The displacement-time graph of an SHM is a sine wave with amplitude 4 cm and period 2 s. At t = 0 the curve passes through zero going upward. Write x(t), v(t) and a(t), and find the velocity at t = 0.5 s. Step 1: A = 0.04 m, T = 2 s → ω = 2π/T = π rad/s. x(t) = 0.04·sin(πt). Step 2: v(t) = dx/dt = 0.04π·cos(πt); a(t) = −0.04π²·sin(πt). Step 3: At t = 0.5 s: v = 0.04π·cos(π/2) = 0. Answer: x = 0.04·sin(πt) m, v = 0.04π·cos(πt) m/s, a = −0.04π²·sin(πt) m/s²; v(0.5 s) = 0 — at t = T/4 the particle is at the extreme +A, momentarily at rest, exactly as the graph shows.