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

2024
Views:

125

Pages:

5
Author:

Jabin F.

# Oscillation of a black suspended from a rod -> Elastic force F = YARA / l = YA * X / l EOM -> mg - YA * X / l = mx'' x'' = - (YA / ml) * (X - mg * l / YA) as x = (w * t + x0) SHM mean position mg * l / YA w = sqrt(YA / ml) T = 2 * pi * sqrt(ml / YA) P-96 A solid vertical rod of length l and cross-section area A is made of a material of Young's modulus Y. The rod is loaded with a mass M and as a result, extends by a small amount Δl in the equilibrium condition. The mass is then suddenly reduced to M/2. As a result, the rod will undergo longitudinal oscillations with an angular frequency ω = ... Solution: Elastic force F = YAΔl / l = mg Now EOM when mass gets reduced to M/2: YA(Δl - x)/l - mg/2 = (M/2)a YAΔl/l - YAx/l - mg/2 = (M/2)a mg - YAx/l - mg/2 = (M/2)a a = - (gYA / ml)(1 - mgd / 2YR) mean position: mgd / 2YR ω = sqrt(RYA / mA) P-96 A monatomic gas of volume V is enclosed in a uniform vertical cylinder, the lower end of which is closed by a rigid wall and the upper by a frictionless piston. The piston is pressed lightly and released. Assume that the gas is a poor conductor of heat and cylinder and piston are perfectly insulating. If the cross-sectional area of the cylinder is A, the angular frequency of small oscillations of the piston about the point of equilibrium is: Solution: Thermodynamic gas: Cp / Cv = 1 + 3/5 = 5/3 = γ equilibrium: PA = mg walls and piston insulating, so no exchange of heat with the surrounding. So process is adiabatic: PV^γ = constant EOM: F_net = ma mg - PA - ma = 0 mg - mg(P + dP)A - ma = 0 mg - mg - (dP)A = ma a = - (A dP / ma) For adiabatic process PV^γ = constant take differentiation dP*V^γ + P*γ*V^(γ-1)*dV = 0 dP = -γ*P*dV = -γ*P*(V_f - V_i) = -γ*PAx/V put this value of dP in eq. (1) a = -γ*A*mg/V ω = sqrt(γ*A*mg/V) P-98 Two bodies A and B of equal mass are suspended from two rigid supports by separate massless springs having spring constant k₁ and k₂ resp. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of amplitude of oscillations of A to B is 1:3 Solution: given that v_max1 = v_max2 ω₁A₁ = ω₂A₂ A₂/A₁ = ω₁/ω₂ = sqrt(k₁/m) / sqrt(k₂/m) = sqrt(k₁/k₂) A particle is executing SHM with amplitude A about x=0. If T be the time period of oscillation, then time taken in going from x=0 to x=A/2 is Solution: equation for displacement x = A*sin(ωt + φ) at t=0, x=0 → 0 = A*sin(0 + φ) → sinφ = 0 → φ = 0 x = A*sin(ωt) at x = A/2, A/2 = A*sin(ωt) → 1/2 = sin(ωt) → ωt = sin^(-1)(1/2) = π/6 t = π/6ω = T/12