A particle of unit mass is moving along the x-axis under the influence of a force and its total ener

Q: Why do we take the negative derivative of potential energy to find force?

In a conservative field, work done by the force equals the negative change in potential energy ($dW = -dU$). Since $dW = F dx$, it naturally follows that $F = -\frac{dU}{dx}$. The negative sign ensures that objects naturally tend to move towards lower potential energy states.

Q: How do I know if an equilibrium point is stable or unstable just from the force equation?

If $F = 0$ at a point, check the force around it. If the force pushes the particle *back* towards the point when displaced (a restoring force), it's stable (a potential well). If the force pushes it *away*, it's unstable (a potential hill). Mathematically, if $\frac{d^2U}{dx^2} > 0$, it's stable.

Q: For option C, why doesn't the exponential term give any equilibrium points?

The function $e^{-x^2/a^2}$ is strictly positive for all real values of $x$. It approaches zero as $x$ goes to infinity, but it never actually equals zero. Therefore, when you set the derivative to zero, only the algebraic factors can provide the roots.

Q: In option D, why does the particle oscillate even though the well is asymmetric?

Oscillation doesn't require a perfectly symmetric well like a parabola. As long as the particle is trapped between two potential barriers and the energy is conserved, it will bounce back and forth between the turning points where its kinetic energy becomes zero.

Q: What happens if the total energy $E$ is exactly equal to the barrier height?

If the energy exactly matches the local maximum (the barrier), the particle will theoretically take an infinite amount of time to reach the top of the hill and will come to rest there. It's a state of unstable equilibrium.

Question

A particle of unit mass is moving along the $x$-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in Column I ($a$ and $U_0$ are constants). Match the potential energies in Column I to the corresponding statements in Column II

List-I:

P. $U_1(x) = \frac{U_0}{2} \left[ 1 - \left(\frac{x}{a}\right)^2 \right]^2$

Q. $U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$

R. $U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp\left[-\left(\frac{x}{a}\right)^2\right]$

S. $U_4(x) = \frac{U_0}{2} \left[ \frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3 \right]$

List-II:

1. The force acting on the particle is zero at $x = a$

2. The force acting on the particle is zero at $x = 0$

3. The force acting on the particle is zero at $x = -a$

4. The particle experiences an attractive force towards $x = 0$ in the region $|x| < a$

5. The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$.

Sigma Percentile · Accepted Answer

### The Master Key: Force and Potential Energy

Welcome to a beautiful problem from JEE Advanced that perfectly blends calculus with physical intuition. We are given four distinct potential energy functions, $U(x)$, and we need to deduce the dynamical behavior of a particle moving under their influence.

The entire problem hinges on one fundamental relationship from classical mechanics: the conservative force acting on a particle is the negative gradient of its potential energy. In one dimension, this is elegantly expressed as:

$$F = -\frac{dU}{dx}$$

This simple equation is our master key. By differentiating the given potential energy functions, we can find the exact force equation for each case. Once we have the force, we can easily find the equilibrium points by setting $F = 0$. Furthermore, by analyzing the sign of the force in different regions, we can determine if the force is attractive or repulsive, and whether a potential well is capable of trapping a particle for oscillation.

### Decoding Option A: The Symmetric Double Well

Let's start with the first potential energy function:

$$U_1(x) = \frac{U_0}{2} \left[ 1 - \left(\frac{x}{a}ight)^2 ight]^2$$

To find the force, we apply the chain rule carefully:

$$F_1(x) = -\frac{dU_1}{dx} = -\frac{U_0}{2} \cdot 2\left[ 1 - \left(\frac{x}{a}ight)^2 ight] \left(-\frac{2x}{a^2}ight)$$

Simplifying this, we get a cubic force equation:

$$F_1(x) = \frac{2U_0}{a^4} x (a^2 - x^2) = -\frac{2U_0}{a^4} x (x-a)(x+a)$$

Setting $F_1 = 0$ gives us three equilibrium points: $x = 0$, $x = a$, and $x = -a$. This immediately confirms statements **p, q, and r**.

Now, is the force attractive towards the origin? Let's check the region $0 < x < a$. Here, $(a^2 - x^2)$ is positive, and $x$ is positive, making $F_1(x)$ positive. A positive force pushes the particle in the $+x$ direction, *away* from the origin. Thus, it is repulsive, and statement **s** is incorrect.

What about oscillation? At $x = -a$, $U_1(-a) = 0$, which is a stable minimum (a potential well). The local maximum (the barrier) at the origin is $U_1(0) = \frac{U_0}{2}$. Since our particle has a total energy of $E = \frac{U_0}{4}$, which is strictly less than the barrier height, it doesn't have enough energy to escape! It will happily oscillate in this well. So, statement **t** is a perfect match.

### Decoding Option B: The Simple Harmonic Oscillator

Moving to option B, we have a simple parabolic potential:

$$U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}ight)^2$$

Differentiating this gives a linear restoring force:

$$F_2(x) = -\frac{dU_2}{dx} = -\frac{U_0}{a^2} x$$

This is exactly Hooke's Law ($F = -kx$)! The force is zero only at the origin ($x = 0$), matching statement **q**. Because it's a restoring force, it always pulls the particle back towards the origin, regardless of whether $x$ is positive or negative. This perfectly matches statement **s**. Since there is no equilibrium point at $x = -a$, it cannot oscillate there, ruling out **t**.

### Decoding Option C: The Gaussian Trap

Option C introduces an exponential term:

$$U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}ight)^2 \exp\left[-\left(\frac{x}{a}ight)^2ight]$$

Don't let it intimidate you. We use the product rule to find the derivative:

$$F_3(x) = -\frac{dU_3}{dx} = -\frac{U_0}{a^4} x (a^2 - x^2) \exp\left[-\left(\frac{x}{a}ight)^2ight]$$

The exponential part is strictly positive and never zero. Therefore, the roots come entirely from the algebraic part. We again find that the force is zero at $x = 0, a, -a$. So, statements **p, q, and r** are matches.

Let's check the direction of the force for $|x| < a$. For $0 < x < a$, the term $(a^2 - x^2)$ is positive, making $F_3(x)$ negative (pointing left towards the origin). For $-a < x < 0$, $x$ is negative, making $F_3(x)$ positive (pointing right towards the origin). So, it is indeed attractive towards the center! Statement **s** is a match.

However, at $x = -a$, the potential $U_3$ is at a local maximum. You can't oscillate on top of a hill! It's an unstable equilibrium, so **t** is incorrect.

### Decoding Option D: The Asymmetric Well

Finally, option D presents a cubic potential:

$$U_4(x) = \frac{U_0}{2} \left[ \frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}ight)^3 ight]$$

Taking the derivative, we get a quadratic force equation:

$$F_4(x) = -\frac{dU_4}{dx} = -\frac{U_0}{2a} \left[ 1 - \left(\frac{x}{a}ight)^2 ight] = \frac{U_0}{2a^3} (x^2 - a^2)$$

Setting it to zero, we find equilibrium points only at $x = a$ and $x = -a$. The force at the origin is $F_4(0) = -\frac{U_0}{2a} 
eq 0$. So, statements **p** and **r** match, but **q** does not.

Is the force attractive towards the origin? For $|x| < a$, the term $(x^2 - a^2)$ is strictly negative. This means $F_4(x)$ is always negative in this region, constantly pushing the particle to the left. It does not pull it towards the origin from both sides. So **s** is out.

But look at $x = -a$. It's a potential well with a minimum value of $U_4(-a) = -\frac{U_0}{3}$. The barrier on the right is at $x = a$ with a height of $U_4(a) = \frac{U_0}{3}$. Our particle's energy is $E = \frac{U_0}{4}$. Since $-\frac{U_0}{3} < \frac{U_0}{4} < \frac{U_0}{3}$, the particle is safely trapped below the barrier. It will oscillate back and forth in this asymmetric well! Statement **t** is a perfect match.

By systematically applying the negative gradient rule and analyzing the physical meaning of the resulting forces and potential wells, we have successfully decoded all four potential functions.

List-I

List-II

Select Matching Pairs:

Visualized Solution

The Force-Potential Relationship

Analyzing $U_{1} (x)$

Oscillations in $U_{1} (x)$

Analyzing $U_{2} (x)$

Analyzing $U_{3} (x)$

Force Direction in $U_{3} (x)$

Analyzing $U_{4} (x)$

Oscillations in $U_{4} (x)$

Final Matrix Matching

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Master Key

Decoding Option A

Decoding Option B

Decoding Option C

Decoding Option D

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