How does the thickness of a string affect the frequency? I've notice that thicker strings have lower frequencies, but is there a specific relationship between theses two? 
 A: A vibrating string is governed by the wave equation:
$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$
Where $y$ is a function $y(x,t)$, $T$ is the string tension and $\rho$ the linear density ($\mathrm{kg/m}$) of the string, and:
$$c=\sqrt{\frac{T}{\rho}}$$
Solved, this equation yields:
$$y(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
For $n=1,2,3,4...$ and with $L$ the string's length.
(the $A_n$ are coefficients yet to be determined from the initial condition (state) of the string)
(My full derivation of the solution can be found here)
Now let $\cos\Big(\frac{n\pi ct}{L}\Big)=\cos\omega t$
with $\omega=2\pi f=\frac{n\pi c}{L}$
For $n=1$, i.e. 'the fundamental' frequency:
$$\boxed{f_1=\frac{1}{2\pi L}\sqrt{\frac{T}{\rho}}}$$
It now becomes apparent that the frequency (pitch) of the fundamental is:
1) inversely proportional to the string's length.
2) proportional to the square root of the tension.
3) inversely proportional to the square root of the string's density.
$$All other things being equal, heavier strings will produce lower pitched notes.**
This is somewhat analogous to the simpler mass-spring system, where higher $m$ also leads to smaller $f$.
$$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$$
A: It's a bit more complicated than "thicker" or "thinner," but if you specify that the strings your are comparing have the same length ($L$), have the same volume density ($\rho$), and have the same tension ($T$), then a thicker string will have a lower fundamental frequency than a thinner one.
Now let's look at this with some mathematical tools:
The fundamental frequency of an ideal string (the real stiffness of a string can affect the frequency slightly) fixed at both ends is
$$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho A}}$$
where $A$ is the cross-sectional area of the string of radius $R$:
$$A=\pi R^2.$$
If we put this area in the fundamental frequency relation we get
$$f_1=\frac{1}{2LR}\sqrt{\frac{T}{\rho \pi}}.$$
This shows us that for strings of same length of the same density experiencing the same tension, the thicker string will have the lower fundamental (and consequently a lower pitch).
