Chapter 12 - Exercises and Problems - Page 222: 29

(a) To find the equilibrium point, we need to find the values of x and y for which the potential energy is minimized. The potential energy is given by: $U(x,y) = a1x - y^2$ To minimize this function, we need to take partial derivatives with respect to x and y and set them equal to zero: ∂U/∂x = a1 = 0 ∂U/∂y = -2y = 0 Solving for x and y, we get: x = 0 y = 0 Therefore, the equilibrium point is at x = 0, y = 0. (b) To determine the stability of the equilibrium, we need to examine the behavior of the potential energy near the equilibrium point. We can do this by calculating the second partial derivatives of U with respect to x and y: $∂^2U/∂x^2 = 0$ $∂^2U/∂y^2 = -2$ The second partial derivative with respect to x is zero, which means that the potential energy is flat in the x-direction near the equilibrium point. This indicates that small displacements in the x-direction will not cause the system to move away from the equilibrium point, but they will also not cause it to return to the equilibrium point. Therefore, the equilibrium point is unstable against small displacements in the x-direction. The second partial derivative with respect to y is negative, which means that the potential energy is concave down in the y-direction near the equilibrium point. This indicates that small displacements in the y-direction will cause the system to return to the equilibrium point, making it a stable equilibrium in the y-direction.

(a) To find the equilibrium point, we need to find the values of x and y for which the potential energy is minimized. The potential energy is given by: $U(x,y) = a1x - y^2$ To minimize this function, we need to take partial derivatives with respect to x and y and set them equal to zero: ∂U/∂x = a1 = 0 ∂U/∂y = -2y = 0 Solving for x and y, we get: x = 0 y = 0 Therefore, the equilibrium point is at x = 0, y = 0. (b) To determine the stability of the equilibrium, we need to examine the behavior of the potential energy near the equilibrium point. We can do this by calculating the second partial derivatives of U with respect to x and y: $∂^2U/∂x^2 = 0$ $∂^2U/∂y^2 = -2$ The second partial derivative with respect to x is zero, which means that the potential energy is flat in the x-direction near the equilibrium point. This indicates that small displacements in the x-direction will not cause the system to move away from the equilibrium point, but they will also not cause it to return to the equilibrium point. Therefore, the equilibrium point is unstable against small displacements in the x-direction. The second partial derivative with respect to y is negative, which means that the potential energy is concave down in the y-direction near the equilibrium point. This indicates that small displacements in the y-direction will cause the system to return to the equilibrium point, making it a stable equilibrium in the y-direction.