University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

221

Pages:

5
Author:

Jabin F.

# Bounded Motion in a given Potential Whether motion of a particle is bounded or not for a given energy actually depends on potential. Meaning of bounded motion in classical physics: If motion of a particle is confined in a region (doesn't extend to ∞), then it is called bounded motion. Condition of Bounded Motion Particle must be lying b/w two potential barriers (may be b/w two potential maxima). In other words, the particle must be in a potential well. Energy of particle must be less than the height of smaller barrier. Extension of Potential Well For bounded motion: Position of particle (x): x1 < x < x2 Energy of particle (E): E ≤ V1, V2 Example: A particle is moving in potential V(x) = x - (x^3)/3. For what max. energy, motion is bounded. Solution: Let us find potential energy at barriers (max). For max: dV/dx |_(x=xo) = 0 => 1 - xo^2 = 0 At xo = ±1 d^2V/dx^2 |_(x=xo) = -2xo At xo = -1, V''(x) = 2 = +ve (min) At xo = 1, V''(x) = -2 = -ve (max) ∴ There is max. potential at xo = 1. V(x) = x - x³/3 V(x) → ∞ as x → ∞ V(x) → -∞ as x → -∞ ∴ Max energy for which motion will be bounded is 1/3. A particle is moving in potential V(x) = x⁴/4 - x²/2. Above what energy motion is unbounded. Solution: V = x⁴/4 - x²/2 ⇒ dV/dx |_(x=x₀) = 0 ⇒ x³ - x = 0 (x - x₀)(x² - 1) = 0 ⇒ x₀ = 0, ±1 d²V/dx² |_(x=x₀) = 3x₀² - 1 At x₀ = 0, V''(x₀) = 1 = +ve (min.) At x₀ = ±1, V''(x₀) = -2 = -ve (max.) At x₀ = ±1, V(x₀) = -1/4 + 1/2 = 1/4. Above 1/4 energy potential, motion is unbounded P-6 Two particles each of mass m are fixed at y = ±a. A particle of mass m is released from x = -√3/2 a, its speed when it reaches x = 0 due to gravitational attraction of fixed masses is: Solution: Apply conservation of energy (as all forces are conservative) (KE + PE)i = (KE + PE)f 0 - 2Gm²/3a = 1/2mv² - 2Gm/a 4Gm/a[1 - 1/3] = 1/2mv² ⇒ v² = 8Gm/3a v = √(8Gm/3a)