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Question

A particle’s displacement is \( x = 5 \sin (2\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at
\( t = 0.5 \, \text{s} \)? (Take \( \sin 60^\circ = \frac{\sqrt{3}}{2} \))

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Explanation

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. Velocity: v = ω A cos (ω t + Φ) . A = 5 m, ω = 2π s⁻¹, Φ = (π/6) . At t = 0.5 : 2π × 0.5 + (π/6) = π + (π/6) = (7π/6) . v = 2π × 5 cos (7π/6) = 10π cos (180° - 30°) = 10π (-(√(3)/2)) ≈ -27.14 m/s . Applying x = A cos(ωt + φ), v

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