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Video 6

Hello and welcome to lecture number 33 in this series on Acoustic Materials and
Metamaterials. So, today is the last lecture on a Membrane Type Acoustic Metamaterials
and this is a tutorial session. So, we will solve a few problems related to the two types of
unit cells that we have studied, so that you can get a more better understanding of how to
design such kind of membrane type metamaterials.
(Refer Slide Time: 00:53)

So, the first numerical if you see here, the problem that is given to us is; you have. A figure
is shown here and you have an acoustic transmission line, which contains two types of unit
cells. It has a unit cell type 1 which is given here and the unit cell type 2 that is given here
and they are connected in series. And the dimensions of the unit cells are uniform
throughout the transmission line.
So, you, so the way it is given is that, so this is like a continuation. So, you have first unit
cell 1, unit cell 2, then unit cell 1, then another unit cell 2. So, alternately these two types
of units cells they are connected together into a long acoustic transmission line or an
acoustic waveguide. So, over here the things that are given to us is that, the tension that is

applied to the membrane it gives it stiffness and that is given by this quantity. So, it is
2000 Newtons per meter. The surface density of the membrane is given as 2 kgs per meter,
square surface density of the center mass is given as 200 kg per meter square.
So, you have to find out here, what is the range of frequencies within which it can act to
control noise or it can act to reduce the sound? So, we will start with this problem here.
Now we know that for both unit cells the range of frequencies where they will reduce the
sound is actually the range where the effective ρ; that is the effective mass density becomes
negative, at is that range within which suddenly the wave propagation stops, because the
propagation vector becomes imaginary. So, let us first deal with unit cell type 1.
(Refer Slide Time: 02:43)

So, for unit cell 1, the frequencies where it can block sounds is equal to the frequency
range of, ρeffective < 0. So, this and we know that ρeffective < 0. So, last class, in the
last few lectures we have studied that, for the first type where there is no mass attached;
then the ρeffective < 0 where the angular frequency is between ω to ω0, where ω0 is the
natural frequency of the unit cell that is given by the stiffness. It is given by the, you can
say the effective stiffness of the membrane; although membrane is not very stiff, but due
to the tension applied it gives some stiffness to the membrane. So, this √
km
M
; where M is
the mass of the membrane plus the mass of the enclosed air. So, this is given to us. So, let
us find out what is this range.

So, if you do this, then what will be the range in the frequency scale? In the linear
frequency scale the range will be: (0 to ω0

). So, the range that we are finding out is

between (0 to ω0

) is going to be this value here. So, we get 1


Km
M
. So, this is the range
of frequencies that we have to find. This is the required frequency range, where unit cell 1
will reduce sounds, will reduce or control sound. So, let us find out this value here this
value. So, Km = 2000 Newtons per meter, ok. And ‘M’, let us find out what is ‘M’. So,
let us find out first what is the mass of the membrane?
So, the mass of the, and the mass of the air; now air here you can see that, the condition
for loading is not given. So, we assume it is being loaded with. So, that is the general
assumption that the transmission line contains air as the medium at room temperature. So,
at room temperature what happens is that. So, at room temperature what you get is the ρ
of air.
So, these are the values which are easily available to you, you can look up into the book.
So, you look books or standard tables even online, so the density and the speed of sound
value. So, the ρ and the c values of air at different temperatures are already pre calculated
and available at various sources. So, ρ of air at room temperature is 1.2041 kg per meter
cube. So, this is the value which we will be using.
So, let us now calculate what is the mass of air? It is going to be this ρ of air multiplied by
the volume of the unit cell, which is going to be 1.2041 multiplied by. So, here you have
area into the length of the unit cell. So, what is the area here it is this is the diameter. So,
diameter d is equal to, the diameter of unit cell is given to us as 0.04. So, radius becomes
half of this which is 0.02 meters, right. And the length is 0.05. So, let us use this value
here. So, what you get is diameter π. So, the area becomes πr

2 = π × 0.022
.

