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

2024
Views:

40

Pages:

5
Author:

Jabin F.

# Normal Mode of Oscillations: When all particles of a system oscillate with the same frequency, then the oscillation is called normal mode. Frequency of normal mode is obtained by solving the determinant equation. V₁₁ - ω² = 0 or ω² - V₁₁ = 0 V₁₂ = V₂₁ is half coefficient of x₁x₂ T₁₁ and T₂₂ are coefficients of x₁² and x₂², respectively. T₁₂₁ is half of the coefficient of x₁x₂. D-19 - 27/8/19 # Normal Modes of Oscillations: Average frequency of normal modes: √((V₁₁ + V₂₂) / 2) ± √((V₁₁ - V₂₂)² / 4 + V₁₂²) P-95 The Lagrangian of a system is given by: L = 1/2 m₁x₁² + 1/2 m₂x₂² - 1/2 kx₁² - 1/2 kx₂² + kx₁x₂ where m₁ and m₂ are positive constants. The frequency of normal mode is: Solution: As L = 1/2 m₁x₁² + 1/2 m₂x₂² - 1/2 kx₁² - 1/2 kx₂² + kx₁x₂, T = 1/2 m₁x₁² + 1/2 m₂x₂² V = 1/2 kx₁² + 1/2 kx₂² - kx₁x₂ For normal modes: |V - ω²T| = 0 | (5k - m₁ω²) -k | | -k (5k - m₂ω²) | = 0 (5k - m₁ω²)(5k - m₂ω²) - k² = 0 5k - m₁ω² - m₂ω² + m₁m₂ω⁴ - k² = 0 m₁m₂ω⁴ - (5m₁ + 5m₂)kω² + 3k² = 0 ω² = (7 ± √(49 - 24)) / (4m₁) = (7 ± 5) / (4m₁) = 2k/m₁, k/2m₁ ω = ±√(3k/m₁), ±√(k/2m₁) P-15 For the coupled system shown in the figure, the normal coordinates are x₁ + x₂ and x₁ - x₂, corresponding to normal frequencies ω₁ = 3ω₀ and ω₂ = ω₀. Solution As x₁ + x₂ and x₁ - x₂ are normal coordinates: x₁ + x₂ = A₁cos(ω₁t) + B₁sin(ω₁t) x₁ - x₂ = A₂cos(ω₂t) + B₂sin(ω₂t) At t = 0, x₁ = A, x₂ = 0 and the velocities are ẋ₁ = ẋ₂ = 0. Therefore: A₁ + A₂ = A A₁ - A₂ = 0 Solving these equations, we get: A₁ = A₂ = A/2 At t = 0, ẋ₁ = ẋ₂ = 0: ω₁A₁sin(ω₁t) + ω₂A₂sin(ω₂t) = 0 ω₁A₁cos(ω₁t) + ω₂A₂cos(ω₂t) = 0 Substituting A₁ = A₂ = A/2: ω₁sin(ω₁t) + ω₂sin(ω₂t) = 0 ω₁cos(ω₁t) + ω₂cos(ω₂t) = 0 At t = 0, ω₁ + ω₂ = 0 Therefore: ω₂ = -ω₁ Substituting ω₂ = -ω₁ in the above equations: ω₁sin(ω₁t) - ω₁sin(-ω₁t) = 0 ω₁cos(ω₁t) - ω₁cos(-ω₁t) = 0 Since sin(-x) = -sin(x) and cos(-x) = cos(x), both equations are satisfied for any value of t.