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Professor Ramamurti Shankar: Okay,
so what did I tell you about waves?
I said let's take a concrete example instead of talking in
generalities. You say wave is the disturbance
of some medium. The medium I chose to study was
a string, under some tension, clamped at both ends.
Then, you pull it and you release it, it's going to be
doing something. This height,
from the normal position, at the point x,
is the actual variable, of ψ of x and
t. So, at every point x
along the string, there's a displacement of the
string in the transverse direction;
that's ψ of x and t.
I showed you that this ψ obeys the following
equation: d^(2)ψ over dx^(2),
equals 1 over v^(2), d^(2)ψ over
dt^(2), where v^(2) is tension
divided by mass per unit length. I'm going to change my notation
into force, because T is also used for time period;
so it's going to be, it's going to get me in trouble
very soon. So, let's just change the
formula to what's used in the books, v^(2) is F
over μ; F is the tension on the
string. So, what--how did we get this?
Let me remind you how I got this, okay?
Let's undo the v^(2) as T over μ.
What we are saying is Td^(2)ψ over
dx^(2), is equal to μ;
d^(2)ψ over dt^(2).
This is very easy to understand, the meaning of the
equation, I claim, is just F = ma.
You realize that this is the a of the string,
moving up and down, this is the acceleration,
and this is the mass per unit length.
So, imagine a tiny segment of width dx.
If you multiply both sides by dx, that is the mass of a
little segment. μ times dx is
the mass; this is mass times acceleration.
The left-hand side is the force on it.
Why is this the force? I tried to explain to you,
that if you took a piece of that string, then the tension is
pulling like that there [pointing to the board]
and like this here, and the two don't cancel
because if you resolve this force into vertical and
horizontal parts, that into the vertical and
horizontal part [pointing to the board], you can see this is
going to be bigger than that. So, the slope itself is
changing and this is the rate of change of the slope.
Another way to think about it is if you remember your
calculus, the second derivative of a function is the curvature
of the function; so that if you draw a graph
like this, that's a function of negative curvature,
that's a function which is positive curvature.
It says, if the string has negative curvature,
the force is down; if the string has positive
curvature that probably means you are below your normal
position; then it's getting a force
upwards. So, it's the curvature of the
string that leads to a net force.
One of you asked me a very good question after class which
is--the question was, "You took care of the vertical
forces as T sin (θ + Δθ) here,
and T sin θ here, and you did all the
cancellations. How about the horizontal part?
That's also not quite balanced, the angles are not equal."
But the point was sin θ was approximated by θ
for small angles, but cos θ,
you guys remember, begins as 1 minus θ^(2)
over 2 and so on; but we are not keeping track of
things more than the first power of θ in the calculation.
So, cos θ is just 1, in this approximation,
and tan θ and sin θ are both equal to θ.
That's a small angle approximation.
That's why the horizontal forces, the fact that they don't
quite match is an error to the next order in θ,
but the vertical forces are proportional to the first power
of θ, and that's what we kept.
That's the origin of the wave equation.
Then, you know you did the cancellation of dx and so
on and you went back to that form.
Then, I said how about solutions to this equation?
Well, if you have a little more time I can tell you how to get
the solution to this equation. I don't have that much time.
So, what we will do--going to do, is to be content with
writing down a solution, then convincing you it's a
solution, analyzing its properties.
So, the solution I wrote down looked like A cos (kx -
ωt). And I said, "Well,
is that a solution of this form?"
Well, take it, put it into the equation and
take all the derivatives. First of all you guys should be
able to see that the A will cancel on both sides.
Do you see that? So, if you put this here,
two derivatives will pull out a k^(2),
two derivatives here will pull out an ω^(2).
Therefore, if you want the two sides to match,
you will want the ω^(2) side divided by v^(2) to
equal the k^(2). That means ω = k
times v.
That's the only condition. If that is satisfied,
if this ω is k times v,
you have a solution.
No, and A stands for anything;
you can have any amplitude you want.
Of course that's a little bit of a fake.
Why is that not really literally true?
Yes? Student: [inaudible]
Professor Ramamurti Shankar: No,
I tell you it's real and positive, and two miles long.
