Lecture 3-1 Part 2. Diffraction and Scattering 1

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From the course by Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

5 оценки

Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

5 оценки

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.

From the lesson

DIFFRACTION AND SCATTERING

Using Kirchhoff-Helmholtz equation that you have learned last week, you will apply into diffraction and scattering problem.

Week 3: Introduction 3:23

Lecture 3-1 Part 1. Review: Kirchhoff-Helmholtz Equation 24:02

Lecture 3-1 Part 2. Diffraction and Scattering 1 17:50

Lecture 3-1 Part 3. Diffraction and Scattering 219:25

Lecture 3-2 Part 1. Diffraction and Scattering 3 13:53

Lecture 3-2 Part 2. Summary of Chapter 4 (1) 27:18

Lecture 3-2 Part 3. Summary of Chapter 4 (2) 12:13

Meet the Instructors

Yang-Hann Kim
Professor
Mechanical Engineering

Okay we move to the diffraction and

scattering problem which is a more difficult obviously.

Okay now, let's see this interesting picture.

This picture was obtained, I forgot when, but a long time ago.

One of my graduate student conducted experiment using

water.

And then, no, no, no, this is computer simulation.

Sorry, sorry about that.

So, as you can see here, when we have a long wavelength compared with

the short wavelength, You will see the sound is propagating.

And the beating some impedance mismatch over here.

And impedance mismatch, it's not constant.

There is a rigid wall impedance mismatch.

Over here free surface, I mean pressure release boundary condition.

So different impedance mismatch.

To understand what's going on over here, so

try to be a fluid particle over here.

Suppose we have three fluid particles over here.

Then this fluid particle, the behavior of this fluid particle and

that fluid particle would be different due to the presence of impedance mismatch

spatially distributed the impedance mismatch.

If there is some excitation the fluid particle over here we'll be excited.

Therefore, the fluid particle over here will radiate sound like the or

depending on the boundary condition.

Therefore, it radiates the sound and

that radiation will be reached over here somehow.

We had experienced this kind of sense when we have a wall.

And when you are here and somebody is shouting or

there is some traffic noise over here you would think that,

you would hear the sound as if there is a sound source over there.

Right?

Suppose you have a wall and you are sitting over here,

then somebody is shouting over there.

Then you would hear sound as if there's some speaker along the wall.

So that's what essentially happened over here.

Okay?

So when we have a long wavelength, the sound is coming,

radiating from this point, it will reach like that.

But when we have a short wave, then the radiation pattern would be different.

So we'll have us some schedule drawn over large

schedule drawn compared with this case.

The fraction simply in the case when we have

this kind of schedule drawn so the case of when we have

a schedule drawn we call his kind of phenomenon as ancient fracture.

Of course in order to have a diffraction we have a scattering over here.

So, observing this kind of physical situation, the one good idea to

reduce the sound coming from this side to there.

We may put some different impedance mismatch over there.

And as a result of this kind of idea, you can see some

interesting sound barrier in the highway.

You will see some sound barrier has its own this kind of shape at the top.

And some sound barrier, you would see that has

some cylinder structure along this side.

And that all comes from the real of this kind of diffraction and

this scattered problem.

So later on we will study how design

the sound barrier based on this physical understanding.

So by seeing this kind of simulation

we can see that a force diffraction really depends on wavelengths,

and the spacial distribution of impedance mismatch.

Let's move to the scattering problem, because a scattering,

you have to understand a scattering to understand a diffraction.

Let's see how we can do it mathematically in the beginning.

Okay, let's do some little mathematics to understand

the scattering which is not complicated.

Okay, here

We have a scatter, that we have a scatter.

And our objective is to,

Is to get scatter the sound field.

And the one observation is P total is composed by two sound fields Pi and

Psc, instant and scattered sound field.

What physical factor will determine this?

Scatter, more specifically what

the physical element of a scatter will determine this?

Impedance on the surface.

Okay, so depending on the shape of the scatter and

depending on the impedance on the surface,

this would be determined.

If this is rigid wall Radiate the scatter in other words.

And that is the most simple case.

Then we can say,

the total sound on

the surface we can obtain.

That is the sound pressure on the surface,

component on the surface.

Would be what?

Okay, knowing that gradient of pressure is related with.

Euler's equation.

And we do know that gradient p.n,

that is the normal vector is

related with P tb dt.n.

If you assume that capital P is

the pressure that depends only

on the spacial, parameter and

we can write at this minus j omega t.

Then the gradient p.n

I can write that as capital

P.n is equal to rogue zero.

And then take a derivative of some velocity which I

assume v vector is u vector explanation minus the omega t and

I have -t omega and then I have the u vector and a dot.

And this is what?

This is the normal velocity of the surface.

So having this relation in mind, we can write that is equal to zero.

Because a rigid wall.

So these two equations essentially determines Psc for given Pi.

This is very straight forward.

I am saying that mathematically,

we are asked to so solve Psc using this equation.

Assuming, pi as given, and

the rigid wall boundary conditions.

Yeah, you can solve it of course.

So let's solve this.

So our plot equation one to two,

this will give simply Pi + Psc,

scattered pressure and

space dot n is equal to zero, that means.

Psc.n is equal to- delta pr.n.

So that is the,

Sort of basis that gives us Psc.

If we know pi.

So let's say assuming that

instant pressure is coming.

With a certain wave number in a certain direction.

So, I say that is exponential j wave number vector k.

The direction of the wave number vector indicates where this coming.

If I draw like that, wave number k has x component and y component.

Dot r, in other words, I am saying, there is a scatter.

This is unit normal vector, normal to the surface,

then I say this is our vector indicating the position.

And a wave is coming arbitrary direction.

And then the coefficient would be failed to be regarded as arbitrary.

Then what we

will have

That means

If I write again this equation.

Equal minus gradient

B explanation

jk.r.n.

All right and what is this?

If B is constant.

That it is constant because we assume that B is the coefficient, arbitrary.

And I put minus and then I have to take a derivative

with respect to if I use a coordinate x, y and

z then I have to take a derivative x, y and z but

if I take a gradient over here, then what will I get?

I will get jk beta.

And then I have more vector

over here and then I have jkr.

What would be the next step then?

This is related with the velocity.

And we got the expression of

velocity in terms of instant wave.

And if you know the velocity, you end.

Then we can obtain the radiated sound.

Due to this velocity, right.

So let's also attempt to get this results.

And also note that,

The gradient of scattered

pressure can be written

as Po, j omega Usc,

because it becomes

By saying U velocity, due to the scattering,

can be expressed as the magnitude velocity and

exponential minus j omega t.

This is simply the result of Euler's equation.

Yeah, Euler equation says that gradient of Psc would be equal to rho zero.

The Usc pt and

that this Psc is equal to

Psc exponentiation minus j omega.

So this is what we got.

Then we plug this over there we can obtain this.

Plug this over there we can obtain this.

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