Practice question
Question
A particle’s displacement is \( x = 5 \sin (\pi t + \frac{\pi}{3}) \) (in m). What is its velocity at
\( t = 0 \, \text{s} \)? (Take \( \cos \frac{\pi}{3} = 0.5 \))
Explanation
**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Velocity: v = ω A cos (ω t + Φ) . A = 5 m, ω = π s⁻¹, Φ = (π/3) . At t = 0 : v = π × 5 cos ((π/3)) = 5π × 0.5 = 2.5π ≈ 7.85 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.