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

2024
Views:

251

Pages:

5
Author:

Jabin F.

# Oscillation of a cylinder in a liquid for equilibrium B₁ = mg V₁g = mg Aρg = mg Aρ = m EOM when cylinder is forced into liquid. Fnet = ma mg - A(h + x)ρg = ma mg - Aρhg - Axρg = ma Aρlg - Axρg = ma Aρg(l - x) = ma (m/Aρ)g(l - x) = ma (l - x) = (a/m) * (m/Aρ) * g (l - x) = (a/Aρ) * g (l - x) = (a²/m) * (m/V) * g (l - x) = (a²/V) * g (l - x) = (a²/πa²h) * g (l - x) = (g/πh) ω² = (g/πh) ω = √(g/πh) P-45 A solid cylinder of height H, radius r, and density ρ floats vertically on the surface of a liquid of density ρ₀. The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is ω. Solution: From the previous article, we have ω = √(g/πh) Now, ρ = m/V = m/πr²h By putting this value, we get ω = √(g/πh) = √(ρ₀g/ρh) Q) In figure shown, the particle is thrown into the tunnel with speed \(v_{0}=\sqrt{\frac{GM}{R}}\). How much time will it take to cross the tunnel. Solution: At the end of the tunnel, speed is zero. So it is not the extreme point on amplitude Therefore, time taken to cross the tunnel is not T. So firstly we have to find out amplitude Now x = A sin(ωt + ϕ) - (1) at t = 0, x = R => R = A sin(ϕ) - (2) also dx/dt = ωA cos(ωt + ϕ) at t = 0, dx/dt = v₀ = ωA cos(ϕ) v₀ = √(GM/R) and ω = √(GM/R³) always = √(GM/R) -R = A cos(ϕ) - (3) [-ve sign for -ve direction of motion] Squaring and adding (2) and (3) 9R² = A² => A = √(9R) Now divide (3) by (2) -1 = (A cos(ϕ)) / (A sin(ϕ)) => tan(ϕ) = -1 => ϕ = tan⁻¹(-1) = (3π/4) t = (3π/4ω) = (3π/4) * √(R³/GM) Putting the value of x = -R, A = √(9R), ϕ = (3π/4) in equation (1) we get -R = √(9R)sin(ωt + (3π/4)) -1 = √(9)sin(ωt + (3π/4)) sin⁻¹(-1) = ωt + (3π/4) (5π/4) = ωt + (3π/4) (5π/4) - (3π/4) = ωt (π/2) = ωt t = (π/2ω) = (π/2) * √(R³/GM) Q) In figure shown, A sticks to B. What is the amplitude of oscillation of Combined system? Solution: apply conservation of momentum during Collision of A and B. (Σf)i = (Σf)f just before just after. A B v₀ + 0 = (m + M)v' m(v₀) = (m + M)v' v' = (mv₀)/(m + M) which is the max. velocity because it is at x = 0 displacement or elongation/compression of spring. Amax = (v₀/ω) = (v₀/√(k/m)) = v₀√(m/k) = (mv₀/√(mk)) = v₀√(m/g) In figure shown, black is oscillating with amplitude A. At mean position a black of mass M is placed on it gently. so that the two blacks move together. k 0000000000 m Aw before M k Calculate new amplitude of oscillation. Solution Apply conservation of momentum just before & just after. k M 0000000000 m Aw after m(ωA) + M(0) = (m + M)(ω'A') m√(k/m)A = (m + M)√(k/(m + M))A' √(m)A = √(m + M)A' A' = (A√(m))/(√(m + M))