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

2024
Views:

44

Pages:

5
Author:

Jabin F.

# Angular Frequency in given Potential V(x) Let x₀ is stable equilibrium point. Then Taylor series expansion of V(x) about x₀ is: V(x) = V(x₀) + (x - x₀) (dV/dx)|x₀ + (1/2)(x - x₀)² (d²V/dx²)|x₀ + (1/3!)(x - x₀)³ (d³V/dx³)|x₀ + ... V'(x₀) = 0 as x₀ is stable equilibrium point ⇒ V(x) = V(x₀) + (x - x₀)² (d²V/dx²)|x₀ + (1/3!)(x - x₀)³ (d³V/dx³)|x₀ + ... Then force becomes: F = -dV/dx = -(x - x₀) (d²V/dx²)|x₀ - (1/2)(x - x₀)² (d³V/dx³)|x₀ - ... If (x - x₀) is small then neglect higher power terms of (x - x₀) in F, then F = -(x - x₀)k where k = (d²V/dx²)|x₀ ⇒ md²x/dt² = -(x - x₀)k ⇒ d²x/dt² + (k/m)(x - x₀) = 0 Let (x - x₀) = x' ⇒ d²x'/dt² + (k/m)x' = 0 This is S.H.M. ω² = k/m ⇒ ω = √(k/m) [Diagram of a potential energy well with a stable equilibrium point at x₀] [\frac{d^{2}x^{\prime}}{dt^{2}}+\frac{k}{m}x^{\prime}=0] whose solution is [x^{\prime}=c_{1}cos(\sqrt{\frac{k}{m}}t)+c_{2}sin(\sqrt{\frac{k}{m}}t)] where [\sqrt{\frac{k}{m}}] is angular frequency of oscillation. Conclusion: If V(x) is given and ω is to be found then use following steps: [\frac{dV}{dx}|_{x=x_{0}}=0] (find x₀) [k=\frac{d^{2}V}{dx^{2}}|_{x=x_{0}}=+\nu_{e}(f)in~d~k] [\omega=\sqrt{\frac{k}{m}}] Note: If V(x) is not given and system is described then first write V(x) to find ω. Note: If k is zero then ω is calculated using time energy relation. [\omega=\frac{2\pi}{T}] or [\omega=3\pi\mathcal{J}] where T is time period of oscillations. Example: A particle of mass m is oscillate about stable equilibrium point of a potential (V(x)=m[\frac{1}{x^{3}}+x^{5}]) , find ω. Solution: [V^{\prime}(x)|_{x=x_{0}}=m[-\frac{2}{{x_{o}}^{3}}+2x_{o}]=0] [x_{\bullet}^{4}=1] [x_{0}=\pm1,\pm i\Rightarrow x_{0}=\pm1] (k=:V^{\prime\prime}(x)|_{x=x_{0}}=m[\frac{6}{x_{o}}{4}}+s]=8\alpha) using values of (x{D}) . [\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{8\alpha}{m}}\underline{Ay}] P-59 A particle of mass m is at the stable equilibrium position of its potential energy V(x) = ax - bx³, where a, b are positive constants. The minimum velocity that has to be imparted to the particle to render its motion unstable is Solution: First we have to find SEP & USEP and Potential energy at those points. V'(x) = a - 3bx² = 0 ⇒ x₀ = ±√(a/3b) V''(x₀) = -3bx₀ at x₀ = √(a/3b), V''(x₀) = -6b√(a/3b) = -Ve (max. USEP) at x₀ = -√(a/3b), V''(x₀) = 6b√(a/3b) = +Ve (min. SEP) Since particle is at SEP, therefore to render its motion unstable extra kinetic energy given to it equal to potential energy difference such that it crosses the boundary. (1/2)mv² = V(x₀ = √(a/3b)) - V(x₀ = -√(a/3b)) (1/2)mv² = (a√(a/3b) - (a/3)√(a/3b)) - (-a√(a/3b) + (a/3)√(a/3b)) (1/2)mv² = (4a/3)√(a/3b) v² = (8a/3m)√(a/3b) = (64a³/97m²b)^(1/2) v = (64a³/97m²b)^(1/4)