Yes? Student: [inaudible]
Professor Ramamurti Shankar: Not only that.
The whole calculation assumed it was a small displacement
problem, right? That's why sin θ and
tan θ are all equal to θ and cos θ is
roughly 1. So, once you make an
approximation and you get an answer, you should not blindly
apply the answer to circumstances that are not
valid. So, even though the answer says
you can have any A that you like, in practice you shouldn't
use it for A, which is so large that the
small angle approximation fails. Alright, so A is anything.
So, let's put in the result I've learned,
namely, that ω = kv, and write it as k times
x - vt. And here's where we were at the
end of the class. What this tells you is that
this guy is some function of x - vt.
It happens to be A cos, but it turns out any function
of x - vt actually satisfies the wave equation.
That's a very interesting exercise for you calculus buffs.
Take f unknown function, whatever you like,
but it's not an arbitrary function in which x and
t appear any way they like,
they appear only in this combination.
So, make up any function you'd like, hyperbolic cos of
x - vt, that's a solution to the wave
equation. Why?
Because if you start taking derivatives, if you call this
variable w, you'll find that partial
derivative of x and partial derivative of,
with respect to T are all connected with derivatives
of w; and then if you do the chain
rule, you'll find it satisfies the wave equation.
Then, I tried to convince you that because of that,
v is actually the velocity of the wave.
Let me tell you why. Take any function you like,
at a given time, t = 0,
it has a certain shape. Our cosine has a certain shape,
but take any shape you like, a t = 0.
Let us say it has peaked here. If you wait a little bit,
if you'd like it to be peaked at x = 6,
for example, if you wait a little bit,
this term minus vt, is subtracting it from it,
you got to add to this the same amount so that the peak now will
be there. The amount by which it has
moved is in fact exactly equal to vt.
In other words, if the vt becomes non
zero, x should become non zero by the same amount,
if you want to ride the crest of that wave.
That means the signal is moving to the right.
So, it turns out you can have x - vt or x + vt,
they're both solutions to the wave equation.
With a minus sign the signal is moving to the right;
with a plus sign it's moving to the left.
That's a summary of what I did last time.
So, last time we derived the wave equation;
we took this particular solution.
Alright. The next thing I want to do is
ask the following question, "What is k and what is
ω?" So, let's take this function's
ψ, of x and t, which is A cos kx -
ωt. See, a function of two
variables is hard for us to visualize.
We are all good at explaining, visualizing a function of one
variable, but this is a function of two variables.
Understand why it's two. Every point on the string has a
variable ψ at that point, and that variable can
vary with time. That's why ψ of
x and t is needed to describe the vibration of an
extended body like a string. But a point particle,
you just need one or two or three coordinates.
Alright. Now, I look at this function
and I say, "Let me at least understand this function at
t = 0." A t = 0, drop that term;
in other words, this is a snapshot of the
function. What does it look like?
If I take a picture at t = 0, then I vary space.
Well, that's cos kx, and you can all plot cos
kx, versus x. It's going to start out at
A, then it's going to oscillate because that's what
the cosine does.
Now, this distance from one peak to the next peak,
we call the wavelength. It's also the distance from one
trough to another trough or from one point to an identical point.
That distance where you repeat yourself is called lambda (λ).
What is the relation of k to λ?
Now, you got to ask yourself, when x was 0,
this angle was 0. I've got to increase x
till it come back to what's essentially is 0.
For A cos, you are back to essentially 0
if this angle is 2π. It can also be 4π or
6π, but the first time you are back to 0 is when it's
2π. Therefore the length λ,
up there, has a property that k times λ is
2π, because if I move to the right
by an amount λ, the angle in the cosine are
changed by 2π. So, this number called k
is related to the number we all understand more intuitively by
this formula. It's the inverse of the
wavelength, up to this factor 2π.
Now you can say, well, you picked time t
= 0; maybe if you picked time
t = 1 second you would have gotten a different picture.
Think about it. If I put t = 1 second,
as long as it's a fixed number, that's some fixed angle
φ; it's just going to shift the
whole pattern by some amount; it won't change the fact that
the peak to peak distance is still λ.
