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Before we really dive into the details of, you know, things like speaker design.

it's really important to understand a little bit more about wave propagation in

particular. And so I've got a, a, a diagram here, I

guess that shows this is, it turns out, this is a pinned boundary being that I

had from another problem I worked sometime ago.

But it could just as easily be a string, fixed and attached at both ends.

And what we want to do is look at the response of this string if it's plucked.

And so, what you see here so here's the boundaries sorry, here, here's the

boundaries, right here, of the string. And if we were to grab hold of the string

and pluck it. particularly if we did so at the center,

you can see this oscillatory behavior here.

That where the string is vibrating up and down basically.

And so, you get a picture that there's a wave propagation in space between the

boundaries of the of the fixed points of the string.

And that's typically what's called a mode of the string or sometimes the standing

wave on the string. the other thing that's important to think

about is the time domain response. Because if you pluck the string, at some

point the oscillations will subside, and if we were to look at the, what we'd,

what I'd label the temporal picture here you can see the time domain response of

the string. And it starts out with a high amplitude,

as you displace it of course, and then basically oscillates until until it damps

out or the, the vibration of the string ceases.

so the sensor that we placed on the string, again right here, gives us a

picture of what happens with the string at that particular point in space.

And we could move that sensor to any other point on the structure, and the

amplitude of the response, the, the peak that you see here would reflect the peak

response that you see on the wave. And so for, for this sensor, it

corresponds to the response of the string through this slice of space, if you will.

If we had put the sensor here, you would see a corresponding response that look

very similar, but with reduced amplitude, because the strings doesn't reach as

large a vibration displacement. So, anyway, just trying to tie together

the fact that there is a spatial response to wave propagation in air or on a

structure string. And then there is the temporal response

that we can measure at any point in time. And, we're going to talk about how those

two things relate to one another. So, for the spatial picture, we tend to

talk about wavelength. wavelength and spatial frequency.

wavelength represented by lambda here. Spatial frequency by new, and a wave

number k, okay? And so, the spatial frequency is related

to the wavelength as such, and shown here.

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And of course, the wave number itself is simply 2 pi times 1, 2 pi times 1 over

the wavelength. or it could be represented as 2 pi times

our spatial frequency, all right. And so, it really describes the wave

number describes the number of wavelengths that occur in 2 pi units.

And so, what I've done here is plotted out, a sign wave.

and of course this is 2 pie, and so it turns out that what I've shown here is

one single wave length over the domain of 2 pie.

Now for the temporal picture, so let's talk about the time domain, remember we

talked about the spatial response that we see, the way propagation in space like

the plucked string, now we can talk about the temporal response.

And the temporal response has a period of oscillation T, all right?

And T is related to, so T is, T is our period of, of oscillation.

F is our frequency of oscillation and we have a circular frequency.

And the frequency of oscillation is related to the period 1 on T.

So, this is the time required for the one complete one complete period of the

cycle. And then, of course, omega is what's

called the circular frequency. And that's equal to 2 pi times 1 on T,

which also, omega equals 2 pi times f. And what I'd like you to to recognize, is

the analogy between the temporal picture and the spatial picture.

If you remember in the spatial picture, and let me just slide back up to the top

of the slide here. we see that k is related to 2 pi on

lambda, which equals 2 pi times nu. And if we go down to the bottom here, we

see omega equals 2 pi on T, and it also equals 2 pi f.

So, the point is, is that omega is analogous, analogous to k, our wave

number. And lambda in the spatial domain is

analogous to T, the period and the time domain.

And then of course, f, our frequency, and the time domain, is analgous to nu here,

alright, in the spatial domain. So, we're describing a wave, basically,

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either in space or time. And we have different different symbols

to represent the two, but they're they're analogous to each other, okay.

And so, I can do the same thing with with in the time domain I can represent that

same [UNKNOWN] response characteristic, and this just happens to be one period of

oscillation. So, the amount of time it takes here is

one period of oscillation for this particular wave, all right, to propagate.

Now, when we move to sound wave propagation in air.

we can easily relate to wavelength to the frequency.

And that's done through the speed of sound.

And so, in air in particular. And so, before we get into those details

we need to define a few basics. So, we need to define our, you know,

we'll do it at room temperature that's what we're going to be working at,

especially around the concept of speaker design and typical acoustics in the

musical range. but speed of sound in air is roughly 340

meters per second 343 at 20 degrees c. And then the density of air, which is

important as well for some of the calculations we make, is 1.21 kilograms

per meter cubed. And you notice that I'm working in the

metric units here. Alright.

So, the relationship between wavelength and frequency of sound and air is fairly

simple. And it's expressed here, lambda, the

wavelength of the frequency. the wavelength, special wavelength, is

related to the speed of sound in air. Remember that's, that's in meters per

second. And frequency is m1 over seconds, so, you

know, if you look at the units on this, you have meters per second divided by 1

over second, which gives you units of meters, all right?

And that makes sense, because the wavelength is going to be represented in

a measure of meters. so let's talk a little bit about the

relationship between, our audible range, and then of course the the wavelength of

sound in the audible range. So, humans actually hear sound in

wavelengths that range from about 20 hertz at the low frequency to 20

kilohertz at high frequency. And you know the older you get actually

you get you, your, your response at higher frequencies decreases.

we know that but, but It's but at birth basically you have a, a range of about 20

hertz to about 20 kilohertz, and it varies from person to person depending on

auditory capabilities. So, you've just computed the wavelength

of sound at 20 hertz. what is the wave length of sound at 20

kilohertz? It's pretty simple all you have to do is

divide that by a thousand. and if you do that you get a wave length

of 1.7 centimeters. So, if our audible range is from 20 hertz

to 20 kilohertz, that's the frequency range over which we hear it corresponds

to wavelengths that range from 17 meters to 1.7 centimeters.

So, the wavelength the sound at 20 hertz is 17 meters long.

Now that's, that's, that's a large wavelength.

If you think about the, look around your surroundings at your room most of you

probably aren't in a room at this point in time that has a dimension of 17 meters

in any given dimension. So, this a really long wavelength,

particularly compared to the size of your head and the distance between your ears.

at high frequency, though at 20 kilohertz for example, wavelengths only 1.7

centimeters long, so it's very short. So, this is important because it's, it's

important in particular in relation to how we actually localize sound.