Practice question

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Acceleration: a = -ω² x . ω = 2π s⁻¹, x = 0.06 cos (2π × 0.25 + (π/4)) = 0.06 cos ((π/2) + (π/4)) = 0.06 cos (3π/4) . cos (3π/4) = -(√(2)/2) , so x = -0.06 (√(2)/2) ≈ -0.0424 m . a = -(2π)² ×