A particle of mass m moves in a one-dimension | vedclass

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(B) The potential energy is given by $U(x) = -ax^2 + bx^4$.To find the equilibrium positions,we set the first derivative to zero:$\frac{dU}{dx} = -2ax

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$2 \sqrt{\frac{a}{m}}$

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(B) The potential energy is given by $U(x) = -ax^2 + bx^4$.To find the equilibrium positions,we set the first derivative to zero:$\frac{dU}{dx} = -2ax + 4bx^3 = 0$$2x(2bx^2 - a) = 0$This gives equilibrium points at $x = 0$ and $x = \pm \sqrt{\frac{a}{2b}}$.To find the minima,we check the second derivative: $\frac{d^2U}{dx^2} = -2a + 12bx^2$.At $x = 0$,$\frac{d^2U}{dx^2} = -2a At $x = \pm \sqrt{\frac{a}{2b}}$,$\frac{d^2U}{dx^2} = -2a + 12b(\frac{a}{2b}) = -2a + 6a = 4a > 0$ (stable equilibrium).The effective spring constant $k$ at the minima is $k = \frac{d^2U}{dx^2} = 4a$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}}$.Substituting $k = 4a$,we get $\omega = \sqrt{\frac{4a}{m}} = 2\sqrt{\frac{a}{m}}$.

$A$ particle of mass $m$ moves in a one-dimensional potential energy $U(x) = -ax^2 + bx^4$,where $a$ and $b$ are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to

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