Two independent harmonic oscillators of equal masses are oscillating about the origin with angular f

Q: Why is the phase space trajectory of a simple harmonic oscillator an ellipse?

Because the position $x(t)$ and momentum $p(t)$ are out of phase by $90^\circ$ (sine and cosine functions). Squaring and adding them eliminates time, yielding the equation of an ellipse: $\frac{x^2}{A^2} + \frac{p^2}{(m\omega A)^2} = 1$.

Q: What physical quantity does the area of the phase space ellipse represent?

The area of the ellipse is $\pi \times \text{semi-major axis} \times \text{semi-minor axis} = \pi A (m\omega A) = 2\pi \frac{E}{\omega}$. This is proportional to the action variable $J = \frac{E}{\omega}$, which is a fundamental quantity in classical mechanics and quantum theory (related to Planck's constant).

Q: How did we get the maximum momentum $p_{\max} = b$ for the first oscillator?

In the phase space plot, the maximum value of momentum $p$ occurs when $x = 0$ (the mean position). Geometrically, this corresponds to the semi-minor axis of the ellipse, which is labeled as $b$.

Q: Why does the radius $R$ cancel out when calculating the angular frequency $\omega_2$ of the second oscillator?

For the second oscillator, the trajectory is a circle, meaning the maximum position (amplitude $A_2 = R$) equals the maximum momentum ($p_{2,\max} = m\omega_2 A_2 = R$). Substituting $A_2 = R$ into $m\omega_2 A_2 = R$ gives $m\omega_2 R = R$, which simplifies directly to $\omega_2 = 1/m$, independent of $R$.

Q: Is the ratio $E/\omega$ always constant for any harmonic oscillator?

Yes, for a given oscillator with a fixed amplitude, the ratio $E/\omega$ is constant. More importantly, if the parameters of the oscillator (like spring constant) are changed very slowly (adiabatically), the ratio $E/\omega$ remains invariant. This is known as an adiabatic invariant.

Question

Two independent harmonic oscillators of equal masses are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$, respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b} = n^2$ and $\frac{a}{R} = n$, then the correct equation(s) is/are

Options:

(A) $E_1\omega_1 = E_2\omega_2$

(B) $\frac{\omega_2}{\omega_1} = n^2$

(C) $\omega_1\omega_2 = n^2$

(D) $\frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

Sigma Percentile · Accepted Answer

### The Magic of Phase Space

Imagine you are trying to describe the complete state of a moving particle.

Just knowing its position is not enough; you also need to know how fast it is moving, and in what direction.

In classical mechanics, we capture this complete state by plotting the particle's **momentum ($p$)** against its **position ($x$)** on a two-dimensional graph.

This abstract space is called **Phase Space**, and the path the particle traces in this space is its phase space trajectory.

For a simple harmonic oscillator, the position and momentum are out of phase by $90^\circ$.

Mathematically, we write:

$$x(t) = A \sin(\omega t + \phi)$$

$$p(t) = m v(t) = m \omega A \cos(\omega t + \phi)$$

If we square both terms and eliminate time $t$, we get the equation of an ellipse:

$$\frac{x^2}{A^2} + \frac{p^2}{(m\omega A)^2} = 1$$

This is a beautiful result! It tells us that the phase space trajectory of any simple harmonic oscillator is always an ellipse.

Let's see how we can use this geometric insight to solve our problem.

### Analyzing the Two Oscillators

Let's look at the two plots given in the problem.

For the **first oscillator**, the trajectory is an ellipse with semi-major axis $a$ along the $x$-axis and semi-minor axis $b$ along the $p$-axis.

Comparing this with our standard ellipse equation, we can immediately identify:

* **Amplitude ($A_1$)** $= a$
* **Maximum Momentum ($p_{1,\max}$)** $= b$

Since maximum momentum is defined as $p_{\max} = m \omega A$, we can write:

$$b = m \omega_1 a \implies \omega_1 = \frac{b}{m a}$$

And the total energy $E_1$ is simply the maximum kinetic energy:

$$E_1 = \frac{p_{1,\max}^2}{2m} = \frac{b^2}{2m}$$

Now, let's look at the **second oscillator**. Its trajectory is a circle of radius $R$.

