Phase Space Momentum Position Graphs of Two Harmonic Oscillators

For a simple harmonic oscillator of mass $m$, displacement $x$, and momentum $p$, the total mechanical energy $E$ is given by the sum of kinetic and potential energies:

$E = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2$

Rearranging this into the standard equation of an ellipse in the phase space $(x, p)$:

$\frac{x^2}{\left(\sqrt{\frac{2E}{m\omega^2}}\right)^2} + \frac{p^2}{\left(\sqrt{2mE}\right)^2} = 1$

From the given phase space diagrams:

Oscillator 1:

The phase space trajectory is an ellipse with semi-major axis $a$ along the $x$-axis and semi-minor axis $b$ along the $p$-axis. $a = \sqrt{\frac{2E_1}{m\omega_1^2}}$ $b = \sqrt{2mE_1}$

Taking the ratio of $a$ to $b$: $\frac{a}{b} = \frac{\sqrt{\frac{2E_1}{m\omega_1^2}}}{\sqrt{2mE_1}} = \frac{1}{m\omega_1}$

Given that $\frac{a}{b} = n^2$, we get: $\frac{1}{m\omega_1} = n^2 \implies m\omega_1 = \frac{1}{n^2} \quad \text{--- (1)}$

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Oscillator 2:

The phase space trajectory is a circle of radius $R$. Thus, the semi-axes along $x$ and $p$ are both equal to $R$: $R = \sqrt{\frac{2E_2}{m\omega_2^2}} = \sqrt{2mE_2}$

Equating the two expressions for $R$: $\sqrt{\frac{2E_2}{m\omega_2^2}} = \sqrt{2mE_2} \implies m\omega_2 = 1 \quad \text{--- (2)}$

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Step 1: Ratio of Angular Frequencies ($\frac{\omega_2}{\omega_1}$)

Using equations (1) and (2): $\frac{\omega_2}{\omega_1} = \frac{m\omega_2}{m\omega_1} = \frac{1}{1/n^2} = n^2$

This confirms that Option (B) is correct.

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Step 2: Relation between Energy and Angular Frequency

From the expression for $b$ and $R$: $E_1 = \frac{b^2}{2m} \quad \text{and} \quad E_2 = \frac{R^2}{2m}$

Now, let's compute the ratio $\frac{E_1}{\omega_1}$: $\frac{E_1}{\omega_1} = \frac{b^2 / (2m)}{\omega_1} = \frac{b^2}{2(m\omega_1)}$

Substituting $m\omega_1 = \frac{1}{n^2}$ into the expression: $\frac{E_1}{\omega_1} = \frac{b^2 n^2}{2}$

We are given the relations:

1. $\frac{a}{b} = n^2 \implies b = \frac{a}{n^2}$
2. $\frac{a}{R} = n \implies a = n R$

Substituting $a = n R$ into $b = \frac{a}{n^2}$: $b = \frac{nR}{n^2} = \frac{R}{n} \implies b n = R$

Substituting $b n = R$ back into the expression for $\frac{E_1}{\omega_1}$: $\frac{E_1}{\omega_1} = \frac{(bn)^2}{2} = \frac{R^2}{2}$

Next, let's compute $\frac{E_2}{\omega_2}$: $\frac{E_2}{\omega_2} = \frac{R^2 / (2m)}{\omega_2} = \frac{R^2}{2(m\omega_2)}$

Using $m\omega_2 = 1$: $\frac{E_2}{\omega_2} = \frac{R^2}{2}$

Comparing the two results: $\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2} = \frac{R^2}{2}$

This confirms that Option (D) is correct.

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Conclusion:

The correct options are (B) and (D).