 Solution

Substitution of the proposed sine and cosine displacements into either linearized equation gives $m{\omega }^2+qB\omega -k=0$. ThusThese are signed angular frequencies, not two positive frequency magnitudes. To first order, with $\epsilon \ll {\rho }_0$,The orbit is therefore a circle of radius $\epsilon$ about $({\rho }_0,0)$ at this order. The positive root runs counterclockwise; the negative root runs clockwise and has the larger magnitude. A positive charge in $B\hat{\mathbf{z}}$ alone circles clockwise. Here the spring permits an additional counterclockwise mode by providing inward force against its magnetic force.