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

2024
Views:

132

Pages:

6
Author:

Jabin F.

# Speed Calculation when V(x) is given -> by using conservation of energy (KE+PE); = (KE+PE)f 1/2 mv^2 + V(x) = 1/2 mv^2 + V(x) Example: A particle of unit mass is released from x=1/2 in the potential V(x) = x - x^3/3. Find speed of particle at x=-1. Solution: V(1/2) = 1/2 - 1/24 = 11/24 V(-1) = -1 + 1/3 = -2/3 Using conservation of energy: 1/2 m(v_i)^2 + V(x_i) = 1/2 m(v_f)^2 + V(x_f) Where v_i = 0 (as SEP) 0 + 11/24 = 1/2 m v^2 - 2/3 m v^2 = 1/4 v = √(3/2) Note: If ± √(3/2) are given in option then in this question we will choose -√(3/2) option because particle is going in -ve direction (1/2 to -1). Q.25 The time period of a particle of mass m, undergoing small oscillations around x=0, in the potential V = Vo cosh(x/L), is: Solution: V = Vo cosh(x/L) => V' = (Vo/L) sinh(x/L) = 0 => x0 = 0 at x = 0; k = V'' = (Vo/L^2) cosh(x/L) = (Vo/L^2) cosh(0) = Vo/L^2 ω = √(k/m) = √(Vo/mL^2) T = 2π/ω = 2π √(mL^2/Vo) Q. In figure shown, if m' is slightly displaced along dotted line then find angular frequency of oscillations. Solution: Potential energy of m' due to two masses M at distance √(d^2+y^2) is given as: V(y) = -2GMm / √(d^2+y^2) dV(y)/dy = -2GMm * (-1/2) * y(d^2+y^2)^(-3/2) = 0 y(d^2+y^2)^(-3/2) = 0 ⇒ y = 0 k = d²V(y)/dy² |_(y=0) = 2GMm / d³ ω = √(k/m) = √(2GM/d³) Q: In figure shown, if q is slightly displaced along the line, find the time period of oscillation. Solution: Potential energy of q due to two charges Q is given as: V(x) = (kQq/4πε₀) * (1/(d+x) + 1/(d-x)) = (kQq/4πε₀) * (2d/(d²-x²)) V'(x) = (kQq/4πε₀) * (2d/(d²-x²)) = 0 => x = 0 k = V''(x) = (kQq/4πε₀) * (2d/(d²-x²)) = (kQq/4πε₀) * (2d/(d³)) = (2kQq/4πε₀d³) ω = √(k/m) = √(2kQq/4πε₀md³) T = 2π/ω = 2π √(m4πε₀d³ / 2kQq)