There are aspects of this demonstration that I am still trying to reconcile with my understanding of woodwind acoustics. A long cylindrical tube that is open at both ends behaves similar to a flute in which the wavelength is 2X the length of the tube which means the standing wave makes a trip down and back. The length of the tube determines the "natural resonant frequency" of a column of air inside the tube when the air molecules are set into vibration. This can be demonstrated by the children's musical toys called "Boomwhackers". When the tube is closed on one end it behaves similar to a clarinet in which the wavelength is 4X the length of the tube which means the standing wave makes 2 complete trips down and back.
When the tube becomes closed on one end at the same time the air is set into vibration, the natural resonant frequency of the air column inside the tube will have a wavelength 4X the length of the tube regardless of whether the tube is struck and closed with a soft surface or a hard surface. It is important to note that the vibrations of the tube material itself, if any, are insignificant. What we hear are the vibrations of the column of air inside the tube similar to a saxophone.
We also know from the study done by Pauline Eveno that the surface of an "uncovered pad" without a resonator does not reflect the sound waves as effectively or efficiently as a hard rigid surface and therefore a portion of the energy of the sound wave is said to be "attenuated". This loss of energy can be the result of a porous surface, a rough surface, or a surface that lacks rigidity. This principle is used when "soundproofing" a room. In "theory" the resultant pitch of the vibrations inside the tube should be the same whether the tube is closed by a hard or soft surface and the air molecules set into vibration. The only difference would be the amplitude of the sound. Perhaps the "material" the tube is made of contributes to this discrepancy, or that the hard surface does not seal the end of the tube completely.
I'm not saying Jim's explanation is wrong, just that some elements of the demonstration are beyond my understanding at this time.
When the tube becomes closed on one end at the same time the air is set into vibration, the natural resonant frequency of the air column inside the tube will have a wavelength 4X the length of the tube regardless of whether the tube is struck and closed with a soft surface or a hard surface. It is important to note that the vibrations of the tube material itself, if any, are insignificant. What we hear are the vibrations of the column of air inside the tube similar to a saxophone.
We also know from the study done by Pauline Eveno that the surface of an "uncovered pad" without a resonator does not reflect the sound waves as effectively or efficiently as a hard rigid surface and therefore a portion of the energy of the sound wave is said to be "attenuated". This loss of energy can be the result of a porous surface, a rough surface, or a surface that lacks rigidity. This principle is used when "soundproofing" a room. In "theory" the resultant pitch of the vibrations inside the tube should be the same whether the tube is closed by a hard or soft surface and the air molecules set into vibration. The only difference would be the amplitude of the sound. Perhaps the "material" the tube is made of contributes to this discrepancy, or that the hard surface does not seal the end of the tube completely.
I'm not saying Jim's explanation is wrong, just that some elements of the demonstration are beyond my understanding at this time.
