Practice question

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. T = 2π √((L/g)) . 1.4 = 2π √((L/9.8)) ⇒ √((L/9.8)) = (1.4/2π) ≈ 0.223 . (L/9.8) = (0.223)² ⇒ L ≈ 9.8 × 0.0497 ≈ 0.487 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.487 m