Relation between frequencies of different harmonies I know the relation between frequencies of different harmonies is $$ f_n = n\times f_1 $$ but I'm heving trouble to develope the equation which proves this equality. can anyone please give me a lead or explain me why is this true? here's a picture from wikipedia to better explain what I mean:

 A: What your wikipedia page shows is that for waves on a string an integer number of half-wavelengths fit into the length: $L=n{\lambda \over 2}$ with $n=1,2,3...$.
Which specifies a rule for the allowed wavelengths:  $\lambda={2 L \over n}$
Now use $c=f \lambda$, where $c$ is the speed of waves on the string (given by $\sqrt{T/\rho}$, though that's not relevant here) and get the allowed frequencies 
as $f={c \over \lambda}=n{c \over 2 L}$. The frequencies are multiples of ${c \over 2L}$
For waves in a pipe the argument is similar except that (1) $c$ is the speed of sound and (2) there is a difference between pipes which are open or closed at both ends (like a flute) and pipes which are closed at one end but open at the other (like a trumpet) for which the pictures are different and the rule is that an odd number of quarter-wavelengths fit into the length. That's why brass instruments have such a distinctive sound; they miss all the even harmonics.
A: The frequency spectrum results from solving the wave equation (here in one dimension):
$$\frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}$$
You can find the full derivation here:
Full derivation (I'm the author of these posts)
And resulting the frequency spectrum:
Frequency spectrum:
$$\cos\Big(\frac{n\pi ct}{L}\Big)=\cos\omega t$$
Thus:
$$\omega=2\pi f=\frac{n\pi c}{L}$$
And:
$$f=\frac{nc}{2L}$$
$$\frac{T}{\rho}=c^2$$
$$f=\frac{n}{2L}\sqrt{\frac{T}{\rho}}$$
$$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho}}$$
Finally:
$$\boxed{f_n = n\times f_1}$$
A: To add to what is here consider the full solution to the 1-dim wave equation, 
y = a sin(kx) + b cos(kx)
where k is the wavenumber, 2*pi/lambda, lambda = wavelength.
The wavelength and frequency are related to the velocity of the waves through the equation, 
v = lambda * f = w/k, where w = 2*pi*f.
For the string with fixed ends at x = 0 and x = L we apply these boundary conditions to the full solution, y(0) = y(L) = 0. 
x = 0 --> y(0) = b = 0.
This leaves you with y = a sin(kx)
x = L --> y(L) = a sin(kL) = 0
This is where the "spectrum" comes from.  We must have kL = n*pi to ensure this term is always zero.  This implies that k = n *pi/L.
Going back to the frequency relation we get, 
w = vk = n(vpi/L).  This is all you need to show that the harmonics of the string with fixed ends are an integer times the lowest frequency.  The wave speed, v, will depend on the tension of the string and of the mass density.  But it suffices to say f1 = (v*pi/L) and f_n = n *f1.
There is a similar situation for air in a tube be the boundary conditions are different since you are dealing with a function that represent the pressure difference or the local particle displacement in a longitudinal wave.  This is critical and you don't want to screw that up.  For an open end of a pipe one has a displacement anti-node, that is the air is free to move.  While for a closed end of a pipe you have a displacement node, no air movement.  For the pressure field these are reversed, open = pressure node, closed = pressure anti-node.  
You will get the same spectrum for the pipe with 2 open ends but not for one open one closed, and the details of the math might be a little different.
Moving on to stiff beams (which you didn't really ask about) the differential equation is different and one needs to specify more than just two BC to get the correct behavior.  This leads to different spectra for the harmonics.  Lastly, for 2 dimensional and 3 dimensional problems you do not get n *f1 for the spectrum but something like, 
f_nm = sqrt(n^2 + m^2) *f1
if the problem has enough symmetry.  I don't want you thinking that all vibrating systems have the same harmonic relationship.
In general one has to go through the following steps.


*

*Find a general solution to the PDE (Partial Differential Equation) for describing the vibrations.

*Figure out the correct BC (Boundary Conditions) for the system (this is the important part).

*Apply the BC to (a) reduce the solution to one that "fits" the system, and (b) find the spectrum.
The BC is where all the magic happens. 
