Reading a Potential Energy Curve — Question 3

Section 8-5 · Physics pages 1096–1097

Question 3 Disconnected allowed regions in a double well Assume one-dimensional classical motion with K = E −U(x) and Fx = −dU/dx, with no dissipation. Let U(x) = (x2 −1)2 J, with x in meters, and E=0.25 J. The particle starts in the left well. x U E = .25 −1 1 1 J A low energy line cuts the double well into two separate allowed regions. (a) Find every allowed x interval. (b) Can it reach the right well without external energy transfer? (c) Locate the stable and unstable equilibria and compare the barrier with E.
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Question 3 - Solution Disconnected allowed regions in a double well x U E = .25 −1 1 1 J (a) Solve the inequality in two stages. Solve the allowed-energy inequality in stages: (x2 −1)2 ≤.25 ⇒|x2 −1| ≤.5 ⇒.5 ≤x2 ≤1.5. x ∈[− √ 1.5, − √ .5] ∪[ √ .5, √ 1.5] m . (b) Keep the trajectory in a connected allowed region. The central barrier is forbidden: U(0) = 1 J > E = .25 J. A left-well trajectory cannot reach the disconnected right allowed interval without additional energy. Allowed endpoint energy alone does not establish reachability. (c) Read extrema and local force directions. Stationary points satisfy U′ = 4x(x2 −1) = 0 ⇒x = 0, ±1, U′′ = 12x2 −4. Thus x = ±1 stable; x = 0 unstable . The barrier is 0.75 J above the trajectory energy.

Question 3

Disconnected allowed regions in a double well

Assume one-dimensional classical motion with K=E−U(x)K=E-U(x) and Fx=−dU/dxF_x=-dU/dx, with no dissipation. Let U(x)=(x2−1)2U(x)=(x^2-1)^2 J, with x in meters, and E=0.25 J. The particle starts in the left well.

Diagram shown in the original SVG worksheet.

  1. Find every allowed x interval.

  2. Can it reach the right well without external energy transfer?

  3. Locate the stable and unstable equilibria and compare the barrier with E.

Question 3 - Solution

Disconnected allowed regions in a double well

Diagram shown in the original SVG worksheet.

(a) Solve the inequality in two stages.

Solve the allowed-energy inequality in stages: (x2−1)2≤.25⇒|x2−1|≤.5⇒.5≤x2≤1.5.(x^2-1)^2\leq.25\Rightarrow|x^2-1|\leq.5\Rightarrow.5\leq x^2\leq1.5. x∈[−1.5,−.5]∪[.5,1.5]m.\boxed{x\in[-\sqrt{1.5},-\sqrt{.5}]\ \cup\ [\sqrt{.5},\sqrt{1.5}]\ \mathrm m}.

(b) Keep the trajectory in a connected allowed region.

The central barrier is forbidden: U(0)=1J>E=.25J.U(0)=1\ \mathrm J>E=.25\ \mathrm J. A left-well trajectory cannot reach the disconnected right allowed interval without additional energy. Allowed endpoint energy alone does not establish reachability.

(c) Read extrema and local force directions.

Stationary points satisfy U′=4x(x2−1)=0⇒x=0,±1,U″=12x2−4.U'=4x(x^2-1)=0\Rightarrow x=0,\pm1,\qquad U''=12x^2-4. Thus x=±1 stable;x=0 unstable\boxed{x=\pm1\text{ stable};\quad x=0\text{ unstable}}. The barrier is 0.75 J above the trajectory energy.