Here is another analysis in Fourier (frequency) space

Consider the sloshing differential equation

$f''(t) + (a + q(t)) f(t) = h(t)$

in frequency space this would be

$(\omega^2 + a) F(\omega)+ Q(\omega) \ast F(\omega) = H(\omega)$

The tricky part is in the second term, which is a convolution.

The Hill or Mathieu factor $q(t)$ is generally a sinusoid. The Fourier transform of a sinusoid is:

$\mathcal{F}[q(t)] = \mathcal{F}[b \cos(\omega_o t)] = b \frac{\delta(\omega-\omega_0) + \delta(\omega+\omega_0) }{2}$

doing the convolution

$Q(\omega) \ast F(\omega) = b \frac{\delta(\omega-\omega_0) + \delta(\omega+\omega_0) }{2} \ast F(\omega) = b \frac{F(\omega-\omega_0) + F(\omega+\omega_0) }{2}$

What this does is bifurcate or splits the spectral peaks.

$(\omega^2+ a)F(\omega) + b \frac{F(\omega-\omega_0)+F(\omega+\omega_0)}{2} = H(\omega)$

This is an unusual looking equation because of what look like "shifted" frequency terms. Ordinarily this is is evaluated recursively, and for a delta forcing, a Mathieu function results for *F* for a *q(t)* that is a single sinusoid.

Otherwise, one can picture in you mind that for a RHS of a QBO frequency corresponding to $2\pi/2.33$ rads/year and a $\omega_0$ corresponding to the Chandler wobble of $2\pi/6.43$, then two sidebands will appear as the sum and difference of these two values.