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Week 10: Plates and Membranes
Aims: To explore 2D vibration and implications for musical
acoustics.
Learning outcomes: An appreciation of modes in more than one
dimension (2D bar example) Using the Laplacian for equations of
motion Using separation of variables to solve (2D Circular Membrane
Example) Using separation of variables to solve (2D Rectangular
plate example
discussed in tutorial)
10.1 A Small Experiment
Two dimensional vibration is an important aspect of musical
acoustics because it not
only helps us describe the sound generation of instruments whose
main oscillators aretwo dimensional such as plates, bars or
membranes but also the resonating surfaces of
sounds boards and bodies belonging to violins, guitars, pianos,
etc.
You could try investigating two dimensional vibration of say a
xylophone by
a) striking a bar with the hammer in different locations
b) trying to damp different locations along a bar with your
finger
As you strike a bar the sound is louder in some locations than
others. Typically the
centre and edges are the loudest suggesting these areas are able
to move the most.
Conversely the sound is least damped when a finger is positioned
above an area that
moves the least (a node).
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If you could view the bar in slow motion or investigate the
phase of the vibration you
find that when the middle of the bar is at its maximum extent
downwards the edges
would be at their maximum extent upwards so it is often useful
to describe this first
mode of vibration in terms of pluses and minuses.
Striking or damping a bar transversely as opposed to
longitudinally reveals similar
results with centre and edges louder than the nodes.
Weve already seen that with one dimensional waveguides there
will be harmonics
superimposed on each other to build more complex timbres. With
two dimensions
there will be added overtones.
So for example the second mode of vibration with combined
horizontal and vertical
vibration will look as follows
10.2 Wave equations in 2D
As with the study of strings or a pipes it is useful to derive
and equation of motion
(i.e. wave equation) for the waveguide and solve it. We have
already seen the 1D
wave equation (for string or pipe) takes the form
2
2
22
2 1
tcx
=
(10.1)
where ( )tx, is the displacement of the string.
It is convenient at this point to introduce a short hand
notation used with manydifferential equations called the Laplacian
operator (or often shown as 2 ). It can
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be used to represent the second order derivate of a term in a
way that is consistent
across different dimensions and coordinate systems.
So for example in one dimension our wave equation the Laplacian
operator is
2
2
x=
in two dimensions
2
2
2
2
yx
+
=
in three dimensions
2
2
2
2
2
2
zyx +
+
=
we can also show our Laplacian operator in two dimensional polar
coordinates is
2
2
22
2 11
+
+
=rrrr
So using the Laplacian operator our 1D wave equation becomes
2
2
2
1
tc
=
(10.2)
Though not this same equation as written here can also represent
our wave equation in
one, two, three, etc dimensions.
So for instance the wave equation in 2D is
2
2
2
1
tc
=
where ( )tyx ,, and the Laplacian 22
2
2
yx
+
=
so expanding gives
2
2
22
2
2
2 1
tcyx
=
+
(10.3)
Similar in 2D polar coordinates is ( )tr ,, and the
Laplacian
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2
2
22
2 11
+
+
=rrrr
Hence our two dimensional wave equation in polar coordinates
is
0111
01
1
2
2
22
2
22
2
2
2
2
2
2
2
=
+
+
=
=
tcrrrr
tc
tc
(10.4)
10.3 Circular Membranes
With circular membranes we are dealing with circular surfaces so
is convenient to use
a polar coordinate system instead a rectangular coordinate
system because our
equations can be simplified to reflect the inherent symmetry of
the system.
Lets assume our membrane is circular and has been struck in the
middle hence our
solution for the membrane has symmetry such that displacement
doesnt change with
. Hence our term is unimportant and the wave equation
becomes
011
2
2
22
2
=+ tcrrr
where ( )tr, (10.5)
Using good old separation of variables when assume a solution of
the form
( ) ( ) ( )tTrRtr =, (10.6)
Hence substituting into the wave equation and apply partial
differentiation gives
011
2
2
22
2
=
+
t
TR
cr
RT
r
T
r
R(10.7)
Next collect all R terms to one side and all T to the other
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2
2
22
2
2
2
22
2
2
2
22
2
2
2
22
2
111
111
11
011
t
T
Tcr
R
rRr
R
R
t
T
cT
r
R
rRT
r
R
R
t
TR
cr
RT
rT
r
R
t
TR
cr
RT
rT
r
R
=
+
=
+
=
+
=
+
(10.8)
Since the left hand side doesnt care about T and the right hand
side doesnt can about
R we can say both sides equate to a constant.
