Two simple pendulum first of bob mass M1 and length L1 and second of bob mass M2 and length L2 are given. If M1 = M2, L1 = 2L2 and the vibrational energy of both is same, then among the following, which one is the correct option?

ω=2πglω= Angular frequency of oscillationl = Length of pendulumg = effective acceleration due to gravityω=12πgl⇒ϖ∝1i⇒ω1ω¯2=l2l1=L22L2⇒w1ω2=12⇒ω2=2ω1⇒ω2>ω1Total energy in simple harmonic motion at any position is the sum of potential energy and kinetic energy which is given asT=12Ka2WhereK=shm⁡ constant =ω2m and a is amplitude.m is the mass of the pendulumEnergy E=12mω2a2⇒a12a22=m2ω22m1ω12 (∵E is same)Given ω2>ω1 and m1=m2⇒a1>a2