So, everything is in SI unit. So, the value that we find is 7.6 × 10−5 kgs. So, that is the
first value.

(Refer Slide Time: 07:47)

Let us find the second value which is the mass of the membrane. And the mass of the
membrane again you have the surface density is given to you. So, surface density of the
membrane is what is the total mass per unit area of that particular membrane? So, you can
take the total mass will be surface density. So, this is the surface density multiplied by the
membrane area.
So, this we can write. So, the surface density given to us is 2 kgs per meter square
multiplied by the area; everything I am writing in SI units and the area is the same as the
area of the unit cell. So, the total mass that you get is going to be 2.513 × 10−3 kgs. So,
as you can see here, mass of membrane is much greater than mass of air. In general mass
of membrane is greater, and sometimes you can approximate the total mass enclosed into
just being the mass of membrane. So, but in this case let us take both the mass we added
together. So, the total mass then M becomes 2.513 this quantity in kgs. So, the total mass
that we are getting over here is 2.59 × 10−3 kgs.
So, this is the mass here. So, let us now find out what is that value for the frequency. So,
the frequency was the range that we have to find is between this to this, that is the required
range. So,
1


2000
2.59×10−3
. So, frequency lies between 0 to this particular value. So,
everything I am putting in SI units. So, the frequency range we end up with is 0 and closed
to 140 hertz. So, this is the range of operation of unit cell 1.

Now for unit cell 2; this is the cell where you have if you look back into the question. So,
this is the type 2 where you have membrane with some dense mass attached on the top of
membrane. So, in that case, the: ρeffective = 0. So, for unit cell 2, ρeffective = 0, when
this:

ω0 < ω < ω0√
m + M
M

So, we have already derived these expressions, we have already learned in the previous
lectures what is the region of the negative density; so this is the region of the negative
density here. So, you can now calculate this value.
So, in this case this implies that, ρeffective = 0 when the frequency lies between:

ω0

< ω <
ω0


m + M
M
So, that is the range we have to find and [ω0 = √
Km
m
] this is the center mass attached to
the membrane and this is the membrane stiffness; membrane stiffness due to tension, due
to applied tension, ok.
So, let us calculate these two values to find out what is the frequency range for the second
case. So, for the type 2, let us first calculate. So, we can write this equation here, this
particular frequency.

(Refer Slide Time: 12:15)

If you put the value of ω0 into this equation what we get is; the overall frequency should
lie between this value here to this value by this.
Now, because the unit cell is the same. So, the membrane is the same and the enclosed air
volume is also the same. So, ‘M’ is the same as the previous case which is this quantity;
this we have already calculated in the previous case for type 1. So, that is ‘M’ and Km is
the same quantity which is 2000 Newtons per meter. So, ω so this first thing here; so let
us call this as quantity A and this as quantity B. So, this implies frequency will lie between
this quantity A and B; and A we calculate as:

1


2000
m

So, here for the center mass some surface density is provided that is 200 kgs per meter
square and the diameter is given to you. So, diameter is given. So, how do you calculate
the center mass? This becomes the surface density of the center mass multiplied by its
area, area of that mass. So, what do you get is 200; everything in SI units 200 kgs per meter
square multiplied by.
So, what you do is, you multiply it by the area and the diameter is 0.01. So, the radius will
be 0.005. So, πr
2
that will be it is area. So, the total mass then comes out to be 0.157 kgs

or you can say 15.7 grams; but we will be using this SI unit kg for the calculation. So, we
put the value of M here. So, the quantity A then becomes 0.157. So, it is:

