= Solution
Taking the <gradient> of the quadratic <potential energy> found in part (ii) resolves the linearized force into identical $x$ and $y$ components:
$$
F_x=-\frac{3q^2}{8\pi\epsilon_0\rho^3}x,
\qquad
F_y=-\frac{3q^2}{8\pi\epsilon_0\rho^3}y.
$$
Each component therefore performs <simple harmonic motion> with
$$
\omega^2=\frac{3q^2}{8\pi\epsilon_0m\rho^3}
=\frac{9\sqrt3}{8\pi}\frac{q^2}{m\epsilon_0a^3}.
$$
Thus the <angular frequency> is
$$
\boxed{\omega=\frac{3^{5/4}}{\sqrt{8\pi}}\frac{|q|}{\sqrt{m\epsilon_0a^3}},
\qquad A=\frac{3^{5/4}}{\sqrt{8\pi}}}.
$$
Back to article page
