Normal mode analysis I'm reading lots of texts about normal modes and I've seen that normal modes are solutions of the wave function produced by separation of variables. However, when most of authors I've read perform the separation of variables, they consider:
$$
\psi (t,x)=\phi (x) e^{i\omega t} \,\, ,
$$
For example. My question is: from where does this exponential dependence come? Why this dependence? It's looking like an \textit{ad hoc} assumption, what do we gain with it? Couldn't it be any other, for example $\exp({-i\omega t})$ ?
 A: $\def\label#1{}$
I think that this can be easily understood from the perspective of Fourier transforms.
We want to calculate the solution $\psi(t,x)$ of the differential equation $\mathcal{L}[\psi(x,t)]=0$.
Let us assume that $\psi(t,x)$ is well behaved in $t$ in the sense that it can be expanded as a Fourier series.
Then,
\begin{align}
\psi(t,x) = \frac{1}{\sqrt{2\pi}} \int d\omega\,
\phi(x,\omega) e^{i\omega t}.
\end{align}
Insert this in the differential equation:
\begin{equation}
\mathcal{L}[\psi(t,x)]=
\frac{1}{\sqrt{2\pi}} \int d\omega\,
\mathcal{L}[\phi(x,\omega) e^{i\omega t}]=
\frac{1}{\sqrt{2\pi}} \int d\omega\,
\mathcal{M}[\phi(x,\omega)] e^{i\omega t}=0,
\end{equation}
where $\mathcal{M}$ is a differential operator containing only derivatives in $x$, because the derivatives in $t$ have already acted on the factor $e^{i\omega t}$.
Then, it follows
\begin{align}
\mathcal{M}[\phi(x,\omega)]=0
\label{eq:M}
\end{align}
for a generic $\omega$.
Imagine that you can solve this last equation and get $\phi(x,\omega)$.
Then, you have $\psi(t,x)$ by making use of the definition of the Fourier transform given above.
As you can see there, the general solution $\psi(t,x)$ is precisely a linear combination of the solutions that you get with the ansatz of your question.
Notice that by doing this, you have converted your original problem, namely solving a differential equation in two variables $x$ and $t$, into a simpler problem: solving a differential equation in just one variable, namely $x$.
The price to be paid is the integration that you have to perform to recover $\psi(t,x)$ from $\phi(x,\omega)$.
I think that this is the reason why the exponential appears in your ansatz.
Moreover, this reasoning allows you to see that there is no loss of generality (that is, the choice of your ansatz is legitimate).
A: The exponential dependence comes about because we are breaking the oscillatory motion into sines and cosines.  Why do we choose sines and cosines?  The reason is that normal modes occur about some equilibrium, and an equilibrium is a local minimum in the energy.  The Hamiltonian or Lagrangian can then be Taylor expanded about this equilibrium point, and the first non-zero term is quadratic.  If we ignore all higher order terms, this is exactly the potential of a simple harmonic oscillator and the motion of a simple harmonic oscillator is given exactly by a sinusoid.  
So for some arbitrary oscillation about an equilibrium point the oscillation will always be approximately sinusoidal as long as the magnitude of the oscillation is not too great. 
A: The wave phenomena we're interested in is usually governed a second order differential equation with respect to time. A spring-mass oscillator or LC circuit for example, are both governed by equations of the form:
$$\frac{d^2 u}{dt^2} = -\omega^2u$$
This equation has sin or cosine solutions, but it's much easier to work with the exponential solutions. You're correct that $\exp(i\omega t)$ and $\exp(-i\omega t)$ are both valid choices. The preferred choice is convention, and changes depending on the field (it can be a pain trying to figure out which convention any particular author's using).
Why exponentials? Because they're eigenfunctions with respect to derivatives, which makes them convenient mathematically. Not only that, but they're orthogonal, which allows us to write any arbitrary time dependence as a sum of complex exponentials.
Another thing worth mentioning is that to solve the wave equation in different geometries, it's often necessary to choose coordinates appropriate to those geometries, and separate using those variables. This will lead to different solutions in the spatial variables. The time dependence of the wave equation is always going to be the same, so you can always choose the same exponential time dependence, without even choosing the boundaries or geometry of the problem.
Edit: To answer why we don't use the general solution $y=Ae^{iwt}+Be^{-iwt}$ or choose the constant $A$ different from the unity, it's useful to consider the rest of the wave equation solution. For our simplest form of the wave equation:
$$\frac{\partial ^2 \psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial ^2 \psi}{\partial t^2}$$
We can separate this into the spatial and time dependent parts. Since this is second order in both time and space, there will be two solutions for each the time and spatial components. The total solution will thus be of the form:
$$\psi = (Ae^{i\omega t} + Be^{-i\omega t})(Ce^{ik x} + De^{-ik y})$$
Multiplying this out will yield four solutions
$$\psi = A'e^{i(\omega t - kx)} + B'e^{i(\omega t + kx)} + C'e^{-i(\omega t -kx)} + D'e^{-i(\omega t -kx)}$$
If we insist that the answer to the wave equation be a real valued quantity, then the $C'$ and $D'$ parts of the solution must be complex conjugates of the $A'$ and $B'$ parts. We can equivalently write the solution as:
$$\psi = \operatorname{Re} \{A''e^{i(\omega t - kx)} + B''e^{i(\omega t + kx)} \}$$
If we're being lazy, we can by convention say that only the real part represents a physical solution, and write:
$$\psi =  A''e^{i(\omega t - kx)} + B''e^{i(\omega t + kx)}$$
Then we can write the solution as:
$$\psi = e^{i\omega t}(A''e^{-ikx} + B''e^{ikx})$$
It turns out that the solution comes out the same as if we simply chose the time dependent part to be $e^{i\omega t}$ in the first place. The constant $A$ gets multiplied into the spatial part anyways, and the $B e^{-i\omega t}$ part gets erased by the convention that we only care about the real part of the solution. Of course, there are other ways to write the solution to the wave equation, so it really boils down to convenience and convention.
