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(A) The force is given by $F = -\frac{dU}{dx}$. Equilibrium points occur where $F = 0$, i.e., $\frac{dU}{dx} = 0$.
For $(A): U_1(x) = \frac{U_0}{2} \

Vedclass · Accepted Answer

$(A) \rightarrow (P, Q, R, S); (B) \rightarrow (Q, T); (C) \rightarrow (P, Q, R, T); (D) \rightarrow (P, R, S)$

Vedclass · Answer

(A) The force is given by $F = -\frac{dU}{dx}$. Equilibrium points occur where $F = 0$, i.e., $\frac{dU}{dx} = 0$.
For $(A): U_1(x) = \frac{U_0}{2} \left[1 - (x/a)^2\right]^2$. 
$\frac{dU_1}{dx} = \frac{U_0}{2} \cdot 2 \left[1 - (x/a)^2\right] \cdot (-2x/a^2) = -\frac{2U_0 x}{a^2} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=0, a, -a$. For $|x| < a$, $F$ is attractive towards $x=0$. 
Thus $(A) \rightarrow (P, Q, R, S)$.
For $(B): U_2(x) = \frac{U_0}{2} (x/a)^2$. 
$\frac{dU_2}{dx} = \frac{U_0 x}{a^2}$. 
$F=0$ at $x=0$. For energy $U_0/4$, $U_2(x) = U_0/4 \Rightarrow (x/a)^2 = 1/2 \Rightarrow x = \pm a/\sqrt{2}$. 
It oscillates about $x=0$. 
Thus $(B) \rightarrow (Q, T)$.
For $(C): U_3(x) = \frac{U_0}{2} (x/a)^2 e^{-(x/a)^2}$. 
$\frac{dU_3}{dx} = \frac{U_0}{2} \left[\frac{2x}{a^2} e^{-(x/a)^2} + (x/a)^2 e^{-(x/a)^2} (-2x/a^2)\right] = \frac{U_0 x}{a^2} e^{-(x/a)^2} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=0, a, -a$. For energy $U_0/4$, it can oscillate about $x=a$ or $x=-a$ if $U_{min} < U_0/4 < U_{max}$. 
Thus $(C) \rightarrow (P, Q, R, T)$.
For $(D): U_4(x) = \frac{U_0}{2} \left[x/a - 1/3(x/a)^3\right]$. 
$\frac{dU_4}{dx} = \frac{U_0}{2a} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=a, -a$. For $|x| < a$, $F$ is negative for $x>0$ and positive for $x<0$, so it is attractive towards $x=0$. 
Thus $(D) \rightarrow (P, R, S)$.

Column $I$	Column $II$
$(A) U_1(x) = \frac{U_0}{2} \left[1 - \left(\frac{x}{a}\right)^2\right]^2$	$(P)$ The force acting on the particle is zero at $x = a$.
$(B) U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$	$(Q)$ The force acting on the particle is zero at $x = 0$.
$(C) U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp \left[-\left(\frac{x}{a}\right)^2\right]$	$(R)$ The force acting on the particle is zero at $x = -a$.
$(D) U_4(x) = \frac{U_0}{2} \left[\frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3\right]$	$(S)$ The particle experiences an attractive force towards $x = 0$ in the region $\|x\| < a$.
	$(T)$ The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$.

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