[NOTES/ME-06002]-Using graph of $V(x)$ to find motion

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You can easily see how several statement about the motion of a particle can be inferred from a graph of potential energy, if the value of energy is known. To be able to use plot of potential energy, we need the concept of a turning point.

Turning Point

As a preparation to the general discussion using plot of potential energy, we explain what is a turning point.

A point $x$ where the velocity is zero is called a {\it turning point}. If $x_0$ is a turning point, using the energy equation $E= \frac{1}{2} mv^2 + V(x)$, we see that turning points are obtained by solving the equation \begin{equation}\label{EQ13} E= V(x_0). \end{equation} It is useful to learn how solution of \eqRef{EQ13} can be obtained graphically.

Solving equations graphically:

Solutions of a nonlinear equation $f(x)=g(x)$ can be obtained graphically. This is done by plotting curves $y=f(x)$ and $y=g(x)$ and finding points of intersection, see \Figref{GraphicSolve}(a). }\end{ColorMiniPage}\\ \fbox{\parbox{14cm}{\FigBelow{20,10}{100}{60}{GraphicSolve}{}}}\\ \noindent \begin{ColorMiniPage}{14cm}{MyLightYellow}{ In \Figref{GraphicSolve}(b), plot of a typical potential is shown. To find graphical solution for turning point we draw line $V(x)=E$ and see where it intersects the graph of the potential. In the present example there are three turning points $x_1, x_2, x_3$. }

1.Qualitative properties of motion in one dimension

We now begin our discussion of general properties of motion in one dimension from the plot of potential energy.

Since $E=$ K.E.+ P.E., and kinetic energy is always positive, we must have $E \ge V(x)$ and the motion is restricted to regions of $x$ where this inequality is obeyed. Thus a particle cannot go to regions where $E \le V(x)$. The region of $x$ where $E>\text{P.E.}$ is said to be {\tt classically accessible.}
At a turning point $E=V(x)$ and the velocity, $\dot{x}= \sqrt{\frac{2}{m}(E-V(x))}$, of the particle becomes zero. The subsequent motion of the particle will be in the direction of the force. Remembering that the force is $F(x)=-\dd[V(x)]{x}$, the force will be in positive $x$- direction if the slope is negative. If the slope of potential energy curve is positive the force will be in the negative $x$- direction. It is apparent that at a turning point, the particle will move in the direction of decreasing potential.
At points where the slope is zero, the force is zero and the particle will continue to move in the direction of velocity.
Let $x=a$ be a point where the potential is maximum or minimum,then the derivative $\dd[V(x)]{x}\big|_{x=a}$ vanishes. Thus the force vanishes. If in addition the energy is equal to the potential energy at that point $V(a)=E$, the velocity of the particle will be zero.\par In this case the particle will be in equilibrium at $x=a$.
At a maximum of potential energy a light 'push' will make it move away from the equilibrium point and we say that the maxima of potential energy correspond to points of unstable equilibrium.
At a minimum of potential energy, a light 'push' will make it move move back towards the equilibrium point and we say that the minima of potential energy correspond to points of stable equilibrium.

2. Examples

We will take up a few examples. Let us look at the motion of a particle under influence of a potential \begin{equation} V(x) = \frac{1}{2}m\omega^2(a x^2-bx^4) \end{equation} We shall be interested in three cases, (i) $a> 0, b>0$ \ (ii) $a<0, b>0$ (iii) $a>0, b<0$.The plot of the potential for the three cases is shown below. \FigBelow{30,10}{90}{60}{3Potentials}{}

Case(i) $a>0,b>0$} In this case the potential has a minimum at $x=0$ and goes to infinity as $x\to \infty$. For any energy $>0$, there are two turning points and the motion is confined to $x$ values between the turning points. Thus the motion is bounded and the particle cannot go to infinity. \FigBelow{40,05}{60}{75}{3PotsA}{Case(i)}

Case (ii) $a< 0, b>0$} In this case the potential is a double well potential with a maximum at $x=0$ and two minima given by $ x_0 = \sqrt{|a/2b|}\equiv x_0$ at minimum the potential takes value $V_0=-\frac{a^2}{2b}$.\\ For $ E>0V_\text{min},\text{and} E<0$ the motion is confined between one of the the two regions $PQ$ or $RS$ of the $x$-axis, see \Figref{3PotsB}. The the initial position of the particle is in the left well, the particle remains confined to $-x_0 <x<0$. if= the= initial= position= is= in= right= well= it= remains= {\it= i.e.}= will= have= restricted= to= $rs$.= for= $e=>0$, there are two turning points and the motion is confined between $MN$. \FigBelow{45,10}{60}{70}{3PotsB}{Case(ii)}

Case (iii) $a> 0, b<0$ } In the case the potential has maxima at $\pm x_0$ and has a double well shape for $|x|< x_0$ and a minimum at $x=0$. Outside this range potential monotonically decreases and tends to $- infty$ as $x\to \pm \infty$. The points $x=\pm x_0$ corresponds to an unstable equilibrium. Given a small push, in either direction, it will continue moving in that direction and go to $\pm\infty$. The points $x=0$ is a point of stable equilibrium and the particle oscillates about the origin if it has $E<0$. For example between $MN$ for $E=E_0$, see \Figref{3PotsC}\\ \FigBelow{40,10}{60}{70}{3PotsC}{Case(iii)}