Velocity and acceleration in SHM

The reference particle moves along the tangent with speed ωA. Projecting that velocity on the x-axis gives v(t) = −ωA sin(ωt + φ), which is also what dx/dt gives.

Its centripetal acceleration ω²A points to the centre. Projecting it gives a(t) = −ω²A cos(ωt + φ) = −ω²x(t), which is also dv/dt.

The acceleration is proportional to the displacement and opposite in sign: when x > 0, a < 0, and when x < 0, a > 0. Whatever x is, the acceleration points towards the mean position.

With φ = 0: x = A cos ωt, v = −ωA sin ωt, a = −ω²A cos ωt. All three have the same period; relative to x, the velocity differs in phase by π/2 and the acceleration by π.

x ranges from −A to A, v from −ωA to ωA and a from −ω²A to ω²A. The velocity amplitude is v_m = ωA and the acceleration amplitude is a_m = ω²A.

Speed is greatest at the mean position and zero at the extremes; acceleration is zero at the mean position and greatest at the extremes.

Eliminating time with sin² + cos² = 1 gives the speed at displacement x: v = ω√(A² − x²).

Worked example: for x = 5 cos(2πt + π/4) in SI units, ω = 2π s⁻¹ and T = 1 s. At t = 1.5 s the phase is 3π + π/4, so x = −3.535 m, the speed is 10π × 0.707 ≈ 22 m s⁻¹ and a = −(2π)² × (−3.535) ≈ 140 m s⁻².