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

2024
Views:

154

Pages:

5
Author:

Jabin F.

# Simple Harmonic Motion Motion of an object is said to be SHM if force/acceleration is proportional to -ve displacement from mean position (equilibrium). F ∝ -x ma = -kx where x = displacement from x = 0 equilibrium position. a = -k/m x ⇒ a = -(ω²)x ⇒ ω = √(k/m) ẍ + ω²x = 0 whose solution is x = c₁cos(ωt) + c₂sin(ωt) x = A(sin(ωt + φ)) For SHM we can also have F = -k(x-x₀) ⇒ disp = (x-x₀) ma = -k(x-x₀) a = -(k/m)(x-x₀) ⇒ d²(x-x₀)/dt² + ω²(x-x₀) = 0 Solution from eq (2) (x-x₀) = A sin(ωt + φ) v = dx/dt = Aω cos(ωt + φ) a = d²x/dt² = -Aω² sin(ωt + φ) A ring is fixed on smooth horizontal plane. In figure shown, the coefficient of friction b/w block and ring is μ. If the block is given initial speed v₀ tangentially, calculate its speed after time t. Solution: Radial eom: f_r = ma_r -N = m(v²/r) Here N is variable as r is variable. N = mv²/r Tangential eom: f_t = ma_t -μN = m dv/dt dv = -μN/m dt ⇒ dv = -(μv²/r) dt ⇒ dv/v² = -μ/r dt ∫(1/v²) dv = ∫-μ/r dt -1/v = -μt/r + C When t = 0, v = v₀ -1/v₀ = C -1/v = -μt/r - 1/v₀ v = rv₀ / (rv₀ + μt) # Test Discussion Q. A particle of unit mass executes S.H.M. about the SEP of potential energy function, V(x) = (x²-1)²/2. Time period of oscillation is: Solution: V(x) = (x²-1)²/2 ⇒ V'(x) = x(x²-1) = 3x³ - 2x = 0 ⇒ x = 0, ±1 V''(x) = 6x²-2 At x = 0, V''(x) = -2 -ve (max) UEP At x = ±1, V''(x) = 4 +ve (min) SEP ⇒ k = 4 ω = √(k/m) = √(4/1) = 2 T = 2π/ω = 2π/2 = π