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Gravitational Potential Energy and Oscillations of a Bead Attracted by Spheres
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[QUOTE="geoffrey159, post: 4977104, member: 532398"] [h2]Homework Statement [/h2] A bead of mass m slides without friction on a smooth rod along the x axis. The rod is equidistant between two spheres of mass M. The spheres are located at x = 0, y = ± a, and attract the bead gravitationally. (a) Find the potential energy of the bead. (b) The bead is released at x = 3a with velocity v0 toward the origin. Find the speed as it passes the origin. (c) Find the frequency of small oscillations of the bead about the origin [h2]Homework Equations[/h2] conservation of energy [h2]The Attempt at a Solution[/h2] (a) The two working forces acting on the bead are the attraction of both spheres: ##\left\{ \begin{array}{} \vec F_{s_+} = \frac{GmM}{(x^2+a^2)^{3/2}} (-x \hat x + a \hat y) \\ \vec F_{s_-} = \frac{GmM}{(x^2+a^2)^{3/2}} (-x \hat x - a \hat y) \end{array}\right. ## Their sum, ##\vec F= -2GmM \frac{x}{(x^2+a^2)^{3/2}} \hat x ##, is clearly a central force, so it is conservative and has a potential function : ##U(x) - U(x_0) = - W_{x,x_0} = 2GmM \ \int_{x_0}^x \frac{r}{(r^2+a^2)^{3/2}} dr = - \frac{2GmM}{\sqrt{x^2+a^2}} + Cst ##. So, without the constants, the potential energy is ## U(x) = -\frac{2GmM}{\sqrt{x^2+a^2}} ## (b) I use conservation of mechanical energy, if (a) is ok, (b) is straightforward (c) The origin is a stable equilibrium point because it is a minimum of the potential function: ##\frac{dU}{dx}(0) = 0## and ##\frac{d^2U}{dx^2}(0) = \frac{2GmM}{a^3} > 0 ##. Now around 0 : ## U(x) \approx U(0) + \frac{1}{2} \frac{d^2U}{dx^2}(0) x^2 ## which is the potential energy of a spring with constant ## k = \frac{d^2U}{dx^2}(0)## Then the frequency of oscillations around equilibrium is ##\omega = \sqrt{\frac{k}{m}} = \sqrt{ \frac{2GM}{a^3}} ## Is that correct ? [/QUOTE]
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Gravitational Potential Energy and Oscillations of a Bead Attracted by Spheres
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