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## Homework Statement

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

## Homework Equations

conservation of energy

## The Attempt at a Solution

(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 ?