Physical interpretation of the relation $\dot{x}(t) \rightarrow i \omega \tilde{x}(\omega)$ If $x(t)$ is some time-dependent real quantity i can interpret its Fourier Transform $\tilde{x}(t)$ as representing, in a generic sense, the frequency components of $x(t)$. What about the FT of $\dot{x}(t)$? Can this quantity be correlated in some intuitive way to the "motion" $x(t)$ itself?
In other words, is there some intuitive picture explaining the fact that $\dot{x}(t) \longrightarrow  i \omega \tilde{x}(\omega)$? This discussion should, of course, not be limited to functions that depend on time (only).
 A: The fact that you can represent a function $f$ (of time) as a Fourier transform $\tilde{f}$ already means that you can imagine the function as a superposition of sinusoids. Focus on any one of those sinusoids
$$s(t) \equiv A \sin(\omega t + \phi).$$
The time derivative (i.e. velocity) is
$$\dot{s}(t) = \omega A \cos(\omega t + \phi) . \quad (*)$$
You can see that $s$ and $\dot{s}$ have the same frequency, but are a quarter cycle out of phase and differ in amplitude by a factor of $\omega$. This is precisely the physical meaning of the $i\omega$ factor which shows up in the frequency domain. Representing the sinusoids in the frequency domain as exponentials
$$\exp[i(\omega t + \phi)]$$
you get a quarter circle phase shift by multiplying by $i$ [1]
$$i \exp[i(\omega t + \phi)] = \exp[i(\omega t + \phi + \pi/2)].$$
The $\omega$ part of $i\omega$ is the same $\omega$ as in $(*)$.
In summary, the physical meaning of $i\omega$ is "scale magnitude by $\omega$ and add a quarter cycle phase shift."
[1] This just comes from the fact that $\exp[i\pi/2]=i$.
