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°. Period: T = (1/v) = (1/0.3) ≈ 3.333 s . T = 2π √((L/g)) ⇒ 3.333 = 2π √((L/9.8)) . √((L/9.8)) = (3.333/2π) ≈ 0.531 ⇒ (L/9.8) = (0.531)² ⇒ L ≈ 2.76 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =