wave phenomenon of sound related question Why in an organ pipe antinodes are not formed from at free end but a starts at a distance e=0.6r?r=radius of pipe
 A: The reason is because the "free end" is not actually free. The motion of the air inside the pipe is coupled to the air outside it.
If you experiment with the longitudinal vibrations of a solid (for example a coil spring, if you want to do an experiment where the frequencies are low enough and the amplitudes are high enough so you can see what is happening without specialized measuring equipment) then the antinode is at the free end, because the amount of coupling between a metal spring and the air is very small. 
But the air just outside the end of the pipe has exactly the same acoustic properties as the air just inside the end - except that its motion is not constrained by the walls of the pipe.
A theoretical explanation of why the numerical factor is $0.6$ in the end correction formula $e = 0.6r$ involves some rather advanced mathematics, but you can show experimentally that the value depends on the conditions outside the pipe, by taking a flat plate (with dimensions comparable with the length of the pipe - it doesn't make much difference whether the plate is round or square), cutting a hole of radius $r$ in the center of the plate, and fixing it around the "free end" of the pipe. This should increase the end correction factor from $0.6$ to about $0.9$. This experiment doesn't need to be "precision engineering" - a "plate" made from cardboard will show the effect.
Another demonstration is to take a similar plate without a hole, and position it at some distance beyond the end of the pipe. (Start the distance roughly equal to the pipe diameter, and reduce it in small increments.) and As the plate is moved closer to the pipe end, the end correction factor will increase. Qualitatively, when the plate is nearer the end of the pipe it reflects more acoustic energy back into the pipe, which is equivalent to partially closing the end of the pipe.
