Match Potential Energies with Force and Motion Conditions

To determine the correct matches between Column I and Column II, we use the fundamental relation between a conservative force $F(x)$ and its potential energy function $U(x)$: $F(x) = -\frac{dU(x)}{dx}$

An equilibrium point occurs where $F(x) = 0$.

• A position $x = x_0$ is a stable equilibrium (allowing oscillations) if $\left.\frac{d^2U}{dx^2}\right|_{x_0} > 0$ (i.e., $U(x)$ has a local minimum at $x_0$).
• An attractive force towards $x = 0$ in a region requires $F(x) < 0$ for $x > 0$ and $F(x) > 0$ for $x < 0$ (i.e., $F(x) \cdot x < 0$ for all non-zero $x$ in that region).
• A particle with total energy $E$ can oscillate about a stable equilibrium $x = x_0$ if $U(x_0) \le E < U_{\text{max}}$, where $U_{\text{max}}$ is the height of the potential barrier enclosing $x_0$.

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Analysis for Function (A):

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

1. Force Expression: $F_1(x) = -\frac{dU_1}{dx} = \frac{2U_0 x}{a^2} \left( 1 - \frac{x^2}{a^2} \right)$

2. Zeros of Force: $F_1(x) = 0 \implies x = 0, \quad x = a, \quad \text{and} \quad x = -a$

   • Thus, $F_1(a) = 0 \implies \mathbf{(P)}$
   • $F_1(0) = 0 \implies \mathbf{(Q)}$
   • $F_1(-a) = 0 \implies \mathbf{(R)}$

3. Check Statement (S): In the region $0 < x < a$, $1 - \frac{x^2}{a^2} > 0 \implies F_1(x) > 0$ (repulsive from $x = 0$). Thus, (S) does not hold.

4. Check Statement (T): At $x = -a$, $U_1(-a) = 0$. Since $U_1(x) \ge 0$ for all $x$, $x = -a$ is a local minimum (stable equilibrium). The adjacent potential barrier at $x = 0$ is $U_1(0) = \frac{U_0}{2}$. Since $U_1(-a) = 0 \le E = \frac{U_0}{4} < U_1(0) = \frac{U_0}{2}$, a particle with energy $E = \frac{U_0}{4}$ is trapped in the potential well around $x = -a$ and can oscillate about $x = -a \implies \mathbf{(T)}$.

Result for (A): Matches to (P), (Q), (R), (T)

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Analysis for Function (B):

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

1. Force Expression: $F_2(x) = -\frac{dU_2}{dx} = -\frac{U_0 x}{a^2}$

2. Zeros of Force: $F_2(x) = 0 \implies x = 0$

   • $F_2(0) = 0 \implies \mathbf{(Q)}$
   • $F_2(a) \neq 0$ and $F_2(-a) \neq 0$, so (P) and (R) do not hold.

3. Check Statement (S): For $x > 0$, $F_2(x) < 0$ (directed towards origin). For $x < 0$, $F_2(x) > 0$ (directed towards origin). Hence, the force is attractive towards $x = 0$ in $|x| < a \implies \mathbf{(S)}$.

4. Check Statement (T): Since $x = -a$ is not an equilibrium position ($F_2(-a) \neq 0$), the particle cannot oscillate about $x = -a \implies \text{Not } \mathbf{(T)}$.

Result for (B): Matches to (Q), (S)

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Analysis for Function (C):

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

1. Force Expression: $F_3(x) = -\frac{dU_3}{dx} = -\frac{U_0 x}{a^2} \left( 1 - \frac{x^2}{a^2} \right) \exp\left[ -\left(\frac{x}{a}\right)^2 \right]$

2. Zeros of Force: $F_3(x) = 0 \implies x = 0, \quad x = a, \quad \text{and} \quad x = -a$

   • $F_3(a) = 0 \implies \mathbf{(P)}$
   • $F_3(0) = 0 \implies \mathbf{(Q)}$
   • $F_3(-a) = 0 \implies \mathbf{(R)}$

3. Check Statement (S): For $|x| < a$, $1 - \frac{x^2}{a^2} > 0$ and $\exp\left[ -\left(\frac{x}{a}\right)^2 \right] > 0$.

   • For $0 < x < a$: $F_3(x) < 0$ (towards $x = 0$).
   • For $-a < x < 0$: $F_3(x) > 0$ (towards $x = 0$). Therefore, the particle experiences an attractive force towards $x = 0$ in the region $|x| < a \implies \mathbf{(S)}$.

4. Check Statement (T): At $x = -a$, $U_3(-a) = \frac{U_0}{2e}$ is a local maximum of $U_3(x)$ (unstable equilibrium). Thus, the particle cannot oscillate about $x = -a \implies \text{Not } \mathbf{(T)}$.

Result for (C): Matches to (P), (Q), (R), (S)

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Analysis for Function (D):

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

1. Force Expression: $F_4(x) = -\frac{dU_4}{dx} = \frac{U_0}{2a} \left[ \left(\frac{x}{a}\right)^2 - 1 \right]$

2. Zeros of Force: $F_4(x) = 0 \implies x = a \quad \text{and} \quad x = -a$

   • $F_4(a) = 0 \implies \mathbf{(P)}$
   • $F_4(0) = -\frac{U_0}{2a} \neq 0$, so (Q) does not hold.
   • $F_4(-a) = 0 \implies \mathbf{(R)}$

3. Check Statement (S): For all $|x| < a$, $F_4(x) < 0$. For $x < 0$, a negative force points away from $x = 0$. Thus, (S) does not hold.

4. Check Statement (T): At $x = -a$: $U_4(-a) = \frac{U_0}{2} \left( -1 + \frac{1}{3} \right) = -\frac{U_0}{3}$ Since $\left.\frac{d^2U_4}{dx^2}\right|_{x=-a} = -\left.\frac{dF_4}{dx}\right|_{x=-a} = \frac{U_0}{a^2} > 0$, $x = -a$ is a local minimum (stable equilibrium).

   The local maximum occurs at $x = a$ with potential energy: $U_4(a) = \frac{U_0}{2} \left( 1 - \frac{1}{3} \right) = \frac{U_0}{3}$ Since $U_4(-a) = -\frac{U_0}{3} \le E = \frac{U_0}{4} < U_4(a) = \frac{U_0}{3}$, the particle is trapped in the potential well around $x = -a$ and can undergo oscillations about $x = -a \implies \mathbf{(T)}$.

Result for (D): Matches to (P), (R), (T)

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Final Summary of Matches:

• (A) matches to (P), (Q), (R), (T)
• (B) matches to (Q), (S)
• (C) matches to (P), (Q), (R), (S)
• (D) matches to (P), (R), (T)