Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $ω_{1} and ω_{2}$ and have total energies $E_{1} and E_{2},$ respectively. The variations of their momenta p with positions x are shown in figures. If $\frac{a}{b} = n^{2} and \frac{a}{R} = n$ , then the correct equation(s) is (are)

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$E_{1} ω_{1} = E_{2} ω_{2}$

$\frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}}$

$ω_{1} ω_{2} = n^{2}$

$\frac{ω_{2}}{ω_{1}} = n^{2}$

answer is B, D.

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Detailed Solution

$\begin{array}{l} x = Asin (ωt); V = Aωcos (ωt) \\ p_{1} = {mA}_{1} ω_{1} = b; A_{1} = a (From ellipse) \\ p_{2} = {mA}_{2} ω_{2} = R; A_{2} = R (From circle) \\ Also, \frac{a}{b} = n^{2} and \frac{a}{R} = n \Rightarrow \frac{R}{b} = n \\ E_{1} = \frac{p_{1}^{2}}{2 m} \Rightarrow E_{2} = \frac{p_{2}^{2}}{2 m}; \\ \frac{E_{1}}{E_{2}} = {(\frac{b}{R})}^{2} = \frac{1}{n^{2}} \Rightarrow \frac{{mω}_{1}^{2} A_{1}^{2}}{{mω}_{2}^{2} A_{2}^{2}} = \frac{1}{n^{2}} \\ \Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{1}{n^{2}} \Rightarrow \frac{E_{1}}{ω_{1}} = \frac{E_{2}}{ω_{2}} \end{array}$