Assuming the mass $m$ stays exactly on the $y$ -axis, the force that it feels due to the two $+Q$ charges is
$\begin{align*}\frac{1}{4 \pi \epsilon_0} \frac{-2 Q q y \hat {\mathbf{y}}}{\left(R^2 + y^2\right)^{3/2}}\end{align*}$
By Newton's Second Law, then
$\begin{align*}m \ddot y = \frac{1}{4 \pi \epsilon_0} \frac{-2 Q q y }{\left(R^2 + y^2\right)^{3/2}}\end{align*}$
We use the approximation that $\left(R^2 + y^2\right)^{3/2} \approx R^3$ in order to convert this into an equation of motion that gives harmonic oscillations. By analogy with a spring with spring constant $k$ , which has angular frequency $\sqrt{k/m}$ , we have that the angular frequency in this case is
$\begin{align*}\omega = \sqrt{\frac{1}{4 \pi \epsilon_0} \frac{2 Q q }{m R^3}}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1992 Problem 65