Why does an OH bond absorb a higher frequency than a CH bond?
First, let me just put down the approximate $\ce{X-H}$ stretching frequencies (where, $\ce{X = O, C}$) just as a reference.

$$\begin{array}{c} 
 &amp; \ce{X-H}\ \text{stretch}\ \mathrm{(cm^{-1})}  \\ \hline
\ce{X = O} &amp; 3600  \\ \hline
\ce{X = C} &amp; 3000 \\ \hline
\end{array}$$

Now, as a simplification I will first discuss a diatomic molecule, thought of as a two balls of some mass $\mathrm{m}$ linked together with a spring. Solving, Schrodinger's equation for this system (called a harmonic oscillator), yields the following relation:

$\mathrm{\bar{\nu}} = \frac{1}{2\pi c}\sqrt{\frac{k}{m}}$

For, dissimilar masses, (say, $\mathrm{m_1}$ and $\mathrm{m_2}$), we can switch to the centre of mass coordinates, and the $\mathrm{m}$ in the preceding equation is replaced with the reduced mass ($\mu$) defined as,
$\frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} $ 

Consequently, $\ce{C-H,O-H}$ bonds in general have much higher stretching frequencies than do corresponding bonds to heavier atoms other than $\ce{H}$.The mass effect becomes evident when deuterated isotopes are examined. The stretching frequency of a free $\ce{O-H}$ bond is ca. $3600\ \mathrm{cm^{-1}}$, but the $\ce{O-D}$ equivalent is lowered to ca. $2600\ \mathrm{cm^{-1}}$. Since deuterium has a mass twice that of hydrogen, the mass term in the equation changes from 1 to 1/2, and the frequency is reduced by the square root of 2.

Naively, one would expect the frequency of the $\ce{C-H}$ bond to be higher, based on the preceding discussion, but there is another factor at play, the force constant $k$. This, is different for $\ce{O-H}$ and $\ce{C-H}$ bonds, and depends on the bond strength (stiffness of the spring)

Bond strengths can be estimated from bond disassociation energies (BDEs), which is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolytic cleavage at $\mathrm{0\ K}$. This values are frequently tabulated, and can be easily found online (here for example; caveat the values here are at $\mathrm{298\ K}$)

$$\begin{array}{c} 
 &amp; \ce{X-H}\ \text{BDE}\ \mathrm{(kJ\ mol^{-1})}  \\ \hline
\ce{X = O} &amp; 428.0\ (21)  \\ \hline
\ce{X = H} &amp; 337.2\ (8) \\ \hline
\end{array}$$

Anyway, this tells you that the $\ce{O-H}$ bond is stronger; Why? Well, simply look at the approximate atomic radii of Oxygen and Carbon. The Oxygen is about $\mathrm{48\ pm}$ while carbon is about $\mathrm{67\ pm}$. The longer atomic radii implies a longer $\ce{C-H}$ bond length, ergo a weaker bond. 

Stronger bond, implies larger $k$, which in turn means greater frequency of vibration.

So, the $\ce{O-H}$ bond is NOT weaker. Remember, when we define bond disassociation energy, we are looking at homolytic cleavage. I really don't understand the point you try to make about electronegativities.