the positive $x$-direction under resultant force $F=12-2x$ N. At $x=0$, its speed is $1\,\mathrm{m\,s^{-1}}$. Since$$2v\frac{\mathrm dv}{\mathrm dx}=12-2x,$$integration between the initial position and $x$ gives$$v^2=1+12x-x^2.$$At $x=3$ m, $v=\sqrt{28}=2\sqrt7\,\mathrm{m\,s^{-1}}$. A turning point requires $v=0$; the positive solution is $x=6+\sqrt{37}$ m.

For an attractive inverse-square force with outward coordinate $x>0$, $F=-k/x^2$. Then$$mv\frac{\mathrm dv}{\mathrm dx}=-\frac{k}{x^2}$$integrates to $\tfrac12mv^2=k/x+C$. The plus sign on $k/x$ after integration is consistent with differentiating $k/x$ to recover $-k/x^2$.

Motion is simple harmonic when acceleration is proportional to displacement from a fixed equilibrium point and always directed back towards that point. With signed displacement $x$ from equilibrium, the defining equation is

To prove SHM in a mechanical system, write a signed equation of motion and simplify it in terms of displacement from equilibrium. Constant force terms must cancel at equilibrium. Reaching $\ddot x=-c x$ with $c>0$ proves SHM and identifies $\omega=\sqrt c$; merely obtaining a periodic-looking expression is not the required force-law proof.

Choose the phase from the initial state. Release from rest at $x=A$ allows $x=A\cos\omega t$; starting at equilibrium in the positive direction allows $x=A\sin\omega t$. For time spent in a region, solve the boundary displacement within one cycle and use the symmetry of the cosine or sine curve without double-counting endpoints.

If $x=0.12\cos(5t)$ metres, then $A=0.12$ m, $\omega=5\,\mathrm{rad\,s^{-1}}$ and $T=2\pi/5$ s. The maximum speed is $0.60\,\mathrm{m\,s^{-1}}$. Starting at the positive endpoint, the first time the particle reaches $x=0.06$ m satisfies $\cos(5t)=1/2$, so $t=\pi/15$ s.

A particle attached to an ideal spring or taut elastic string can perform SHM along the element's direction. Measure displacement from the equilibrium position: doing so makes the constant weight term cancel and exposes the restoring force.

For a vertical element, let its equilibrium extension be $e$. Equilibrium gives $ke=mg$. If downward displacement from equilibrium is $x$, the extension is $e+x$ and$$m\ddot x=mg-k(e+x)=-kx.$$Gravity shifts the equilibrium position but does not appear in $\omega$ after this cancellation.

A $0.5$ kg particle oscillates vertically on an elastic element with stiffness $18\,\mathrm{N\,m^{-1}}$. Then $\omega=\sqrt{18/0.5}=6\,\mathrm{rad\,s^{-1}}$ and $T=\pi/3$ s. If its amplitude is $0.08$ m, its speed at displacement $0.04$ m is$$v=6\sqrt{0.08^2-0.04^2}=0.416\,\mathrm{m\,s^{-1}}.$$If the element is a string, this calculation applies for the whole oscillation only after checking that the minimum extension remains non-negative.