In a simple pendulum executing SHM, what ensures that the period is independent of the initial angular displacement?
**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. For small angles, the approximation sin θ ≈ θ holds, making the period T = 2π √((L/g)) independent of amplitude, a feature of linear systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The small angle approximation follows, reflecting SHM dependence on amplitude A, ω and system parameters.
Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM
