A particle free to move along the X-axis has potential energy given by U(x) = k[1 - exp(-x^2)] for -

Q: Why is the force equal to the negative derivative of potential energy?

In conservative systems, work done by a conservative force is stored as potential energy ($W = -\Delta U$). For an infinitesimal displacement, $F \cdot dx = -dU$, which gives $F = -\frac{dU}{dx}$. The negative sign indicates that conservative forces always act in a direction that decreases the potential energy of the system.

Q: How did we approximate $\exp(-x^2)$ to $1$ for small $x$?

The Taylor series expansion of $e^y$ is $1 + y + \frac{y^2}{2!} + \dots$. Substituting $y = -x^2$, we get $\exp(-x^2) = 1 - x^2 + \frac{x^4}{2} - \dots$. When $x$ is extremely small ($x \ll 1$), terms like $x^2$ and higher powers are negligible compared to $1$, so we can approximate $\exp(-x^2) \approx 1$.

Q: What happens to the motion if the displacement is large?

If the displacement is large, the linear approximation $F \approx -2kx$ no longer holds because the higher-order terms in the expansion of $\exp(-x^2)$ become significant. The motion remains periodic as long as the total energy $E < k$, but it will not be simple harmonic (the time period will depend on the amplitude).

Q: Can the particle escape the potential well?

Yes! Since the potential energy asymptotically approaches $k$ as $x \to \pm\infty$, if the particle is given a total mechanical energy $E \ge k$, it will have enough kinetic energy to overcome the potential barrier and escape to infinity, performing non-periodic motion.

Q: Why is the effective spring constant $2k$ instead of just $k$?

The constant $k$ in the potential energy expression $U(x) = k[1 - \exp(-x^2)]$ is an energy scale parameter, not a physical spring constant. When we differentiate and approximate the force for small displacements, the coefficient of $-x$ turns out to be $2k$, which acts as the effective force constant ($k_{\text{eff}}$) for the harmonic oscillations.

Question

A particle free to move along the $X$-axis has potential energy given by $U(x) = k[1 - \exp(-x^2)]$ for $-\infty \le x \le +\infty$, where $k$ is a positive constant of appropriate dimensions. Then,

Options:

(A) at points away from the origin, the particle is in unstable equilibrium

(B) for any finite non-zero value of $x$, there is a force directed away from the origin

(C) if its total mechanical energy is $k/2$, it has its minimum kinetic energy at the origin

(D) for small displacements from $x = 0$, the motion is simple harmonic

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine a particle trapped in a smooth, symmetric valley. Mathematically, this valley is described by the potential energy function: $$U(x) = k[1 - \exp(-x^2)]$$ Here, $k$ is a positive constant that sets the scale of the energy. Let's analyze the behavior of this potential energy function at extreme points: 1. At the origin ($x = 0$): $$U(0) = k[1 - \exp(0)] = k[1 - 1] = 0$$ This is the lowest possible value of the potential energy because the exponential term $\exp(-x^2)$ is always between $0$ and $1$ for all real $x$. 2. As the particle moves infinitely far away ($x o \pm\infty$): $$U(x) o k[1 - 0] = k$$ This tells us that the potential energy increases symmetrically on both sides of the origin, asymptotically approaching a maximum value of $k$. This is a classic **potential well**. ### The Force-Potential Relationship To understand how the particle moves, we must find the force acting on it. In a conservative field, the force $F(x)$ is the negative gradient of the potential energy: $$F(x) = -\frac{dU}{dx}$$ Let's differentiate $U(x)$ with respect to $x$ using the chain rule: $$\frac{dU}{dx} = \frac{d}{dx} \left( k[1 - \exp(-x^2)] ight)$$ $$\frac{dU}{dx} = k \left( 0 - \exp(-x^2) \cdot (-2x) ight) = 2kx \exp(-x^2)$$ Now, substituting this back into our force relation: $$F(x) = -2kx \exp(-x^2)$$ This equation is the master key to unlocking the particle's dynamics. ### Analyzing Equilibrium and Stability An equilibrium position is a point where the net force acting on the particle is zero: $$F(x) = 0 \implies -2kx \exp(-x^2) = 0$$ Since $k > 0$ and the exponential term $\exp(-x^2)$ can never be zero for any finite value of $x$, the only way this product can vanish is if: $$x = 0$$ Thus, the **origin is the unique finite equilibrium position**. To determine the stability of this equilibrium, let's look at the direction of the force when the particle is displaced: - If we displace the particle to the right ($x > 0$): The force $F(x) = -2kx \exp(-x^2)$ is negative ($F < 0$), meaning it points back to the left (towards the origin). - If we displace the particle to the left ($x < 0$): The force $F(x)$ becomes positive ($F > 0$), meaning it points back to the right (towards the origin). Since the force always acts to restore the particle back to its equilibrium position, $x = 0$ is a **stable equilibrium** point. This immediately disproves options (a) and (b). ### The Small Displacement Approximation (SHM) What happens if we gently nudge the particle near the bottom of the well? For very small displacements ($x \ll 1$), we can use the Taylor series expansion of the exponential function: $$\exp(-x^2) \approx 1 - x^2 \approx 1$$ Substituting this approximation into our force equation: $$F(x) \approx -2kx$$ This is a remarkable result! For small displacements, the restoring force is directly proportional to the displacement: $$F \propto -x$$ This is the exact mathematical definition of **Simple Harmonic Motion (SHM)**, with an effective force constant of $k_{ ext{eff}} = 2k$. Therefore, for small displacements from $x = 0$, the motion is simple harmonic, which perfectly matches **Option (d)**. ### Checking the Energy Aspect Let's also address option (c) to be absolutely thorough. If the total mechanical energy of the particle is $E = k/2$, the conservation of energy states: $$E = K(x) + U(x) \implies K(x) = E - U(x)$$ At the origin ($x = 0$), the potential energy is minimum ($U = 0$). Therefore, the kinetic energy at the origin must be: $$K(0) = E - 0 = E = \frac{k}{2}$$ Since $U(x)$ is minimum at the origin, the kinetic energy $K(x)$ must be at its **maximum** at the origin, not minimum. This disproves option (c). Thus, we confidently conclude that **Option (d)** is the correct choice.

Animated Solution for Physics - Oscillations: A particle free to move along the $X$ -axis has potential energy given by $U (x) = k [1 - exp (- x^{2})]$ for $- \infty \leq x \leq + \infty$ , where $k$ is a positive constant of appropriate dimensions. Then,

Select Answer:

Visualized Solution

Visualizing the Potential Energy Curve

The Force-Potential Relation

Calculating the Force Function

Finding Equilibrium Points

Analyzing Stability of Equilibrium

Approximation for Small Displacements

Confirming Simple Harmonic Motion

Evaluating the Options

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Force-Potential Relationship

Analyzing Equilibrium and Stability

The Small Displacement Approximation (SHM)

Checking the Energy Aspect

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