Because what's happening as a function of time,
we have seen very generally is that this whole pattern is going
to slide to the right. So, I want to combine this
picture with a dynamic picture by saying, if you now vary time,
imagine this sliding by. If you are sitting somewhere
here, or some point, that'll be the height for you,
and as time goes by that'll be the height and that'll be the
height and that'll be the height and that'll be the height;
all these things will go past where you are.
So, let us now ask, "What does it look like to a
person sitting at a certain location, as a function of
time?" In other words,
the waves are being manufactured in the ocean.
Let's say they're being sent towards the shore.
You stand in one place; the wave's going to go past you.
So, you will bob up and down, and that's what I want to
understand. So, I pick some location.
I pick x = 0 for convenience, then ψ at
x = 0, as a function of time looks
like A cos ωt.
I didn't miss the minus sign; for a cosine,
it doesn't matter if you change the angle to minus the angle.
So, plot the same graph now this time as a function of time,
and ask what happens. Well, you start with a maximum
and you bob up and down. This is a picture at one point
where the water or the string goes up and down.
So we like to call this, from one maximum to the next
maximum, the time period [T]. That's the time over which the
wave repeats itself. Well it follows then,
that ω times the time period is 2π,
because when I put little t equal to time period,
I've gone from this maximum to that maximum.
Therefore, ω is 2π over T;
in other words k and ω, which we put into the
wave, are related to the time period and the wavelength
through this reciprocal formula.
So, if you like you can write the wave as A cos
(2πx/λ - 2πt/T).
It's strictly equivalent, you write it any way you like.
This makes it clear that when x changes by λ,
nothing happens to the cosine because when x changes by
λ you are adding a 2π to the cosine.
Here, it's clear that if you add to little t an amount
of time big T, nothing happens,
because 2πs don't matter.
So, this makes very clear the periodicity in space and time;
but this is more compact because if we don't want to
carry 2π's and λs in this complicated
form, we give a new name to
2π over λ and 2π over T.
Okay. What about the velocity of the
wave? Remember, it satisfied this
condition, so ω over k.
But now, let's write this in terms of those new guys,
2π over T divided by 2π over
λ, is λ over T.
So, the velocity of this--This wave is called a sine wave or A
cosine wave or a plane wave. You got to understand waves in
water and waves in string always have a velocity.
If you pluck the string and let it go, a little blip will travel
from left to right. But not every wave has a
wavelength or a time period. This is a special function that
has periodicity in x and t.
In other words, if you go to a lake and drop a
rock, it'll send out some waves and after awhile they'll spread
out. It's not going to happen over
and over and over again. But if you want a wave to be
present for all time, for all space,
you should keep on agitating the water.
That's what we imagine here is the source of this wave is to
keep manufacturing these waves from left to right.
So, you must remember that in general every wave on a string
or in a medium has a velocity but not every wave has a
wavelength or a time period. Any pulse that travels from
left to right with speed v is a possible
displacement of the medium. This is a particularly simple
situation we study. For that simple situation,
the velocity is connected to wavelength and time period in
this fashion, or you may also like to write
it as λ times the frequency.
So, that's a very easy thing to remember, because if you're
trying to get a velocity, it's going to be some distance
over some time, the only distance you can think
of for a periodic wave is a wavelength;
the only time you can think of is the time.
It turns out that λ over T is in fact simply
the velocity. Here's another way to
understand this. Suppose I have a string that
goes on to infinity, and I'm catching it at this end
and I start wiggling this string.
When I start wiggling this, these pulses will travel--let
us say they'll come up to this point.
Let's wait one more second. In one more second I would have
manufactured a few more of these pulses and the wave will look
like that. How many pulses would I make
per second? f;
how long is each one, λ? So, that's how much the wave
has advanced in one second; that's why it's the velocity.
You're making λ-sized objects, f at a time,
f per second, and pushing them out;
so the front of the wave advances a distance λf
in one second, that is the velocity,
that's one way to understand why λf is equal to
v.
Okay, so now, I consider another aspect of
these waves, and that's the energy in the wave.
