Two particles execute SHM with x₁ = 5·sin(10t) cm and x₂ = 5·sin(10t + π/3) cm. Find (a) the phase difference and (b) the time lag of particle 2 behind particle 1. Step 1: (a) Δφ = φ₂ − φ₁ = π/3 − 0 = π/3 rad (60°). Step 2: (b) A phase difference Δφ corresponds to time lag Δt = Δφ/ω = (π/3)/10 = π/30 ≈ 0.105 s. Step 3: Check: period T = 2π/10 = 0.628 s; π/3 is 1/6 of 2π, and 0.105 s is 1/6 of T. Consistent. Answer: (a) π/3 rad, (b) ≈ 0.105 s lag. Particle 2 does everything particle 1 does, one-sixth of a cycle later.