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

2024
Views:

390

Pages:

5
Author:

Jabin F.

# Generalized Velocity (q̇): Total time derivative of generalized coordinate is called generalized velocity. It may be linear velocity or angular velocity or may be something else. For electric circuit, current is generalized velocity. P-15 The motion of a particle in 1-d is described by Lagrangian: L = (1/2) * (ẋ² - x²) The value of the action along the classical path from x = 0 at t = 0 to x = x₀ at t = t₀ is: Solution: L = (1/2) * (ẋ² - x²) Firstly, we have to express ẋ and x in terms of t. So, EOM: d/dt (∂L/∂ẋ) - (∂L/∂x) = 0 ẋ + x = 0 This is a 2nd order differential equation. Given: ẋ₀ = 0, x₀ = 0 Solution: x = C₁cos(t) + C₂sin(t) At t = 0, x = 0 → C₁ = 0 At t = 0, ẋ = 0 → C₂ = 0 Therefore, x = x₀sin(t) Now, Action: S = ∫₀ᵗ L dt = ∫₀ᵗ (1/2)(ẋ² - x²) dt S = (1/2) ∫₀ᵗ (x₀²cos²(t) - x₀²sin²(t)) dt S = (x₀²/2) ∫₀ᵗ (cos²(t) - sin²(t)) dt S = (x₀²/2) ∫₀ᵗ cos(2t) dt S = (x₀²/2) * (1/2)sin(2t) |₀ᵗ S = (x₀²/4) * (sin(2t) - sin(0)) S = (x₀²/4) * sin(2t) 2) A particle of mass m is dropped from height H near earth's surface. Calculate action integral over time of fall. Solution: KE = (1/2)mv², PE = -mgx ⇒ L = (1/2)mv² + mgx (PE = 0 at the initial position) L = (1/2)m(u + at)² + mg(ut + (1/2)at²) a = g ⇒ L = (1/2)mgt² + mg(ut + (1/2)gt²) ⇒ L = mgt² + (1/2)mgt² = (3/2)mgt² Time of fall T = √(2H/g) Action = ∫₀ᵀ L dt = ∫₀^(√(2H/g)) (3/2)mgt² dt = (mg/3) * (2H/g)^(3/2) = (mg/3) * (2H)^(3/2) / g^(3/2) = (2/3)mgH^(3/2)