Why does the weaker C–H bond have a higher wavenumber than the C=O bond?
A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object:

$$\omega = \sqrt{\frac{k}{m}}$$

We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $\mu$, defined by:

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$

where $m_1$ and $m_2$ are the masses of the two bodies. So, for a C–H bond, we have

$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$

and for a C=O bond, we have

$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$

From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $k$ for the C=O bond in formaldehyde is approximately $2.46$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).

So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $\bar{\nu}$, is directly proportional to the (angular) frequency $\omega$.

$$\begin{align}
\frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &amp;= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\
&amp;= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\
&amp;= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\
&amp;= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\
&amp;= 1.74
\end{align}$$

This is not too far from the experimental value of $\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$ (using values in the middle of the typical ranges for alkanes and aldehydes).