Why there's a displacement node at the closed end of an organ pipe? Every textbook mentions that there's a displacement node at the closed end of an organ pipe.
But the particles in air rapidly strike (or collide) with the closed end this can induce longitudinal waves in SOLID material of the organ pipe. Therefore, at the closed end the displacement of the particles shouldn't be zero. 
Also, when there's a discontinuity in the medium the reflected and the transmitted waves are given by their respective reflected and transmitted coefficients. 
Now velocity of sound in solid is significantly greater than air, so solid is rarer medium compared to air, the reflected wave is NOT inverted (w.r.t incident wave) when wave goes from air (denser) to solid (rarer) medium. Clearly the displacement at the closed end (given by sum of the reflected and incident wave) is not zero.
Am i missing something. Please help!
 A: 
Now velocity of sound in solid is significantly greater than air,
  so solid is rarer medium compared to air, [...]
Am i missing something?

Yes, you are missing something. You are correct, that
speed of sound in a solid is significantly greater than in air.
But you are wrong with the densities. A solid is certainly
denser, not rarer than air.
The speed of sound $c$ depends not only on the density $\rho$,
but also on the stiffness $K$ of the medium
(see also Speed of sound - Equations):
$$c=\sqrt{\frac{K}{\rho}}$$
From this formula you see two things:


*

*As you already know:
The speed of sound is higher for rare materials (small $\rho$)
than for dense materials (large $\rho$).

*But you missed:
The speed of sound is higher for hard materials (large $K$)
than for soft materials (small $K$).


Take for example air and steel (using data from
Density - Various materials and 
Bulk modulus - Selected values).
Air has small density $\rho=1.2\text{ kg/m}^3$
and very low stiffness $K=142\text{ kPa}=1.42\cdot 10^5\text{ Pa}$.
This gives the speed of sound $c=\sqrt{\frac{K}{\rho}}=340\text{ m/s}$.
Steel has high density $\rho=8600\text{ kg/m}^3$ and
very high stiffness $K=160\text{ GPa}=1.6\cdot 10^{11}\text{ Pa}$.
This gives the speed of sound $c=\sqrt{\frac{K}{\rho}}=4300\text{ m/s}$.
So the result is: The speed of sound is much higher in steel than in air.
And therefore, at the closed end of a steel pipe
the reflected sound wave is inverted.
