To determine The location and stability of equilibrium points with a = − 8 aJ / nm 2 and b = 1 aJ / nm 4 . Explanation Refer Example 12.4. Write the potential energy function. U ( x ) = a x 2 − b x 4 Differentiate the above equation with respect to x to find the points of equilibrium. d U ( x ) d x = d ( a x 2 − b x 4 ) d x = 2 a x − 4 b x 3 = 2 x ( a − 2 b x 2 ) (I) Equate d U ( x ) d x to zero. 2 x ( a − 2 b x 2 ) = 0 Solution of above equation gives the equilibrium points. Write the solution of above equation. x = 0 x = ± a 2 b Conclusion: Consider the case when x = 0 . The equilibrium point x = 0 is a point of local minimum in the potential energy curve. It means that x = 0 is a metastable point. Consider the case when x = ± a 2 b . To find the type of equilibrium point, find the second derivative of potential energy function. Differentiate equation (I) with respect to x again. d 2 U ( x ) d x 2 = d ( 2 a x − 4 b x 3 ) d x = 2 a − 12 b x 2 Rewrite the above expression by substituting ± a 2 b for x

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Chapter 12, Problem 26E

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The location and stability of equilibrium points with a=−8 aJ/nm2 and b=1 aJ/nm4.

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Chapter 12, Problem 24E

Chapter 12, Problem 27E

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