Lecture 5-1 Part 3. Analysis Methods for a Closed Space

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En provenance du cours de Korea Advanced Institute of Science and Technology(KAIST)

Introduction to Acoustics (Part 2)

13 notes

Korea Advanced Institute of Science and Technology(KAIST)

Introduction to Acoustics (Part 2)

13 notes

Learners might have learned the basic concepts of the acoustics from the ‘Introduction to Acoustics (Part 1).’ Now it is time to apply to the real situation and develop their own acoustical application. Learners will analyze the radiation, scattering, and diffraction phenomenon with the Kirchhoff –Helmholtz Equation. Then learners will design their own reverberation room or ducts that fulfill the condition they have set up.

À partir de la leçon

WAVE PROPAGATION IN SPACE / DUCT ACOUSTICS

For the last, you will learn Helmholtz Resonator and the phenomenon about the duct acoustics

Week 5: Introduction 2:12

Lecture 5-1 Part 1. Direct and Reverberant Field 1 15:14

Lecture 5-1 Part 2. Direct and Reverberant Field 2 29:09

Lecture 5-1 Part 3. Analysis Methods for a Closed Space 17:46

Rencontrer les enseignants

Yang-Hann Kim
Professor
Mechanical Engineering

We have a room, like this.

And we would like to, we would like to, predict the sound at any point inside.

What is p r can be.

If the sound, I mean in the surface enclosed by this kind of

impedance mismatching surface.

There will be as indicated this picture, the sound will go and

reflect and this one will go and reflect like that.

Okay?

But we know that this room.

[BLANK_AUDIO]

Acoustics of this room can be expressed by, the normal mode, of this room, okay?

Then we can express, the sound field inside of the room like this.

Psi,l, m, n is the normal mode, in x, y, z direction, like this.

For rigid wall

case.

And a, l, m, n is in the contribution of each mode.

lmn mode.

For example, if l is one, m is one, n is one,

then the modes over here, is when l is one, that is cosine pi x Lx.

What it'll look like?

Okay, when X equals zero.

And X equal zero, it has this pressure, one.

Okay, when X equal Lx, what is it?

That is cosine pi, and

cosine 2 pi, is zero.

No, no, no of cosine pi over 2 is 0 so when l equal one and

x equal Lx that is cosine pi, and a cosine pi

is minus 1, is it correct?

cosine pi is minus one. So this is this kind of mode.

Ok?

Because there is a rigid wall.

Ok?

The first mode that satisfy this boundaries condition

would we like that.

So, x direction, this mode, and y direction, same mode.

And y direction same mode, and z direction.

Same mode.

Right?

There are many many modes involved in this room.

Right?

When I excite a certain frequency.

And this mode can be excited, some other mode can be

excited, some other mode can be excited in XYZ direction.

And modal participation of this kind of room is very, will be extensively high.

Okay anyway.

So we can say that the, at any point of the,

at the room, when r equal, for example, when r equals some

value that indicate your position, sound pressure, we can say the,

would be, expressed by the superposition of normal mode of the room.

And how can you obtain this?

[BLANK_AUDIO]

Hmm?

How can you obtain this?

Depending on excitation, right?

Depending on excitation.

What I shout over here.

then you have one

kilohertz excitation right?

This is actually what you can obtain.

I am exciting the medium with one kilohertz at r equal r0,

okay?

Then you have to, you have to be able to express

the position at r equal r0 where I'm exciting by using

normal mode.

And this is this term.

And what you hear at your position has to be expressed by normal mode.

So this is that term.

Ok?

Of course the, it depends, so inversely proportional to the volume.

And P r, of course, depends on,

I mean proportional to the strengths of the source, right?

And then.

If there, if the position, I mean, frequency, has some resonance

of one of the, one of room, one of this room, there will be high

peak, and this term expresses that kind of effect.

Of course k is equal to omega over C, and omega equal to 2 pi f.

So when you have a resonance at certain mode, and it will

magnify the contribution, right?

So this is very similar with the value you can obtain for the case of vibration.

And this is one zero mode, zero one mode, one one mode, etc.

Okay.

Okay.

Up to now, we consider the sound propagation in large room.

