TITLE: Physical meaning behind Frequency domain?
QUESTION [5 upvotes]: I understand its usage and why is it important because It transforms differential equations to algebraic ones.. 
But I can't get the physical meaning of the new form of the equation and the meaning of this transformation.. and also what does it mean to change the domain of the function ?

REPLY [3 votes]: Laplace transformation in its usual form is applied to onset-processes (OP's for short)
$$f:\ {\mathbb R}_{\geq 0}\to{\mathbb C}\ ,\qquad t\mapsto f(t)\quad(t\geq0)\ .$$
The result ${\cal L}f$, defined by
$${\cal L}f(s):=\int_0^\infty f(t)e^{-s\, t}\ dt\ ,$$
has no physical interpretation whatever; so nobody has ever looked at the graph of an ${\cal L}f$. But the transformation $f\mapsto{\cal L}f$ has interesting formal properties which make it useful in applications: Differentiation with respect to $t$ is transformed into multiplication with $s$, etcetera. Above all ${\cal L}$ is injective. This implies that knowing the Laplace transform of some unknown OP $f$ it is in principle  possible to get back $f$. 
As a rule, Laplace transformation and its inverse  is not applied to data, but to finite analytic expressions, using a set of rules and catalogues.
Contrasting this, Fourier transformation is applied to time signals (TS's)
$$f:\ {\mathbb R}\to{\mathbb C},\qquad t\mapsto f(t)\quad(-\infty<t<\infty)\ .$$
The result
$$\hat f(\xi):=\int_{-\infty}^\infty f(t)\, e^{-i \xi t}\ dt$$
has an interesting physical interpretation: The value $|f(\xi)|$ tells you the amplitude of the frequency $\xi$  in the considered TS. So looking at the graph of $\hat f$ when $f$ is, e.g., an analog audio signal, you get interesting information about the kind of music that is being played.
Fourier transform is applied to "abstract" functions in theoretical considerations, to "analytic expressions" in many applications where "solutions in finite terms" are desired, but also in a large extent to data coming from sampled TS's.