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(A) The force acting on the particle is given by $F = -\alpha x^3 - \beta x^4$.Since $F = -\frac{dU}{dx}$,we have $\frac{dU}{dx} = \alpha x^3 + \beta

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in stable equilibrium

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(A) The force acting on the particle is given by $F = -\alpha x^3 - \beta x^4$.Since $F = -\frac{dU}{dx}$,we have $\frac{dU}{dx} = \alpha x^3 + \beta x^4$.At $x = 0$,$\frac{dU}{dx} = 0$,which implies the particle is in equilibrium.To determine the stability,we examine the derivatives of the potential energy $U$ at $x = 0$:$\frac{d^2U}{dx^2} = 3\alpha x^2 + 4\beta x^3$,which is $0$ at $x = 0$.$\frac{d^3U}{dx^3} = 6\alpha x + 12\beta x^2$,which is $0$ at $x = 0$.$\frac{d^4U}{dx^4} = 6\alpha + 24\beta x$,which is $6\alpha$ at $x = 0$.Since $\alpha > 0$,the first non-zero derivative at $x = 0$ is of an even order ($4^{th}$ order) and is positive. This indicates that $U$ has a local minimum at $x = 0$.Therefore,the particle is in stable equilibrium.

Column-$I$	Column-$II$
$(1)$ Conservative force	$(a)$ Frictional force
$(2)$ Non-conservative force	$(b)$ Gravitational force
	$(c)$ Internal force

$A$ particle is acted upon by a force given by $F=-\alpha x^3-\beta x^4$,where $\alpha$ and $\beta$ are positive constants. At the point $x=0$,the particle is

Similar Questions

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Column-$I$ Column-$II$

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