The highlight of the sound propagation in large room

was seen

in terms of the interesting measure, reverberation

period as well as radius of reverberation.

Let's move to the case when we have small space.

Suppose I have a duct, and there is a

sound source that is propagating.

At any point,

the impedance is rho0 C.

If the wavelength

is large compared with cross sectional dimension of the duct,

so this case, so lambda is very large compared with

the square root of cross-sectional dimension.

Okay?

There is low propagation in this

direction, only propagation in that direction.

But suppose I put some resonator over here.

What's going to happen?

The

wave is propagating over there,

and then this wave will excite

the fluid particle over here, okay?

If this length is small compared with the wavelength,

then the whole mass of this bottle will move back and forth in phase.

Yeah.

So, this part of the neck, will behave as one mass.

Of course, the depending on the boundary condition over here,

some of the additional mass will move back and forth, and

somehow the addition of mass will back and forth with

the motion of the particle fluid particle inside of the neck.

If this one is pushing this bottle, I

mean this resonator, then what's going to happen over here?

The fluid particle will be compressed,

whole of the fluid particle inside of the resonator also

move back and forth in phase, therefore inside of the particle will

behave like spring constant, linear spring.

So actually, this resonator

would have some natural frequency omega n

square, that is k over m.

And our objective is to find out, what is effective mass and what is effective

stiff, spring constant, that expresses what I just, what we just describe.

Okay?

At this resonant frequency, what's going to happen?

What's going to happen is.

At resonance, resonance means what?

This fluid particle move back and forth, very fast, very

fast means the velocity over here will be very fast.

In other words, the velocity at the neck, is very fast.

What does it mean by, what we have been used consistently

to represent the acoustics.

Impedance, right?

So, therefore the impedance at this point, at this area

would go down, very

abruptly, compare with the impedance over there, that is rho0 C.

So therefore, the wave coming to, coming here, and propagating through the pipe.

We'll see rho0 C up to here and then, see Z,

zero, over here, because U is very high, okay?

So in this system, it's like I have a pipe.

There is impedance mismatch Z, that is zero.

And the there's the Z rho0 C.

And rho0 C for air normally.

415, and the 415 impedance is changing abruptly to value of zero.

Therefore, what's going to happen?

Because there is a big impedance mismatch, sound will be reflected.

Very little transmission, right?

Therefore, what you can obtain by using this kind of resonator, or impedance

mismatch, you can reduce sound or noise coming from this side.

And that we can use this kind of resonator to filter out at certain frequency.

Suppose we have the engine over here, and it has this kind of

unwanted sound coming from the source, or fan maybe.

Okay, that is f zero, then we can design a certain resonator that has

resonant frequency at f equal zero, then we can cut or filter out this

sound energy.

Therefore what we obtain over here would be like this, pretty quiet noise.

Especially the, this is a very effective for, for reducing fan noise.

Very typical

resonant frequency for famous Korean soju,

is about 180 hertz.

I

just measured.

Before I came in using my app, okay, blowing a

common soju bottle, I found that it is approximately 1, it is 178.

So, this is the number I want to give you today in the coming lecture.

Where we'll see how we can get this number, by,

investigating rather the physics related with small space.

So let us briefly see what we learned today.

What I'm, what I'm talking about, acoustically small space.

So as I said before, okay,

when I excite pressure over here by certain propagating wave.

The pressure outside and the pressure inside, there will

be some difference in pressure outside and pressure inside.

According to the Newton's law, the difference of

this pressure would move the, the acoustic particle over here.

That could be one equation.

And this pressure will push this volume.

And the

[BLANK_AUDIO]

when I push and compress the air inside there

is, some change of temperature as well as volume.

That sort of things will provide what is K, equivalent K over there.

OKay, let's see how we can do it.

The change of pressure will change of the volume according to the

And also this proportionality will be related with, acoustic compliance.

And.

We will talk about this acoustic compliance next lecture, and using the

conservation of mass relation, and then the previous result, we can get the

equation like that and the finally we have a relation

of the the natural frequency that is related with compliance,

and neck length and density in the neck area, okay.

And we will see how we can get this result.

In the next lecture, in the next lecture we will estimate 180

natural frequency of the soju bottle roughly seeing the this value.'

Kay.

[BLANK_AUDIO]

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