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Question

A particle in SHM has \( x = 4 \cos (2t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \,
\text{s} \)? (Take \( \sin 60^\circ = \frac{\sqrt{3}}{2} \))

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Explanation

**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 = 4 m, ω = 2 s⁻¹, Φ = (π/3) . At t = 0.5 : 2 × 0.5 + (π/3) = 1 + (π/3) ≈ 2.047 rad ≈ 117° . v = -2 × 4 sin 117° ≈ -8 sin (180° - 63°) ≈ -8 × 0.838 ≈ -6.7 m/s

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