A circle is just a special case of an ellipse where both axes are equal! Therefore:

* **Amplitude ($A_2$)** $= R$
* **Maximum Momentum ($p_{2,\max}$)** $= R$

Using the same logic, we find its angular frequency $\omega_2$:

$$R = m \omega_2 R \implies \omega_2 = \frac{1}{m}$$

And its total energy $E_2$ is:

$$E_2 = \frac{R^2}{2m}$$

### Connecting the Scales

The problem provides two scaling relationships to connect these two systems:

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

$$\frac{a}{R} = n \implies a = R n$$

Equating these two expressions for $a$ gives:

$$b n^2 = R n \implies R = b n$$

This is our master key! We can now express all our physical quantities in terms of just $b$, $m$, and $n$.

Let's rewrite $\omega_1$ using this relation:

$$\omega_1 = \frac{b}{m a} = \frac{b}{m (b n^2)} = \frac{1}{m n^2}$$

### Verifying the Options

Now, let's test each option one by one with our simplified expressions.

#### Option (b): Ratio of Angular Frequencies

Let's calculate the ratio $\frac{\omega_2}{\omega_1}$:

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

This matches Option (b) perfectly! Therefore, **Option (b) is correct**.

#### Option (c): Product of Angular Frequencies

Let's calculate the product $\omega_1 \omega_2$:

$$\omega_1 \omega_2 = \frac{1}{m n^2} \cdot \frac{1}{m} = \frac{1}{m^2 n^2} 
eq n^2$$

Thus, **Option (c) is incorrect**.

#### Option (d): Ratio of Energy to Frequency

Let's calculate the ratio $\frac{E}{\omega}$ for both oscillators.

For Oscillator 1:

$$\frac{E_1}{\omega_1} = \frac{b^2 / (2m)}{1 / (m n^2)} = \frac{b^2 n^2}{2}$$

For Oscillator 2:

$$\frac{E_2}{\omega_2} = \frac{R^2 / (2m)}{1 / m} = \frac{R^2}{2}$$

Since $R = b n$, we have $R^2 = b^2 n^2$. Substituting this into the expression for Oscillator 2:

$$\frac{E_2}{\omega_2} = \frac{b^2 n^2}{2}$$

They are exactly equal! Therefore, **Option (d) is correct**.

#### Option (a): Product of Energy and Frequency

Let's calculate the product $E \omega$ for both oscillators:

$$E_1 \omega_1 = \frac{b^2}{2m} \cdot \frac{1}{m n^2} = \frac{b^2}{2 m^2 n^2}$$

$$E_2 \omega_2 = \frac{R^2}{2m} \cdot \frac{1}{m} = \frac{R^2}{2 m^2} = \frac{b^2 n^2}{2 m^2}$$

These are clearly not equal unless $n=1$. Thus, **Option (a) is incorrect**.

### A Deeper Physical Insight: Adiabatic Invariance

Why did the ratio $\frac{E}{\omega}$ turn out to be so beautifully symmetric?

In physics, the area enclosed by a phase space trajectory is given by:

$$	ext{Area} = \pi 	imes 	ext{semi-major axis} 	imes 	ext{semi-minor axis}$$

For our first oscillator, this area is:

$$	ext{Area}_1 = \pi a b = \pi A_1 (m \omega_1 A_1) = 2\pi \left( \frac{\frac{1}{2} m \omega_1^2 A_1^2}{\omega_1} ight) = 2\pi \frac{E_1}{\omega_1}$$

This ratio, $J = \frac{E}{\omega}$, is known as the **Action Variable**.

In classical mechanics, if you slowly (adiabatically) change the parameters of an oscillator (like slowly shortening the string of a pendulum), the energy $E$ and frequency $\omega$ will both change, but their ratio $\frac{E}{\omega}$ will remain **absolutely constant**!

This is known as an **adiabatic invariant**, and it laid the foundation for the old quantum theory, where Bohr quantized this very action variable in units of Planck's constant $h$.

So, by solving this JEE Advanced problem, you have actually touched upon one of the deepest bridges connecting classical mechanics to quantum physics!

Select Answer:

Visualized Solution

Understanding Phase Space Plots ( $p$ - $x$ curves)

Connecting Geometry to Physics

Analyzing Oscillator 1

Analyzing Oscillator 2

Utilizing the Given Geometric Relations

Expressing $ω_{1}$ in terms of $n$

Finding the Ratio of Angular Frequencies (Option b)

Checking Option (c)

Analyzing the Energy-to-Frequency Ratio (Option d)

Checking Option (a)

Summary and Conclusion

The Sigma Insight: Force and Energy Method in SHM

The Magic of Phase Space

Analyzing the Two Oscillators

Connecting the Scales

Verifying the Options

# Option (b)

# Option (c)

# Option (d)

# Option (a)

A Deeper Physical Insight

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