Correct Answer - B::D
For first oscillarator , For second oscillator
`b = maomega_(1)`
`(a)/(b) = (1)/(momega_(1)) = n^(2) , (1)/(momega_(2)) = 1`
`(omega_(2))/(omega_(1)) = n^(2)`
`E_(1) = (1)/(2)m omega_(1)^(2) a^(2) , E_(3) = (1)/(2)momega_(2)^(2) R^(2)`
`(E_(1))/(E_(2)) = (omega_(1)^(2))/(omega_(2)^(2)) xx n^(2) = (omega_(1)^(2))/(omega_(2)^(2)) xx (omega_(2))/(omega_(1)) , (E_(1))/(E_(2)) = (omega_(1))/(omega_(2)) rArr (E_(1))/(omega_(1)) = (E_(2))/(omega_(2))`