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**Unformatted text preview: **1 Physics 2210. Homework assignment 7. T&M 3.1. (a) = 2 = 1 2 k m = 1 2 10 0.1 = 1.59 Hz. And =1/ =0.628 sec. (b) E total =1/2kA 2 =0.5x10x0.03 2 =0.0045 J. (c) Maximum speed occurs when all the energy turns into kinetic energy. E total =1/2mv 2 max or v max = 2 E total m = 0.009 0.1 = 0.3 m/s. T&M 3.7 At the equilibrium, the buoyancy force is balanced by the weight of the body. Therefore, the weight of the body is the weight of the displaced water at the equilibrium: mg= Vg, where V is the volume of the displaced water at the equilibrium and m is the mass of the body. Let positive x direction be upward, and the origin of x at the equilibrium (i.e., the dotted line). Assume that the body is perturbed upward by the amount of x (>0) (i.e., the difference between the centers of mass before and after it is perturbed. Now the buoyancy is F b = g(V-Ax) and is positive because it points upward. The net force is F net =-mg+ F b =- gV+ g(V-Ax)=- gAx, and is negative (pointing down). Therefore, we have: F net = m x or m x + gAx = . where m is the mass of the body and is V. We can write the equation as x + gA V x = . Therefore, = gA V and = 2 = 2 V gA . For the numbers given here, = 2 0.8 10-6 10 10-4 = 0.178 sec. T&M 3.9. Let x be the equilibrium position of the block without the constant force F. Here x is measured from the ceiling (i.e., the ceiling is the origin of x coordinates, see the figure on the right). It is clear that at the equilibrium, the spring is already stretched by h=mg/k, compared to when the spring is free....

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