AS for physical meaning of Hardy's inequality for p=2. Consider the Schroedinger operator $$H = -\frac{d^2}{dx^2} - \frac{c}{x^2}$$ on $(0,\infty)$ with Dirichlet boundary condition at $0$. A natural question is: When is this operator positive on $L^2(0,\infty)$ and the answer is just Hardy's inequality, since $$\langle \psi, H \psi \rangle \geq 0$$ is equivalent to $$\int_0^{\infty} |\psi'(x)|^2 dx \geq + c \int_0^{\infty} \left|\frac{\psi(x)}{x}\right|^2 dx$$ an inspection of Hardy's inequality now shows that $c = \frac{1}{4}$ is critical.
One can actually use this reasoning to prove Hardy's inequality, since the ODE $-u''(x) = c u(x)/x^2$ is explicitly solvable.