Electric Potential Energy — Question 7

Section 24-1 · Physics pages 3804–3805

Question 7 A double-well electric energy distinguishes local stability from barrier crossing A constrained charged particle has prescribed electric energy U(x) = U0[(x/a)2 −1]2 with U0, a > 0. Its mass is m and no other forces do work. x U −a a Schematic; labels, not drawing lengths, specify the data. (a) Find all equilibria and classify them using the second derivative of U. (b) Find the small-oscillation frequency about x=a and the threshold launch speed from x=-a for crossing toward the other well.
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Question 7 - Solution A double-well electric energy distinguishes local stability from barrier crossing x U −a a (a) Method and calculation U′ = 4U0x a2 ( x2 a2 −1 ) = 0 ⇒x = 0, ±a. U′′ = 4U0 a2 ( 3x2 a2 −1 ) : U′′(0) < 0, U′′(±a) = 8U0/a2 > 0. (b) Interpretation and further calculation ω = √ U′′(a)/m = √ 8U0/(ma2). 1 2mv2 crit = U(0) −U(−a) = U0 ⇒vcrit = √ 2U0/m. Finite-time crossing requires a speed strictly above the threshold. Physical check Stable small oscillations about one minimum do not rule out crossing to another minimum when enough energy is supplied.

Question 7

A double-well electric energy distinguishes local stability from barrier crossing

A constrained charged particle has prescribed electric energy U(x)=U0[(x/a)2−1]2U(x)=U_0[(x/a)^2-1]^2 with U0,a>0U_0,a>0. Its mass is m and no other forces do work.

Diagram shown in the original SVG worksheet.

  1. Find all equilibria and classify them using the second derivative of U.

  2. Find the small-oscillation frequency about x=a and the threshold launch speed from x=-a for crossing toward the other well.

Question 7 - Solution

A double-well electric energy distinguishes local stability from barrier crossing

Diagram shown in the original SVG worksheet.

(a) Method and calculation

U′=4U0xa2(x2a2−1)=0⇒x=0,±a.U\prime=\frac{4U_0x}{a^2}\left(\frac{x^2}{a^2}-1\right)=0\Rightarrow x=0,\ \pm a.U′′=4U0a2(3x2a2−1):U′′(0)<0,U′′(±a)=8U0/a2>0.U\prime\prime=\frac{4U_0}{a^2}\left(\frac{3x^2}{a^2}-1\right):\quad U\prime\prime(0)<0,\quad U\prime\prime(\pm a)=8U_0/a^2>0.

(b) Interpretation and further calculation

ω=U′′(a)/m=8U0/(ma2).\omega=\sqrt{U\prime\prime(a)/m}=\sqrt{8U_0/(ma^2)}.12mvcrit2=U(0)−U(−a)=U0⇒vcrit=2U0/m.\tfrac12mv_{\mathrm{crit}}^2=U(0)-U(-a)=U_0\Rightarrow v_{\mathrm{crit}}=\sqrt{2U_0/m}.Finite-time crossing requires a speed strictly above the threshold.

Physical check

Stable small oscillations about one minimum do not rule out crossing to another minimum when enough energy is supplied.