What is the physical interpretation of dividing $2\pi$ by a variable? Looking at the angular wavenumber eqn: 
$$k = \frac{2\pi}{\lambda} = \frac{2\pi\nu}{v_p} = \frac{\omega}{v_p}$$
I'm curious what it means to divide $2\pi$ by the wavelength and why $2\pi$ was chosen. 
 A: The wave number $\kappa$ is defined to be the number of wavelengths of a wave per unit length.  So, in SI units, to get $\kappa$, you simply count the number of peaks that occur in a one-meter segment of the wave train.  (That makes the $\kappa=\lambda^{-1}$.)  This wave number is analogous to the frequency, which measures the same kind of thing, except in the time domain.  The frequency $\nu$ is the number of peaks that pass by every second.  The wave number and frequency are very useful in many contexts.
However, mathematically, the two quantities $\kappa$ and $\nu$ are not the most useful.  If you write out the equation for a sine wave (with the argument of the sine measured in radians),
$$\psi(x,t)=A\sin(2\pi\kappa x-2\pi\nu t+\phi),$$
you see that there are some inopportune factors of $2\pi$.  To get rid of these, we just define new quantities—the angular wave number $k=2\pi\kappa$ and angular frequency $\omega=2\pi\nu$—to simplify the expression for the wave to
$$\psi(x,t)=A\sin(kx-\omega t+\phi).$$  These quantities $k$ and $\omega$ have the added advantage that they come out as multiplicative factors when taking derivatives,
$$\frac{\partial\psi}{\partial x}=kA\cos(kx-\omega t+\phi).$$
A: The mathematical function $\sin(x)$ has a periodicity of $2\pi$, i.e. we have $\sin(2\pi) = \sin(0)$.
The wavelength in physics corresponds to this periodicity. We don't want to have a wavelength of $2\pi$ (which doesn't even have the dimension of length) but an arbitrary wavelength $\lambda$. In order make the wavelength tuneable, we use the function $y(x) = \sin(kx)$, where $x$ now has the dimension of length and $k$ is a constant with dimension of inverse length which needs to be determined.
To do so, note that the wavelength $\lambda$ has the property $y(\lambda) = y(0)$ (see the picture from wikipedia below). Rewriting as $\sin(k \lambda) = \sin(0)$ and comparing with the simple sine function above, we get $k\lambda = 2\pi$.
So $k=\frac{2\pi}{\lambda}$ and $\sin(kx) = \sin\left(2\pi\frac{x}{\lambda}\right)$.

