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Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies `omega_(1)` and `omega_(2)` and have total energies `E_(1)` and `E_(2)` repsectively. The variations of their momenta `p` with positions `x` are shown in figures. If `(a)/(b) = n^(2)` and `(a)/(R) = n`, then the correc t equation(s) is (are) : A. `E_(1)omega_(1) = E_(2)omega_(2)`B. `(omega_(2))/(omega_(1)) = n^(2)`C. `omega_(1)omega_(2) = n^(2)`D. `(E_(1))/(omega_(1)) = (E_(2))/(omega_(2))` |
Answer» 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))` |
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