First of all, you've got to understand a
vibrating string has some energy compared to a string which is
not vibrating. Okay, you have an intuitive
feeling the vibrations are a cause of energy.
I want to calculate now the energy in a string vibrating
according to this formula here [pointing to the board].
Now, if it's an infinitely long string, the energy in it is
infinite, so you don't define the energy of an infinitely long
string, you define the energy of a
piece of a string; in fact, you talk about energy
per unit length. So, if you take a string,
it's been vibrating for a long time, and the waves have been
manufactured on the left, they've gone out far to the
right, we don't care how far. Take a portion of the string,
of length dx, and ask, "How much energy does
it have?" We have already seen that if
you sit at one point, the string just moves up and
down. It executes simple harmonic
motion. When the string has gone out
all the way fully, it's like a spring which has
been extended maximally. That's when the string stops,
turns around and comes back; when the string has come to the
horizontal position, that's the normal equilibrium
position for a mass and a spring,
that's when it's moving the fastest, and overshoots and goes
to the other side and goes back and forth.
So, every part of a string is just undergoing simple harmonic
motion--that's pretty obvious from this.
So, we just need to write down the energy.
So, how do you write the energy? The energy is going to be
either kinetic or potential, or the sum of the two.
Now, you can find the energy whenever you like because it's a
constant. Let's find it when the velocity
is a maximum, like for harmonic oscillators,
it's like finding the velocity on a mass that's just swinging
past the normal position, x = 0.
There the entire energy is kinetic energy,
it's ½, μdx is the mass of the segment I'm looking
at. What's the maximum velocity?
You guys remember from harmonic oscillators, what's the maximum
velocity that a harmonic oscillator has,
whose motion is given by this? Yep?
Student: [inaudible] Professor Ramamurti
Shankar: Yes, ωA.
This is--You're getting a Yale education.
So, that's the difference between 40,000 dollars and
10,000 dollars is that this is not w,
but ω [laughter]. Okay?
By the way, I could not get any of my kids to say ω;
they keep saying w. I'm trying to have more success
with this class than with my own offspring.
But if you go to the outside world, if you talk to any
physicist, if you say w, they'll think you're talking
about the width of something. So, you have to say ω,
ok so the Greek alphabet is essential to communication.
So, this is Aω is the maximum velocity and I've got to
square that.
Now, this is the energy of a segment of a length dx.
So, what we normally define is a quantity called little
u, and that is the energy per unit length.
That means just divide by dx, and you get ½
μ, A^(2), ω^(2).
It is just the kinetic energy of a segment of unit length.
It's the total energy, but I chose to compute it when
it's all kinetic. Yes?
Student: [inaudible] Professor Ramamurti
Shankar: µ is the mass per unit length.
You understand? You take the string,
you find its mass and you divide by length.
There's so much mass per centimeter or per meter,
that's μ. That's why if you go back to
the derivation of the wave equation we were thinking in
terms of μ. Okay.
Now, we can ask another question, "What is the power
sent into this wave by the person exciting the medium?"
See, the string is tied to a point at infinity.
But let's say it started oscillating a while ago,
and this is the part of the string that has started moving,
and I'm still shaking it here. Let me wait one second.
If I wait one second, this part of the string is now
activated, and what's the energy contained in the extra segment?
It's the energy per unit length, ½ μa^(2)ω^(2),
times the velocity. That is the power I have to
give in to produce this wave continuously.
Do you guys follow that? If I wait one second,
a segment of length v, has now started moving,
in the front end, because the wave is being
produced by me and sent out into the medium.
The distance it travels in one second is v,
and all of that extra segment is now vibrating,
with this energy per unit length.
So, this is the extra energy per second I have to give,
and that's the power. That power is provided by the
person shaking the string. A special property of life in
one dimension is that this wave goes undiminished in amplitude.
You can go 10 miles, 100 miles from me,
the amplitude is still the A that I produced here.
That's because all the energy goes along this line.
It's sort of atypical. What typically happens is when
you live in, when you take into account vibrations such as
sound, in two or three dimensions,
if you have a tower on top of which you put a speaker,
just making noise, the energy radiates out in
concentric circles. And the power at the source is
now spread over bigger and bigger spheres.
