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)) . 2 = 2π √((L/9.8)) ⇒ 1 = π √((L/9.8)) . √((L/9.8)) = (1/π) ⇒ (L/9.8) = (1/π²) ⇒ L = (9.8/π²) ≈ 1 m (using π² ≈ 9.87 ). Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²