A simple pendulum of length 1m is oscillating with an angular frequency  10 rad s-1. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad s-1 and an amplitude of 10 -2m . The relative change in the angular frequency of the pendulum is best given by(take acceleration due to gravity g=10m/s2)

ω=10 rad s-1,g=10 m s-2gmax=g + a gmin=g - a difference of above two equations is Δg, Δg=2a=( 2)acceleration,and a=ωsupport2A, here A is amplitude, Δg=2ωs2A---(1) ωsupport=ωs=1 rad.s-1 A=10-2m ω=geffl differentiate above equation ⇒Δωω=12Δgg Δω=12Δgg×ω substitute equation (1), Δω=12×2ωs2Ag×ω ∴ Δω=1×10−210×10=10−2rad s-1 Δω change in angular frequency of pendulum