Two independent harmonic oscillators of equal | vedclass

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Question

(A) For a simple harmonic oscillator,the equation of the trajectory in the phase space $(p-x)$ is given by $\frac{p^2}{2mE} + \frac{x^2}{2E/m\omega^2}

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$(B, D)$

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(A) For a simple harmonic oscillator,the equation of the trajectory in the phase space $(p-x)$ is given by $\frac{p^2}{2mE} + \frac{x^2}{2E/m\omega^2} = 1$,which represents an ellipse $\frac{p^2}{b^2} + \frac{x^2}{a^2} = 1$,where $a$ is the amplitude and $b$ is the maximum momentum $p_{max} = m\omega a$.For the first oscillator:$E_1 = \frac{1}{2} m \omega_1^2 a^2$ and $b = m \omega_1 a$. Thus,$\frac{a}{b} = \frac{1}{m \omega_1}$.For the second oscillator:$E_2 = \frac{1}{2} m \omega_2^2 R^2$ and the trajectory is a circle,so $p_{max} = x_{max} \Rightarrow m \omega_2 R = R \Rightarrow m \omega_2 = 1$.Substituting $m \omega_2 = 1$ into the expression for $\frac{a}{b}$:$\frac{a}{b} = \frac{1}{m \omega_1} = \frac{\omega_2}{\omega_1} = n^2$ (Option $B$ is correct).Also,$E = \frac{1}{2} m \omega^2 A^2$. For the first oscillator,$E_1 = \frac{1}{2} m \omega_1^2 a^2$. For the second,$E_2 = \frac{1}{2} m \omega_2^2 R^2$. Since $m \omega_2 = 1$,$E_2 = \frac{1}{2} \omega_2 R^2$. Given $\frac{a}{R} = n$,then $a = nR$. From $\frac{a}{b} = n^2$,$b = \frac{a}{n^2} = \frac{nR}{n^2} = \frac{R}{n}$.Since $b = m \omega_1 a$,$\frac{R}{n} = m \omega_1 (nR) \Rightarrow m \omega_1 = \frac{1}{n^2}$.Now,$\frac{E_1}{\omega_1} = \frac{\frac{1}{2} m \omega_1^2 a^2}{\omega_1} = \frac{1}{2} m \omega_1 a^2 = \frac{1}{2} (\frac{1}{n^2}) (nR)^2 = \frac{1}{2} R^2$.And $\frac{E_2}{\omega_2} = \frac{\frac{1}{2} m \omega_2^2 R^2}{\omega_2} = \frac{1}{2} m \omega_2 R^2 = \frac{1}{2} (1) R^2 = \frac{1}{2} R^2$.Thus,$\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$ (Option $D$ is correct).

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