Practice question
Question
A particle in SHM has \( x = 4 \sin (4t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.25 \,
\text{s} \)? (Take \( \cos 60^\circ = 0.5 \))
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 cos (ω t + Φ) . A = 4 m, ω = 4 s⁻¹, Φ = (π/6) . At t = 0.25 : 4 × 0.25 + (π/6) = 1 + (π/6) ≈ 1.523 rad ≈ 87° . v = 4 × 4 cos 87° ≈ 16 × 0.052 ≈ 0.832 m/s . Applying x = A cos(ωt
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