Solution to 1992 Problem 65

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.

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