For those who doubt that my Helmholtz tuning method with a grid of tubes can tune a room here, what i call "mechanical equalization", here are some simple basic observations to read from a MASTER book i just discovered 5 minutes ago ...
I cannot reproduce images of the text here, nor the equations correctly, but the text say it all and give some basic understanding about an important fact in my grid of tubes : The more absorbing tubes and the more diffusing one, or the open/closed tubes and the open/open tubes and the case in between these 2 types, with mutiples littltes tubes inside a bigger one, and also some tube with a cloth fabric filtering device at one end and the other end open... These are the 4 types of tubes of various size at different locations i used in my "mechanical equalizer"...
«2.5.3 Sinusoidal wave excited by a velocity source in an acoustic
tube.
In the last subsection sound waves were shown to be excited by a sound pressure
source at the entrance x = 0 of an acoustic tube. Another type of exciting source is a
velocity source.
Fig. 2.16 shows an image of an acoustic tube with an open end at x = L, and a
sinusoidal velocity source at the entrance x = 0.
The boundary condition can be formulated such that
p(x,t)|x=L = 0 (Pa) and ∂p(x,t)∂x ����
x=0
= −iωρV eiωt, (Pa/m) (2.54)
where
v = V eiωt (m/s) and − ∂p
∂x
= ρ
∂v
∂t
= iωρv (Pa/m) (2.55)
hold [1][7][13] between the sound pressure (Pa) and the particle velocity v (m/s) for
sinusoidal plane waves, and ρ (kg/m3) denotes the volume density of the medium.
Assuming the general solution given by Eq. (2.43) once more, then the solution
must satisfy
Ae−ikL + BeikL = 0 (Pa) and Ak − Bk = ωρV (Pa/m) (2.56)
under the boundary condition specified by Eq. (2.54). Solving the simultaneous equation in Eq. (2.56) for A and B, then the general solution can be rewritten as
p(x,t) = iρcV
sink(L − x)
coskL
eiωt. (Pa) (2.57)
Comparing the solution given by Eq. (2.57) with that by Eq. (2.46) for the sound
pressure source, the poles of the solution for the velocity source are located at lower
frequencies:
cosknL = cos
ωn
c
L = 0 or ωn =
c
2L
(2n − 1)π. (2.58)
The fundamental (n = 1) is lower than that for the pressure source by 1/1−octave.
The pressure response to the velocity source is purely imaginary. In addition, the
poles are composed of odd harmonics without the even harmonics. The difference in
the poles between the pressure and velocity sources can be interpreted as the difference in the boundary conditions for the acoustic tubeFig. 2.17 displays the two types of boundary conditions for acoustic tubes: openopen and open-closed end conditions [17]. The open-open tube represents the case
for the pressure source that makes the condensation or dilation of the medium at the
entrance. Assuming the impulsive condensation at the left end, then the pulse-like
positive pressure travels inside the tube. The pulse-like pressure wave is reflected at
the right end (open end) by the negative magnitude. The return of the negative pressure wave to the left end changes the sign of the pressure to positive, which goes
toward the right end. The periodic traveling yields the fundamental frequency.
In contrast, the propagation wave from a velocity source can be interpreted as
the traveling waves in an acoustic tube under the open-end boundary conditions, as
shown in the right panel of Fig. 2.17. Velocity excitation might be made by piston
motion of a “plate” at one end of the tube. As happens in the open-open tube, the reflected pressure wave with the negative sign comes back to the source end. However,2.5
Acoustic tubes with open-open and open-closed ends, where pulse-like waves propagate
between the two ends; from Figs. 4.1 and 4.5 [17].
no sign change occurs at the end because the end is closed by the piston plate, which
reflects the wave without a sign change. This traveling of the pulse-like wave is periodic; however, the period is two times longer than that for the open-open tube. The
longer period makes the fundamental lower by 1/1− octave than that for open-open
tube [17].
The two types of boundary conditions may represent the conceptual models for
a flute by the open-open condition, and a clarinet by the closed-open condition. The
difference of the boundary conditions might explain the difference in the fundamental
frequencies with their harmonics. However, the boundary condition for a clarinet
might be mixed rather than purely open-closed [5]. The harmonics can be composed
of even and odd harmonics even for a clarinet.»
Acoustic Signals and hearing by Mikio Tohyama 2020