So, in three dimensions we have the notion of intensity,
which is the power per unit area.
What I'm saying is, if you've got a speaker there
that's sending out sound waves, I take a one meter by one meter
window and hold in front of me and I ask how much power crossed
this in one second?, That's intensity.
You don't have the notion in one dimension because you cannot
hold a window in one dimension. Three dimensions,
you take a little square, one by one meter,
and you define intensity to be power per unit area.
Of course, it doesn't have to be one meter big.
You can take a tiny square, provided you divide by the area
of the tiny square, that is this.
So, what do you think will happen if there's a speaker
mounted on top of a platform that's radiating power at the
rate P? The intensity you get where you
are, I hope you understand, will be simply [P over]
4πr^(2), where r is the radius of
this sphere and you are a distance out from the center;
and the power is going out uniformly in all directions,
and your share for the whole sphere was P;
the share for the unit area on the sphere is P divided
by 4πr^(2). And that's measured in watts
per meter squared. Yes?
Student: [inaudible] Professor Ramamurti
Shankar: This one? What else do you think it
should have? T for time period, why?
Student: [inaudible] Professor Ramamurti
Shankar: It's energy per length.
But if I wait one second, the wave has gone a distance
v times one second. And so, the extra energy I've
given every second has to pay for v more meters.
Student: [inaudible] Professor Ramamurti
Shankar Yes, this is the energy per second,
that's the power, that's correct.
So, if you check your units you will find it's got energy per
second. Okay, so this is called the
intensity. And for sound we have another
way to measure intensity. We define something called beta
(β), which is 10 times the logarithm of the intensity
divided by a standard intensity. The standard intensity has a
value, 10 to the minus 12 watts per meter square.
And this is called the number of decibels.
So, you measure the loudness of a sound in decibels.
You must have heard that everywhere.
You look at your amplifier will have so many dBs in it.
So, what's the dB talking about? The dB says your speaker is
making some--sending out some energy;
stay wherever you want, at a certain distance from the
speaker, see how much power is passing past the unit area,
that's the numerator, divided by the standard number.
People all agreed on dividing by the standard;
then, take the logarithm of that to the base 10 and multiply
by 10. That's called the number of
decibels. So, for example,
if the power coming to me was say 10 to the minus 11 watts per
meter square, then I over
I_0 will be 10 to minus 11, divided by 10 to
the minus 12, which will be 10 to the 1;
and the log of 10 to the 1 is just 1, and β will be 10
dB. So, a 10 decibel sound is a
factor of 10 bigger than the standard sound.
But here's the interesting thing.
Suppose the intensity is 10 to the minus 8 watts per meter
square. That is a thousand times more
intense than this one. But let's look at β.
β becomes 10 log of 10 to the 4 now;
because 10 to the minus 8 divided by 10 to the minus 12 is
40. So, when intensity goes from
some number to something a thousand times bigger,
the decibel goes from 10 to 40. The decibel grows much more
slowly than the actual intensity;
in fact, the decibel is keeping track of how many zeros there
are in the power, it grows logarithmically.
So, every time I add an extra 0 to the power,
you go from 10 to 100,100 to a 1000, the decibel goes up just
by, from 10 to 20 to 30 to 40. That's why 80 decibels can be
incredibly loud, 80,90, 100;
100's a very painful number of decibels.
And I believe people use decibels because the human ear
is able to take a whole range of intensity and hear it.
The ability of the ear to listen to intensity -- you might
think there's a very narrow band you can hear -- you can hear a
very wide band, where the intensity ranges over
several powers of 10. In that case,
instead of dealing with numbers that go all the way from 10 to
the minus 5 to 10 to the plus something,
if you deal with the logarithm of the number then it's more
manageable. So, you can imagine trivial
problems involving finding β, or given β
finding the intensity and so on. Okay, so now I'm finished with
basic properties of waves. Decibels is a property just for
sound, you don't apply that to light.
And the whole standard intensity
I_0--intensity is well defined for everything;
even electromagnetic waves have intensity;
it's the power coming per unit area.
