Introduction
In SHM, energy oscillates between kinetic and potential forms. Total mechanical energy is conserved and proportional to amplitude squared, providing powerful methods for analyzing oscillating systems.
Kinetic Energy
K = 1/2mv2 = 1/2mA2ω2sin2(ωt + φ). Maximum at equilibrium (sin2 = 1): K_max = 1/2mω2A2 = 1/2kA2. Zero at turning points. Average over cycle: ⟨K⟩ = 1/4kA2.
Potential Energy
U = 1/2kx2 = 1/2kA2cos2(ωt + φ). Maximum at turning points: U_max = 1/2kA2. Zero at equilibrium. Average over cycle: ⟨U⟩ = 1/4kA2. For spring: U = 1/2kx2. For pendulum: U ≈ 1/2mglθ2.
Total Energy
E = K + U = 1/2kA2 = constant. Total energy depends only on amplitude, not time. At any instant, E = 1/2mv2 + 1/2kx2. Energy conservation provides alternative solution method: solve for velocity at any position.
Energy Diagram
Parabolic potential U = 1/2kx2. Horizontal line at E = 1/2kA2 represents total energy. Intersection with U at x = ±A are turning points. K is vertical gap between E and U curves. Width of allowed region increases with E.
\nIntroduction
In SHM, energy oscillates between kinetic and potential forms. Total mechanical energy is conserved and proportional to amplitude squared, providing powerful methods for analyzing oscillating systems.
Kinetic Energy
K = 1/2mv2 = 1/2mA2ω2sin2(ωt + φ). Maximum at equilibrium (sin2 = 1): K_max = 1/2mω2A2 = 1/2kA2. Zero at turning points. Average over cycle: ⟨K⟩ = 1/4kA2.
Potential Energy
U = 1/2kx2 = 1/2kA2cos2(ωt + φ). Maximum at turning points: U_max = 1/2kA2. Zero at equilibrium. Average over cycle: ⟨U⟩ = 1/4kA2. For spring: U = 1/2kx2. For pendulum: U ≈ 1/2mglθ2.
Total Energy
E = K + U = 1/2kA2 = constant. Total energy depends only on amplitude, not time. At any instant, E = 1/2mv2 + 1/2kx2. Energy conservation provides alternative solution method: solve for velocity at any position.
Energy Diagram
Parabolic potential U = 1/2kx2. Horizontal line at E = 1/2kA2 represents total energy. Intersection with U at x = ±A are turning points. K is vertical gap between E and U curves. Width of allowed region increases with E.
\nVirial Theorem
For SHM: ⟨K⟩ = ⟨U⟩ = 1/2E. Equal time-averaged kinetic and potential energy is characteristic of harmonic oscillators. This relates to equipartition theorem in statistical mechanics.
Solved Example: Energy in Oscillator
A 2 kg mass on a spring (k = 500 N/m) oscillates with amplitude 0.2 m. Find: (a) Total energy, (b) Maximum velocity, (c) Velocity at x = 0.1 m, (d) Energy distribution at x = 0.1 m. Solution: (a) E = 1/2kA2 = 1/2 × 500 × (0.2)2 = 250 × 0.04 = 10 J. (b) From E = 1/2mv_max2: v_max = √(2E/m) = √(20/2) = √10 = 3.16 m/s. Or v_max = Aω = 0.2 × √(500/2) = 0.2 × 15.8 = 3.16 m/s ✓. (c) At x = 0.1 m: E = 1/2kx2 + 1/2mv2 → 10 = 1/2 × 500 × (0.1)2 + 1/2 × 2 × v2 → 10 = 2.5 + v2 → v2 = 7.5 → v = 2.74 m/s. (d) At x = 0.1 m: U = 1/2kx2 = 2.5 J (25% of total), K = 7.5 J (75% of total). At equilibrium: U = 0, K = 10 J. At turning points: U = 10 J, K = 0.
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