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

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Из курса от партнера Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

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Korea Advanced Institute of Science and Technology

Introduction to Acoustics (Part 2)

6 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.

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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

Познакомьтесь с преподавателями

Yang-Hann Kim
Professor
Mechanical Engineering

We have total pressure that is composed by two components,

instant wave and scattered wave.

And the boundary condition reached the wall boundary condition that says Pt,

gradient Pt on the wall has to be 0, will give this result.

And we can calculate and

this is what we saw in the last lecture and this and that one.

Okay zeta is the angle from this direction so the first term.

This is the velocity in surface node.

So when theta is 0, this is 1.

When theta is pi over 2, this is 0.

When theta is pi, this is minus so

the velocity on this point sorry.

The velocity on this point, say, is 1, and

the velocity in this point is minus.

And minus meaning that the fluid is moving that direction and

plus 1 meaning that fluid is moving that direction.

Therefore, the cosine zeta is describing

the velocity fluctuation of scattered light.

Like monopole radiation.

And if you expand this, in other words, when ka is small enough to be linearized,

meaning that I want to express each jka

cosines at one plus j k a cosine zeta.

Okay then the first term describes the contribution of monopole and

second term in other way describes contribution of

dipole because cosine set on multiply.

Cosine set there is a cosine square set, meaning this point is

moving like that, A and B.

So that makes sense, if you have some scatterer.

There are two contribution, one is monopole contribution, and

the other one is dipole contribution.

The first order approximation.

And now we move on to the finite barrier case, and

we obtain this is rather interesting result and

the graphical expression of this was, look like that.

Okay, this is the DB scale, meaning that 30 DB.

It is what I can or we can get by using sound barrier like that.

And it is depends on A plus B divided by D.

Okay, that we call number.

Over there, Where is the okay.

Is A plus B minus D, divided by half a wavelength.

They simply see the, the past difference,

and scale them by half wavelengths, okay.

That is pretty reasonable, because physically what you hear from here,

Is the sound radiating from the top, right?

You don't know where the sound is because you cannot see it.

The sound over here is pretty much related with the distance from the source

to there like 1 over r as we saw before.

So definitely that has to be related with A and B, and what is this?

This is the sound, the distance you will get if you don't have a barrier.

So this is pretty useful graph.

So one thing, I mean, so did DB reduction.

Then you need to have number, for

example over here, I got 10, 20, 30, 40.

So how to get to the number 40.

Let's see.

Let's see, I want to get number 40.

And a rational number is A plus B minus D

divided by n over 2 okay.

I am here and

the car is moving over here making some noise.

And this is A and then this is B.

And this is D.

You look at the table and you decided to so

the DB reduction is at three [INAUDIBLE] and

using famous app, which I do have okay.

Sound scope and you measure the traffic noise and it look like that.

Traffic noise look like in frequency domain.

Okay, and this is you measure 100db and

we found that this is 343 Hz.

For example, I mean to get a simple mass [INAUDIBLE].

And you know that these peaks.

You don't have to consider because DB scale addition and

subtraction always dominated by peak DB.

So you don't have to consider this in the beginning, okay.

The wavelengths correspond in this frequency.

What is this?

One meter, okay.

So wavelength you've got is 1 meter.

So I have A + B- d, one divided by 2.

So I have 2.

A plus B minus D.

That has to be 40 somehow.

So I have A plus B minus D has to be

20, okay?

And you know the nearest position of car.

Okay, is say 20 meter.

And then you bought the land

away from this position, like 100 meter.

So, D is in this case 120 meters, right?

So I have d=120m and

therefore, A + B is 140 meter.

Okay this is 20 and then this is 100, okay.

All right, and I have 120 meter over.

Right?

Right?

I think I'm okay.

Did I make some mistake?

That is 120.

A plus B has to be 140, so

how high we should have?

We need some calculator, right?

If it is 120 and 60.

Then B, if it is, sorry.

Okay, this is a 20,

and this is 100.

This is A and this is B.

And we'd like to estimate H.

In such a way that A plus B should be 140.

If it is 100,

then this should be square root of 100.

That is 140.

So 100 is too high right.

Maybe 50 meter that is too high because it's too expensive.

And maybe 20 meter.

20, 20 and I have here 20 multiplied 1.4 that is 28.

And 20, 100 that would be approximately 100.

So 20 meter is approximately good.

So you can say okay.

And you may bravely say to your wife.

We want to buy this because I can reduce the noise.

By using this acoustic barrier or sound barrier by 30 dB,

so we will get 70 dB over there.

This is still very lousy.

You say using the window that

has transmission laws 40 dB,

I can reduce sound at our house 30dB and

that could be nice.

You could feel some noise when you go outside.

But you feel very comfortable inside of house by just to using 20

meter height sound barrier which is not trivia,l I guess that cost a lot to, okay?

Using that kind of calculation you can design or

determine the height of sound barrier, okay?

That's good.

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