Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω₁ and ω₂ and have total energies E₁ and E₂, respectively. The variations of their momenta p with positions x are shown in figures. If ab = n² and aR = n, then the correct equation(s) is (are) E₁ω₁ = E₂ω₂, ω₂/ω₁ = n², E₁ω₁ = E₂ω₂, ω₁ω₂ = n²

First oscillator:

p²/2m + 1/2 mω₁²x² = E₁

Comparing this with equation of ellipse having semi-major and semi-minor axes a and b respectively,
x²/a² + p²/b² = 1

b² = 2mE₁
a² = 2mE₁/ω₁²

=> ab = √(2mE₁)(2mE₁/ω₁²) = 2mE₁/ω₁ = n²

Second oscillator:

p²/2m + 1/2 mω₂²x² = E₂

b² = 2mE₂
a² = 2mE₂/ω₂²

=> ab = √(2mE₂)(2mE₂/ω₂²) = 2mE₂/ω₂ = n²

From (1) and (2) we have:

2mE₁/ω₁ = 2mE₂/ω₂

=> E₁ω₂ = E₂ω₁

Also, aR = n

=> √(2mE₁/ω₁²) * √(2mE₁/ω₁²) = n

=> 2mE₁/ω₁ = n²

Similarly,

2mE₂/ω₂ = n²

From the above two equations,

2mE₁/ω₁ = 2mE₂/ω₂

=> E₁ω₂ = E₂ω₁

Also, a = n and b = n²

Therefore from the above equations:

E₁ω₂ = E₂ω₁

And ω₁ω₂ = n² is also satisfied.