When a particle is performing linear `SHM` its `KE` is two times its `pE` at a position `A` and its `PE` is two times its `KE` at another position `B`. Find ratio of `KE_(A)` to `KE_(B)`

Explanation

`TE_(A) = KE_(A) +PE_(A)`
But `KE_(A) = 2PE_(A)`,
`TE_(A) = KE_(A) +(1)/(2)KE_(A) = (3)/(2) KE_(A)`.
Similarly, `TE_(B) = KE_(B) +PE_(B)`.
But `PE_(B) = 2KE_(B)`
`TE_(B) = kE_(B) +2KE_(B) = 3KE_(B)`.
By the principle of conservation of energy
`TE_(A) = TE_(B), (3)/(2) KE_(A) = 3KE_(B)`
`(KE_(A))/(KE_(B)) = (2)/(1)`