But when applied to sound, if you take the log and
multiply it by 10, that's the decibel level of
that sound. Okay, now for another totally
different property of wave propagation, which I'm going to
again apply to sound, is the familiar Doppler effect.
So, the Doppler effect is the well known phenomenon that if
you have a source of definite frequency,
like a siren in a fire truck, when the fire truck is coming
towards you, you hear a higher frequency,
and when it's gone past you, you hear a lower frequency,
and we just want to know why, and by how much.
So, to proceed with this, I want to remind you this
relation, λ times f is v. So we're going to bear that in
mind. So let's take a source that is
sitting still, and you are standing here,
listening to the sound. The waves go by you and you see
a certain wavelength. That's the distance from one
crest to the next. Now suppose the source is
moving to the right, at a certain speed,
u; u is the speed of the source.
So what does the pattern look like?
It's not too hard to visualize that having emitted one crest,
it's moved to the right before it emits a second crest.
So the pattern will get crunched in a forward direction
like this.
So the waves will get squashed because the spacing will have
been λ, but now it's less than λ because it's moved to the
right to emit the second crest; so spacing from crest to crest.
The new λ, which I call λ prime, will be λ minus the
distance the source travels in that time.
In the time it takes to emit one more pulse.
That is the time period. So let me write it as λ minus
u over f, because T is 1 over f. So, what is the new frequency
that I will hear? The new frequency I will hear
will be the velocity divided by the new wavelength.
By the way, you've got to understand that even though the
source is moving to the right, the velocity of sound is not
altered by that process. A moving truck does not emit
sound at an increased speed in the forward direction or
decrease. The speed of sound is
controlled by the medium in which it travels.
If the medium is air it can travel only at that speed.
So you don't have a new velocity when the thing begins
to move. The velocity is the same in
air, as long as the air is not moving.
So let's write it as v over λ minus u over f;
multiply top and bottom by f. So we write it as f times
something, which is v divided by λ f, minus u,
and we fiddle with that and write it as v divided by v minus
u, because λ f is v;
and divide top and bottom by v. This is something you'll find
in every possible textbook. So I don't want to take too
long to go over the details. The main trick is in writing
this equation. Once you've got this,
the rest of it is just algebra. The point is to realize that
the waves get crunched, because when you emit a crest
and you rush towards that crest, and emit a second crest,
you've reduced the spacing by your velocity,
multiplied by the time between such emissions,
which is the time of vibration of the siren.
So this is the formula. The frequency is the normal
frequency divided by 1 minus something.
So one minus something is less than 1.
So the frequency will go up. If you want to see what happens
here, behind the thing, you have the option of doing
the same calculation behind, and I think it's very obvious
what you will find, you will get 1 plus something.
So we'll write both solutions as minus or plus.
In one case, the frequency will be
increased; other case the frequency will
be decreased. So one way to think about it is
if the source is coming towards you, use the minus sign;
if the source is going away from you, you use the plus sign.
But you should know intuitively at least what sign to pick.
You've got to know if the source is coming towards you the
frequency will go up, and if it's going away from,
you it'll go down. So, you pick the sign using
that common sense. Somebody had a question in the
back. Yes?
Student: With relativistic speeds does that
change for light? Professor Ramamurti
Shankar: Ah. Relativistic speeds,
a lot of things happen, that's correct.
When you handle light you got to be very careful.
The velocity of light will be the velocity of light,
but the frequency will change for many reasons.
First of all, the ambulance clock and your
clock no longer agree, so there is something called a
transverse Doppler shift, where even if it's not coming
towards you or going away from you, or it is going in a circle,
you're sitting in the middle. It's going tangentially,
so it's not coming towards you or away from you,
you will still feel a Doppler shift.
That's because it's a moving clock and therefore it'll slow
down. So, relativistic problems are
more complicated. Yes?
Student: How does this deal with sonic booms?
Professor Ramamurti Shankar: Okay,
sonic boom comes in, in the following thing:
f prime is f divided by 1 minus u over
v. Let me embellish the formula by
writing use of s, where s means source.
Okay, when the source is moving, this is the formula.
