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

2024
Views:

394

Pages:

5
Author:

Jabin F.

# Object thrown upward with speed √(2gh) a = -g + 8t - g(8t)² ⇒ u = √(2gh) ⇒ If an object is thrown up with speed √(2gh) then it will go up to a height h. If speed = √(5gh) then object will go up to height (5/4)h. v = u - gt 0 = √(2gh) - gt ⇒ t = √(2h/g) The time for upward and downward motion is the same in the presence of gravity if there is no air drag. If there is air drag then t_up ≤ t_down particle doesn't return with the same speed. The time evolution of a coordinate x of a particle is described by the equation: (d⁴x/dt⁴) + 9Ω³(d³x/dt³) + (Ω⁴ - A⁴)x = 0 For Ω > A, the particle will: (a) eventually come to rest at the origin (b) eventually drift to infinity (c) oscillate about the origin (d) eventually come to rest at Ω/A Solution: (d⁴x/dt⁴) + 9Ω³(d³x/dt³) + (Ω⁴ - A⁴)x = 0 (D⁴ + 9Ω³D³ + Ω⁴ - A⁴)x = 0 (D³ + Ω³)² + (Ω⁴ - A⁴) = 0 D³ + Ω³ = ±i√(A⁴ - Ω⁴) D = ±i√(Ω² - A²), ±i√(Ω² + A²) x = C₁e^(i√(Ω² - A²)t) + C₂e^(-i√(Ω² - A²)t) + C₃e^(i√(Ω² + A²)t) + C₄e^(-i√(Ω² + A²)t) Image Analysis Visual Elements: Text: The primary content of the image consists of handwritten mathematical text, including equations, derivations, and calculations. Lines and Arrows: These elements are used to connect different parts of the calculations and indicate the flow of the solution. Symbols: Mathematical symbols like "m", "F", "t", "x", "ω", "γ", "D", "C", "P", and "E" are used throughout. Textual Content: Problem: A particle of mass m is subjected to a force F = Fo Sin(ωt) (time-dependent force). As the particle moves, a restoring force -mw²x acts on it, and also experiences a damping force -γmẋ. Write the equation of motion and solve it. Solution: EOM: Fo Sin(ωt) - γmẋ - mw²x = ma = mẍ ⇒ ẍ + γẋ + ω₀²x = (Fo/m) Sin(ωt) For C.F. auxiliary equation: D² + γD + ω₀² = 0 D = (-γ ± √(γ² - 4ω₀²)) / 2 For C.F.: x = C₁e^((-γ + √(γ² - 4ω₀²))t/2) + C₂e^((-γ - √(γ² - 4ω₀²))t/2) For P.I.: D² + γD + ω₀² = 0 ⇒ (D² + γD + ω₀²)x = (Fo/m) Sin(ωt) ⇒ x = (Fo/m) * (1 / (D² + γD + ω₀²)) * Sin(ωt) = (Fo/m) * (1 / ((ω₀² - ω²)² + 4γ²ω²)) * (4γ²ω² + (ω₀² - ω²)²) / ((ω₀² - ω²)² + 4γ²ω²) * Sin(ωt) = (Fo/m) * ((ω₀² - ω²)sin(ωt) + 4γ²ω²cos(ωt)) / ((ω₀² - ω²)² + 4γ²ω²)