1. Jan 15, 2015

### geoffrey159

1. The problem statement, all variables and given/known data

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
(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

2. Relevant equations
conservation of energy

3. 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 ?

2. Jan 15, 2015

### haruspex

All looks right.
You could get (a) more easily by just summing the two potentials.
Alternatively, having found the force in (a), the quickest route to (c) is to write down the expression for acceleration and approximate it to $-\frac{2GMx}{a^3}$

3. Jan 15, 2015

### geoffrey159

I see, thank you very much !