Practice question
Question
A particle in SHM has \( x = 4 \cos (2t - \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \,
\text{s} \)? (Take \( \sin 30^\circ = 0.5 \))
Explanation
**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 2 s⁻¹, Φ = -(π/6) . At t = 0.5 : 2 × 0.5 - (π/6) = 1 - (π/6) ≈ 0.476 rad ≈ 27.3° . v = -2 × 4 sin (27.3°) ≈ -8 × 0.46 ≈ -3.68 m/s
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