2
2
2
22
2 111k
t
T
Tcr
R
rRr
R
R=
=
+
(10.9)
Hence we have two ordinary differential equations
01 2
2
2
2=+
kt
T
Tc(10.10)
011 2
2
2
=+
+
kr
R
rRr
R
R(10.11)
The first differential equation (11.10) can be written as
0222
2
=+
Tkct
T(10.12)
and has the standard solution
( ) ( )kctBkctAT sincos += (10.13)
The second differential equation (11.14) can be written as
01 2
2
2
=+
+
Rkr
R
rr
R
Lets assume krs = hencek
sr= and
k
dsdr= .
Substituting in gives
02
=++ RkdskdR
s
k
dsds
kkdRdR
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01
2
2
=++ Rds
dR
sds
Rd
022
22 =++ Rs
ds
dRs
ds
Rds (10.14)
This a Bessels equation being of a form
( )ynxyxyx 222 ++ where 0=n (10.15)
Hence the solution consists of Bessel functions of the first and
second kind.
( ) ( )xYAxJAy 0201 += (10.16)
Where the Bessel function of the first kind is
( )( )
( )
nm
m
m
n
x
nmmxJ
+
=
++
=2
0 21!
1(10.17)
and the Bessel function of the second kind (Neumann function)
is
( )( ) ( ) ( )
( )
n
xJnxJxY nnn
sin
cos = (10.18)
Dont worry too much about the equations (11.17) and (11.18) as
resulting values can
usually be access from tables or computer functions such as
Malabs besselj().
One can already see that if 0=n ( )xYn is infinite, which doesnt
make much sensegiven the displacement of the drum will be a small
finite value, so well ignore it.
So lets bring our solutions (11.13) and (11.6) together to
represent drum
displacement.
( ) ( )( ) )(sincos)()( 0krJcktBcktArRtT +==
(10.19)
Hence lets try to simplify and make sense of this solution by
applying realistic
boundary conditions. Typically drum membranes are fixed to a
circular rim so our
solution can only be valid if ( ) 000 =krJ where 0r = the radius
of the drum surface.Lets try plotting 0J with respect to 0kr see
where the zeros occur.
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We can see 00 =J when 0kr =2.40, 5.52, 8.65, 11.75, 14.93 etc.
Hence oursolution will be the sum of discrete values
( ) ( )[ ] ( )
=
+=1
00sincosn
nnnnnrkJtckBtckA (10.20)
Considering the initial conditions the drum has no shape so
clearly all values of 0=A
because the ( )cktcos is non zero when 0=t
Hence
( )[ ] ( )
=
=1
00sinn
nnnrkJtckB (10.21)
So our we are looking at a set of sinusoids of the form ( )tnsin
oscillating at discrete
frequencies n where clearly ckn = and
01 /40.2 rk = , 02 /52.5 rk = , 03 /65.8 rk = , 03 /75.11 rk = ,
03 /93.14 rk = , etc
These frequencies do not form a harmonics series so the timbre
might not have as
discernable a pitch as one dimensional waveguides. Its
interesting to try hearing
some of the discrete frequencies together. Matlab is ideal for
this, where by adding
sinusoid at these Eigen frequencies one can producing a range of
timbres from gong-
like and drum-like sounds depending on values for tension,
density and the relative
amplitudes of the harmonics.
Cut and paste following Matlab the material properties chosen
arent that realistic.
fs=11000;SurfaceTension=10;
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SurfaceDensity=0.003;
DecayConstant=7; % try 1 for gong
r0=0.3;
B1=1.0; B2=0.7; B3=0.5; B4=0.3; B5=0.1;
c=sqrt(SurfaceTension/SurfaceDensity)
w1=2.40*(c/r0)w2=5.52*(c/r0);
w3=8.65*(c/r0);
w4=11.75*(c/r0);
w5=14.93*(c/r0);
t=linspace(0,5,fs*5);
y=B1*sin(w1*t)+B2*sin(w2*t)+B3*sin(w3*t)+B4*sin(w4*t)+B5*sin(w5*t);
y=0.1*y.*exp(-DecayConstant*t);
sound(y,fs)
