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

2024
Views:

414

Pages:

5
Author:

Jabin F.

# Stable Equilibrium Point (SEP): The equilibrium point to which a particle returns on being slightly displaced is called SEP. F = -dV/dx = -slope = -tanθ tanθ = +ve; θ < 90° tanθ = -ve; θ > 90° At SEP, potential energy is minimum. d²V/dx² > 0 (+ve) P-98 A solid spherical cork of radius R and specific gravity 0.5 floats on water. The cork is pushed down so that its center of mass is at a distance h (where 0 < h < R) below the surface of water and then released. The volume of the part of the cork above the water level is: πR³(1/3 - cosθ + 1/3cos³θ) where θ is the angle as shown in figure. At the moment of release, the dependence of the upward force on the cork on h is: (A) h/R - 1/3(h/R)³ (B) h/R + 1/3(h/R)³ (C) -h/R + 1/3(h/R)³ (D) -h/R - 1/3(h/R)³ Solution: Specific Gravity = Density of object / Density of water 0.5 = ρo / ρw → ρo = 0.5ρw B = V_object * ρ_liquid * g = πR³[1 - (1/3 - cosθ + 1/3cos³θ)]ρw * g = πR³[2/3 + cosθ - 1/3cos³θ]ρw * g ∴ B = πR³[2/3 + h/R - 1/3(h/R)³]ρw * g is upward force. Now downward force, mg = Vρg = (4/3)πR³(0.5ρw)g = (2/3)πR³ρw * g Fnet = Fup - Fdown = πR³[2/3 + h/R - 1/3(h/R)³]ρw * g - (2/3)πR³ρw * g Fnet = πR³[h/R - 1/3(h/R)³]ρw * g Fnet ∝ h/R - 1/3(h/R)³ P-90 A particle of mass, m = 4, moves along the x-axis under the influence of the potential V(x) = 9 [e^(-3x/4) - 2e^(3x/4)]. If the particle oscillates with small amplitude around the minimum potential, what is the period of oscillation? Solution: V(x) = 9 [e^(-3x/4) - 2e^(3x/4)] V'(x) = 9 [-3/4 e^(-3x/4) + 3/2 e^(3x/4)] = 0 ⇒ -e^(-3x/4) + 2e^(3x/4) = 0 ⇒ e^(-3x/4) = e^(3x/4) ⇒ -3x/4 = 3x/4 ⇒ x₀ = 0 k = V''(x₀) = 9 [9/16 e^(-3x/4) + 9/4 e^(3x/4)] |ₓ₀ = 9 [9/16 + 9/4] = 9 ⇒ w = √(k/m) = √(9/4) = 3/2 ⇒ T = 2π/w = 9π/4 = 8π = 8.3776 Ans