Practice question
Question
A particle’s displacement is \( x = 8 \sin (2\pi t - \frac{\pi}{3}) \) (in m). What is its velocity at
\( t = 0.25 \, \text{s} \)? (Take \( \cos 30^\circ = \frac{\sqrt{3}}{2} \))
Explanation
**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Velocity: v = ω A cos (ω t + Φ) . A = 8 m, ω = 2π s⁻¹, Φ = -(π/3) . At t = 0.25 : 2π × 0.25 - (π/3) = (π/2) - (π/3) = (π/6) . v = 2π × 8 cos (π/6) = 16π × (√(3)/2) ≈ 43.54 m/s . Applying x = A cos(ωt + φ), v =
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