Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 12 - Exercises and Problems - Page 224: 44

Answer

As long as a is greater than $\sqrt{12}$ or less than $-\sqrt{12}$, there will be two equilibria.

Work Step by Step

To find the equilibria, we take the derivative of the potential energy function and set it equal to zero. Thus, we find: $0 = U_0(\frac{3x^2}{x_0^3}+\frac{2ax}{x_0^2}+\frac{4}{x_0})$ We only care about the determinant of the quadratic formula, so we find: $\sqrt{(2a)^2 - 4(3)(4)}$ Here, we see that as long as a is greater than $\sqrt{12}$ or less than $-\sqrt{12}$, there will be two equilibria. We take the second derivative of the original function to find: $\frac{dU^2}{d^2x}=U_0(\frac{6x}{x_0^3}+\frac{2a}{x_0^2})$ Whether an equlibria is stable or unstable depends on the value of $x_0$. However, we know that if $\frac{dU^2}{d^2x}$ is positive, the eqilibrium will be stable, and if $\frac{dU^2}{d^2x}$ is negative, the equilibrium will be unstable.
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