A particle in SHM has \( x = 3 \cos (2\pi t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)?
**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 2π s⁻¹, Φ = (π/3) . At t = 0.5 : 2π × 0.5 + (π/3) = π + (π/3) = (4π/3) . v = -2π × 3 sin (4π/3) = -6π sin (180° - 60°) = -6π (-(√(3)/2)) ≈ 16.31 m/s . Applying x =
Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance
