Oscillations of a tetra-atomic molecule

Let us displace the charges along diagonals by a infinitely small distance $x$ . Let U be the potential energy of system.Let $a = \frac{l}{\sqrt{2}}$

$U = 4 \frac{kq^2}{l} + \frac{kq^2}{2(a + x)} + \frac{kq^2}{2(a - x)}$

= $4 \frac{kq^2}{l} + \frac{kq^2}{a}(1 + \frac{x^2}{a^2})$

$\frac{dU}{dx} = \frac{2kq^2x}{a^3} = \frac{4\sqrt{2}kq^2x}{l^3}$

$F = \frac{1}{4}(-\frac{dU}{dx}) = - \frac{\sqrt{2}kq^2x}{l^3}$

$\Rightarrow T = 2\pi \sqrt{\frac{ml^3}{\sqrt{2}kq^2}} = \fbox{0.0528}$

Suppose the distance from 2 opposite charges to the center of the system is $a$ , from 2 other charges to the center is $b$ . Hence, $a^2+b^2=l^2$ .

Therefore, $2a \Delta a+2b \Delta b=0$ .

When the system is in equilibrium, $a=b=\frac{l}{\sqrt{2}}$ . Therefore, with small oscillation, $\Delta a+\Delta b=0$ or the displacement of all the charges are the same.

Suppose each charge is displaced by $x$ .

The energy of the whole system is:

$E=4\frac{mx'^2}{2}+4\frac{kq^2}{l}+\frac{kq^2}{l\sqrt{2}+2x}+\frac{kq^2}{l\sqrt{2}-2x}$

$=2mx'^2+\frac{4kq^2}{l}+\frac{2\sqrt{2}lkq^2}{2l^2-4x^2}$

$\approx 2mx'^2+\frac{4kq^2}{l}+\frac{2\sqrt{2}lkq^2}{2l^2}(1+\frac{4x^2}{2l^2})$

$= 2mx'^2+\frac{4kq^2}{l}+\frac{\sqrt{2}kq^2}{l}(1+\frac{2x^2}{l^2})$

Since the energy of the system is constant, $\frac{dE}{dt}=0$

Hence, $4mx'x''+\frac{\sqrt{2}kq^2}{l} \frac{4xx'}{l^2}=0$

$x''+x \frac{\sqrt{2} kq^2}{ml^3}=0$ .

Therefore, the period of the small oscillation is: $T=2\pi \sqrt{\frac{ml^3}{\sqrt{2}kq^2}}=0.0528(s)$ .

In general, the four particles form a rhombus. Suppose that the lengths of the two diagonals of the rhombus are $2x$ and $2y$ , where $x^2 + y^2 = \ell^2$ . Then the kinetic energy of the system is $\tfrac12m\dot{x}^2 + \tfrac12m\dot{x}^2 + \tfrac12m\dot{y}^2 + \tfrac12m\dot{y}^2 \; = \; m(\dot{x}^2 + \dot{y}^2)$ and the potential energy of the system is $\frac{kq^2}{2x} + \frac{kq^2}{2y}$ Only the distances between the opposite pairs of particles change (the other four pairs of particles are all distance $\ell$ apart), and hence we only need to add their contribution to the potential energy equation. Thus $m(\dot{x}^2 + \dot{y}^2) + \frac{kq^2}{2x} + \frac{kq^2}{2y} \; = \; c$ for some constant $c$ , and hence $2m\dot{x}\ddot{x} + 2m\dot{y}\ddot{y} - \frac{kq^2}{2x^2}\dot{x} - \frac{kq^2}{2y^2}\dot{y} \; = \; 0$ Linearising this equation near the point of equilibrium we have $x \approx \tfrac{\ell}{\sqrt{2}} + u$ , $y \approx \tfrac{\ell}{\sqrt{2}} - u$ , where $\begin{array}{rcl} \ddot{u} + \tfrac{kq^2\sqrt{2}}{m\ell^3}u & \approx & 0 \\ \ddot{u} + (\sqrt{2}\times10^4)u & \approx & 0 \end{array}$ giving SHM with approximate period of $2\pi/\sqrt{\sqrt{2}\times10^4} = 0.0528$ seconds.

One can use the concept of instantaneous axis of rotation along one rod, and then restore the torque about instantaneous axis of rotation.