Practice question

**Pendulum motion** exhibits isochronism for small amplitudes, period depends only on L and g. Using g = 9.8 m/s², T calculation requires √(L/g), frequency reciprocal of period. Angular frequency directly √(g/L), e.g., L = 0.25 m gives T ≈ 1.0 s. T = 2π √((L/g)) = 2π √((1.5/10)) = 2π √(0.15) ≈ 2 × 3.14 × 0.387 ≈ 2.43 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.43 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.