Introduction
Potential energy curves U(x) visualize how potential energy varies with position, revealing crucial information about force, motion, equilibrium, and stability. These diagrams are powerful tools for understanding complex systems without solving detailed equations of motion.
Force from Potential Energy Curve
Force is the negative slope of the potential energy curve: F(x) = -dU/dx. Where slope is positive, force is negative (toward left). Where slope is negative, force is positive (toward right). Where slope is zero, force is zero (equilibrium). Steeper slope means stronger force. The shape of U(x) determines all force information.
Equilibrium Points
Equilibrium occurs where F = 0, i.e., where dU/dx = 0 (flat points on U curve). Types: (1) Stable equilibrium - minimum of U (small displacement produces restoring force); (2) Unstable equilibrium - maximum of U (small displacement produces force away from equilibrium); (3) Neutral equilibrium - flat region (no force anywhere).
\nIntroduction
Potential energy curves U(x) visualize how potential energy varies with position, revealing crucial information about force, motion, equilibrium, and stability. These diagrams are powerful tools for understanding complex systems without solving detailed equations of motion.
Force from Potential Energy Curve
Force is the negative slope of the potential energy curve: F(x) = -dU/dx. Where slope is positive, force is negative (toward left). Where slope is negative, force is positive (toward right). Where slope is zero, force is zero (equilibrium). Steeper slope means stronger force. The shape of U(x) determines all force information.
Equilibrium Points
Equilibrium occurs where F = 0, i.e., where dU/dx = 0 (flat points on U curve). Types: (1) Stable equilibrium - minimum of U (small displacement produces restoring force); (2) Unstable equilibrium - maximum of U (small displacement produces force away from equilibrium); (3) Neutral equilibrium - flat region (no force anywhere).
\nStability Analysis
Stability is determined by the second derivative: d2U/dx2 > 0 at minimum → stable (restoring force); d2U/dx2 < 0 at maximum → unstable; d2U/dx2 = 0 → need higher-order analysis. Near stable equilibrium, potential is approximately quadratic: U ≈ U_min + 1/2k_eff(x - x_eq)2, giving effective spring constant k_eff = d2U/dx2 at minimum.
Turning Points and Allowed Motion
For total energy E, kinetic energy K = E - U(x) must be non-negative. Motion is allowed only where U(x) ≤ E. Turning points occur where E = U(x) (K = 0, v = 0), where the particle reverses direction. Bound motion occurs when E < U(±∞) trapping particle in a potential well. Unbound motion when E > U at large distances.
\nExamples of Potential Curves
Harmonic oscillator: U = 1/2kx2 (parabola, single stable equilibrium). Pendulum: U = -mglcosθ (periodic, stable at θ = 0, unstable at θ = π). Lennard-Jones potential (intermolecular): attractive at long range, repulsive at short range, equilibrium at intermediate distance. Nuclear potential: combines Coulomb repulsion with strong nuclear attraction. Gravitational potential well around massive objects.
\nSolved Example: Particle in Potential Well
A particle moves in potential U(x) = 4x2 - x4 (in J, x in m). Find: (a) Equilibrium points, (b) Stability, (c) Turning points if total energy E = 3 J. Solution: (a) F = -dU/dx = -(8x - 4x3) = 4x(x2 - 2) = 0 at x = 0, x = ±√2 ≈ ±1.41 m. (b) d2U/dx2 = 8 - 12x2. At x = 0: d2U/dx2 = 8 > 0 → stable. At x = ±√2: d2U/dx2 = 8 - 24 = -16 < 0 → unstable. (c) Turning points where U(x) = E = 3: 4x2 - x4 = 3 → x4 - 4x2 + 3 = 0. Let y = x2: y2 - 4y + 3 = 0 → (y-1)(y-3) = 0 → y = 1 or 3. So x = ±1 and x = ±√3 ≈ ±1.73 m. Particle with E = 3 J oscillates between x = -1 and x = +1 m (inside potential well centered at origin), or escapes if it gets past the barrier at x = ±√2.
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