Reading a Potential Energy Curve — Question 7

Question 7

Approaching the top at exactly the barrier energy

For $-1\le x\le1$ m, a 2 kg particle has $U(x)=4-x^2$ J and E=4 J. It starts at x=$-1$ m moving right. Use $v=dx/dt$ and the positive velocity branch before reaching zero.

1. Find velocity as a function of x for x<0.

2. Solve the resulting first-order equation for x(t), taking t=0 at release.

3. Does it cross the maximum in finite time? What changes if E is slightly above 4 J?

Question 7 - Solution

Approaching the top at exactly the barrier energy

(a) Use the energy gap with its correct square root.

The energy gap is $K=x^2$ in the supplied SI numerical form. For $m=2$ kg and rightward motion, $$v=\sqrt{2K/m}=|x|/(1\ \mathrm s)\Rightarrow\boxed{v=-x/(1\ \mathrm s)}\quad(x<0).$$

(b) Integrate the approach equation.

Separate variables, writing $\tau=1$ s: $$\frac{dx}{x}=-\frac{dt}{\tau}\Rightarrow\ln\frac{|x(t)|}{1\ \mathrm m}=-\frac t\tau.$$ $$\boxed{x(t)=-(1\ \mathrm m)e^{-t/(1\ \mathrm s)}}.$$

(c) Check the time limit.

The approach time diverges: $$t=\tau\ln\frac{1\ \mathrm m}{|x|}\longrightarrow\infty\quad\text{as }x\to0^-.$$ At $E=4$ J the particle never crosses in finite time. For $E>4$ J, $v(0)=\sqrt{2(E-4\ \mathrm J)/m}>0$, allowing passage across the maximum.