A particle in SHM has \( x = 2 \sin (4t + \frac{\pi}{3}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)?
**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 = 4 s⁻¹, x(0) = 2 sin (π/3) = 2 × (√(3)/2) = √(3) ≈ 1.732 m . a = -4² × 1.732 = -16 × 1.732 ≈ -27.71 m/s² . Applying x = A cos(ωt + φ), v = -ωA
Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM