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

2024
Views:

408

Pages:

5
Author:

Jabin F.

# Constraints (Kinematicel Conditions): Types of constraint -> 1) Holonomic -> If constraint can be expressed as algebraic equation in terms of co-ordinates then it is called holonomic. 2) Non-Holonomic -> If constraint is expressed as inequality or non- solvable differential equations then it is called non-holonomic 3) Scleronomic -> If constraint relation doesn't contain time explicitly then it is called scleronomic. 4) Rheonomic -> Opposite of scleronomic. Constraint contains time explicitly. Q2 A particle of mass m is executing SHM with angular frequency ω and amplitude A. Calculate average KE and average Power during oscillation. Solutions: y = A sin(ωt + φ) v = Aω cos(ωt + φ) a = -Aω² sin(ωt + φ) KE = 1/2 mv² = 1/2 mω²A² cos²(ωt + φ) = 1/2 mω²A² = 1/2 mω²A² * 1/2 = 1/4 mω²A² = = = <-mAω² sin(ωt + φ)(Aω cos(ωt + φ))> = <-mA²ω³ sin(ωt + φ)cos(ωt + φ)> = -mA²ω³ * 1/2 = 0 Q3 A is attached to spring, B is released from position as shown in figure. If B sticks to A calculate amplitude of oscillation of A & B together. Solution: Since (9m) is oscillating ... ω = √(k/(9m)) When B reaches to A its speed is v0 = √(9g/k) = √(9gmg/k) = 9g√(m/k) Now apply conservation of momentum just before and after the collision m0v0 + mv0 = (m0 + m)v For oscillation: x = A sin(ωt + φ) at t = 0, x = mg/k [from equilibrium mg = kx by black A on spring] Now 0 = A sin φ at t = 0, v = Aω cos φ = √(k/m) * A * cos φ = √(k/9m) * A * cos φ Squering and adding: (mg/k)² + (√(k/9m) * A * cos φ)² = A² (mg/k)² + 9(mg/k)² = A² => A = √10 * mg/k P-30 Hamiltonian of a system with generalized coordinate and momentum (q, p) is H = p²q³. A solution of the Hamiltonian equation of motion is: (a) p = Be^(-At), q = Ae^(-At) (b) p = -Ae^(-At), q = Ae^(-At) (c) p = Ae^(At), q = Ae^(-At) (d) p = -Ae^(At), q = Ae^(At) where A and B are constants. Solution: H = p²q³ ∂H/∂q = 3p²q² ∂H/∂p = 2pq³ q̇ = ∂H/∂p = 2pq³ ṗ = -∂H/∂q = -3p²q² => 1/q dq/dt = 1/p dp/dt => ln q = ln c - ln p => ln q = ln(c/p) => q = c/p using this relation in Hamilton's eom q̇ = ∂H/∂p = 3p²q² = 3(c/q)²q² = 3cq² ṗ = -∂H/∂q = -3p²q² = -9(c/q)²q² = -9cp dq/dt = 3cq² dp/dt = -9ct ∫dq/q = ∫3cdt ∫dp/p = -∫9cdt ln q = 3ct + ln c₁ ln p = -9ct + ln c₂ ln q/c₁ = 3ct ln p/c₂ = -9ct q = c₁e^(3ct) p = c₂e^(-9ct) These results will only match with option (a)