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(C) The potential energy of the particle is given by $U(x) = \frac{x^4}{4} - \frac{x^2}{2}$.To find the equilibrium points,we differentiate $U(x)$ wit

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between $x = -1.0 \, m$ to $x = 0.0 \, m$

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(C) The potential energy of the particle is given by $U(x) = \frac{x^4}{4} - \frac{x^2}{2}$.To find the equilibrium points,we differentiate $U(x)$ with respect to $x$ and set it to zero:$\frac{dU}{dx} = x^3 - x = x(x^2 - 1) = 0$.This gives critical points at $x = 0$ and $x = \pm 1$.Using the second derivative test: $\frac{d^2U}{dx^2} = 3x^2 - 1$.At $x = 0$,$\frac{d^2U}{dx^2} = -1$ (local maximum).At $x = \pm 1$,$\frac{d^2U}{dx^2} = 2$ (local minima).The potential energy graph shows two wells separated by a barrier at $x = 0$. Since the total mechanical energy $E$ is small and negative (specifically,$E Given that at $t = 0 \, s$,the particle is at $x = -0.5 \, m$,it is located in the left potential well. Because the total energy is less than the potential barrier at $x = 0$,the particle cannot cross the barrier and will remain confined between the turning points in the region $x = -1.0 \, m$ to $x = 0.0 \, m$.

$A$ particle has a total mechanical energy that is small and negative. It is under the influence of a one-dimensional potential $U(x) = \frac{x^4}{4} - \frac{x^2}{2} \, J$,where $x$ is in meters. At time $t = 0 \, s$,it is at $x = -0.5 \, m$. Then,at a later time,it can be found:

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