1


2000
0.0157

So, when you calculate this quantity what you get is; it is coming out to be somewhere
approximately 57 hertz. And the value B you can calculate, it will be this quantity
multiplied by this; so it will be 57 hertz multiplied by small m is going to be. Let us do it
in, let us take both the numerator and denominator in grams and see; because that will
make our calculations easier, we can reduce some powers. So, it will be 15.7 plus the value
of capital M was given to was calculated as this. So, it is 2.59 grams, so we have taken
10−3
and removed it divided by 2.59.
So, what do you get this value becomes approximately when you calculate it becomes as
151 hertz. So, now, the range of so, for unit cell 1, noise reduced in between 0 to 140 hertz
that we had calculated earlier and for unit cell 2 the noise is reduced between 57 hertz to
151 hertz. Now because they are connected in series; so for series connection, you will
simply combine the two frequency ranges. The total frequency range where noise is
reduced will be range 1 union range 2. So, this will give us from 0 to 151 hertz. Now this
is the range 1 and this is the range 2.
So, to think of it in this way, you have one unit cell and then you have another and so on.
So, when the sound passes through the first one; then 0 to 140 hertz are already reduced.
Then it passes through the second one, then further reduction between 57 to 151 hertz. So,
overall reduction will be a combination of all these ranges. So, this gives us the value
which we were looking for. So, that is the answer.

(Refer Slide Time: 17:33)

So, the acoustic transmission line can reduce sounds in the frequency range of 0 to 151
Hertz, ok.
(Refer Slide Time: 17:39)

Let us study another problem that is problem 2. So, in this problem we are given design a
unit cell. So, this is a design problem. So, you have, you already know what should be the
range of operation, how should it reduce the sound; and then based on that you have to
design the dimensions and the value of the masses etcetera. So, here what is given to us is;

we have to design a unit cell with a stretched membrane and a center mass attached to the
membrane. So, we have the unit cell type 2.
So, this is now loaded with air at room temperature, so that it can operate as a barrier
material. So, we have to design this particular unit cell type 2 loaded with air, so that this
unit cell can act as a barrier material in the range of 50 to 200 hertz. So, we are designing
it, so that it can act as a perfect barrier material and it can block the sounds between 50 to
200 Hertz.
And the type of material used for membrane and the external mechanism used for applying
tension to the membrane is such that; so based on what type of material you used and how
much of the applied tension it can withstand. It is given that, a total stiffness of 3000
Newtons per meter can be created. So, this is a design constraint that is given that, you
have to design, so that these stiffness do not exceed this value.
(Refer Slide Time: 19:17)

So, the stiffness, so the constraint here becomes, so here the design constraint let us say is
that; the stiffness must be smaller than or equal to 3000 Newtons per meter. So, what we
will now look for as a designer that, it can easily operate at 3000. So, if that is the last
allowable stiffness, the nearest take that stiffness and because we know that the less
stiffness you have then; because the range of the operation is directly proportional to

ω = √
Km
m

So, if you increase the stiffness, then mass has to be increased proportionately to keep that
ω same. So, the more the stiffness you apply; sorry the more stiffness you apply if you
want to get a bigger range with a smaller mass also. If you increase the k value, then the
ω0 value will increase; because of the increase in the k value. However, if your k value is
small, then in that case you will have to reduce the mass even further.
But anyways let us just use the limiting case. So, let us use the limiting case of 3000
Newtons per meter. So, it is given that, this rather than this equality then; let us make this
as equality. So, the design constraint is given that, this is the kind of stiffness which can
be created by using an external mechanism to apply the tension. So, we are already given
Km value.
So, now we have for type for unit cell containing a stretched membrane with a center mass
ρeffective < 0, it acts. Let us just put it this way that, this particular unit cell with a
stretched membrane and the center mass will act as a barrier material or it will block
sounds when ρeffective < 0. And therefore, we have to find.
So, therefore, the range in which this will be acting as a barrier material is when
ρeffective = 0 implies the frequency lies somewhere between

ω0

< f <
ω0


m + M
M

So, this is the range that we have to find. This is the range which is given to us, where m
is the center mass and capital M is the mass of membrane plus the mass of air, which we
can approximate. We know that, mass of membrane is much greater in magnitude than
mass of air; so for ease of calculation here what we can do is we can simply approximate,
this M as mass of the membrane itself.
So, m becomes the center mass small m and the capital M we are taking it as the mass of
the membrane. So, let us now find out this range here that we have been given in equation
1. Now here:

ω0 = √
Km
m

So, this is the range here. So, this ultimately you can reduce this equation even further and
what you get is:

1


Km
m
< f <
1


Km
m

m + M
M

So, this becomes the entire expression.
So, this is the range of frequency of operation. So, let us give it as equation 2. Now it is
given from equation 2. Now we know that from equation 2 it is given that, this range is
actually between 50 hertz to 200 hertz. So, this is what is given to us and this is what is
the analytical expression. So, if you comparing equation 2 and 3; if you compare the two
equations, then what you get is.
(Refer Slide Time: 24:29)

So, let us do it here. So, what you get is 1

. So, from equation 2 and 3 this is what you get
from equation 2 and 3; this value should be 50 hertz and we know that Km = 3000
Newtons per meter and small m is what we have to find.
So, when you put this value here what do you get is this. If you solve this equation, you
squared this quantity up; this should be the answer, it is going to be this one from this
particular equation. So, when you solve this and you put the values various values here,
what you get is it comes out to be 0.0304 kgs. So, this is all in SI unit. So, which means

around 30.4 grams is the is equal to the center mass. So, one design parameter we have
obtained that is the center mass should be 30.4 grams.
Now, let us find again we compare from equation 2 and 3. So, let us compare the other
end of the frequency. So, here this quantity should be equal to 200 hertz, if you see
equation 2 and 3. So, this entire thing becomes 200 hertz and we already know that this is
50 hertz;

1


Km
m
= 50 Hertz

So, from 2 and 3; 50 multiplied by this quantity should give us 200 hertz, and we know m
is equal to 0.0304. So, putting everything in grams; what do you get should. So, in grams,
what we get is this thing. So, this comes out to be 4 here. So, we solve this equation further.
So, what we get is this thing.
So, which means that M should come out to be 30.4; so here the M goes it becomes 15 M,
so it is:
30.4
15
= 2.03 grams. So, this is what you should be getting. So, when you solve this,
the answer that you get is 2.03 grams. So, this is the membrane mass. So, that is the.
(Refer Slide Time: 27:35)

So, we have found the two design parameters which I am showing you here. So, the unit
cell should be designed such that the mass of membrane is about 2.03 grams which we

obtained and the center mass attached to it is about 30.4 grams, so that the frequency range
of operation is 50 to 200 Hertz.
Some other things to explore are that see the dimensions. So, here this is the value of the
mass now; what should be the dimensions of the unit cell? There are innumerable options
to make the dimensions of the unit cell; the only thing is that, the dimension should be sub
wave length is in nature. So, whatever is our frequency range we are dealing with, so the
minimum wavelength that we are dealing with; the dimension should be much smaller in
magnitude, then the minimum wave length that it targets.
So, this is what is given here, that all the dimensions they should be sub wave length. So,
smaller than much smaller than the lambda minimum, and the minimum lambda is when
the frequency is the maximum. So, what is the maximum frequency of operation here; it
is 200 Hertz.
So, the λ minimum becomes:
λmin =
c
fmax
= 1.7 meters

So, as you can see here, the minimum wave length we are dealing with; because we are it
is a low frequency noise control, so sub wave length dimension is not that a problem. So,
the minimum wave length that we get is 1.7 meters and the order of magnitude of the
various dimensions of the unit cell would should be much smaller than this.
So, in general the practices to take the dimensions at least by 10λ. So, if you do by 10 you
get about one point about 17 centimeters and so on. So, let us take all the dimensions
within 10 centimeters range. So, one choice, one good choice could be 2 centimeter as the
radius and length as a 5 centimeter. So, this is just one option. So, this is just one possible
option; many such dimensions can be chosen with the constraint at which are less than
equal to let us say 10 centimeters, something within 10 centimeter range should work fine.
So, we have solved these two problems here and with this we would like I would like to
end the lectures on membrane type acoustic metamaterials.
Thank you.