In relativity, it was chock full of formulas
like this, but we never worried about what happens when the
denominator vanishes because you just cannot approach the speed
of light. You can approach it but you
cannot reach it. But the speed of sound is not
an upper limit at all, and you can have an ambulance,
or a plane, that eventually travels faster than this.
So that's when, imagine the waves getting
crunched till all the crests are piled right on top of each
other, and that's when there is the sonic boom.
And if you go faster than that, the way the wavelets emerge
from you, so that you leave behind a trail,
there's nothing in front of you.
So, if a jet plane is coming towards you, faster than the
speed of sound, you won't have time to get out
of the way because it'll hit you before the sound waves hit you.
Okay, so sound is not the absolute limit.
So, there are possibilities of, I mean, every time you go on
the Concord, well, not any more but when you did,
you would be crossing this thing pretty soon after takeoff,
because you'd go several times faster than the speed of sound.
Okay, the last Doppler thing I want to do is the opposite one
where the siren is emitting nice spherical waves,
but you are rushing to meet the siren.
You are going with the speed u, sub-observer
[u_0]. Then, what happens?
So, let's see what to do. What's the frequency at which
you will see crests? If you sat there and did
nothing, you'll see them at the normal frequency.
But now you are, these waves are spaced at
distance λ apart, but your velocity,
relative to sound is now u_0 + v,
your relative velocity. Sound is moving to the right,
you are moving to the left. So, with respect to these
crests you are zipping at a speed u_0 + v.
And the crests are spaced at a distance λ apart;
there's nothing wrong with the spacing between them;
they are just like shown here. You are just rushing to meet
them now, from your side. So, how many will you intercept
in a second as this; this is the relative speed and
that's λ. So let's write this,
u_0 + v; λ = v divided by
f. So, I'll bring the f
upstairs. And that you can write as
f times 1 + u_0 over
v.
So, when you move towards the source, the correction factor
appears in the numerator; when the source moves towards
you, correction factor appears in the denominator.
You got to use common sense to see if it's plus and minus and
so on. It's pretty obvious if you're
rushing to meet the waves it's plus;
if you're running away from the waves it's minus.
Then, you can go on and on and ask what happens if you are
moving and the ambulance is also moving;
then, you can combine the formulas.
But I don't want to do that, that's not the whole course,
dedicated to theory of sound, but this is just to tell you
how to reason in the two special cases, source moving and
observer moving. Okay, so now,
the rest of the class is going to be focused on one essential
property of waves. So, the whole time I'm going to
do just this one important property of waves that has to do
with interference of waves. Look at the wave equation I
wrote down. It's a linear equation.
You see that? Don't be fooled by this
d^(2)ψ. That doesn't make it quadratic;
it's the first power of ψ on both sides.
Then, do the following exercise in your mind.
ψ_1 is a solution, ψ_2
is a solution. Add the two equations.
You can check that ψ_1 +
ψ_2 is also a solution.
So, these waves also have the property that--what that really
means is the following: If you make some sound and
you're sending some waves, they make a certain pattern
that travels through space. Then, I turn you off and I turn
on another speaker here; let's say the speaker makes a
sound. If you and the speaker are
making sound at the same time, then the air displacement where
I am will be simply the sum of the two.
So, if one cause produces one effect, a second cause produces
a second effect, then both are turned on;
the wave they produced will simply be the sum.
You can simply add them because that respects the--satisfies the
wave equation. So, all I'm going to do from
now until the end is take a variety of situations where I'm
going to be adding to waves. So, the simplest problem is the
following. Pick a certain spot.
You just sit there and listen to two waves.
So, you're not going to look at the wave as a function of space.
It's a function of time; somebody is making the waves
and sending them to you. The first wave looks like
ψ_1 = A cos ω_1t.
That's the second source of sound, which is A cos
ω_2t. I've chosen the two to have the
same amplitude, but not necessarily the same
frequency. You want to add these two,
and you want to see what you hear.
So, what you will hear will be ψ_1 +
ψ_2, which will be A cos
ω_1t + cos ω_2t.
Now you can do the trigonometric identities and
figure out what has happened, but you have to think a little
bit about what's going to happen.
Think about these two waves. If they have the same
frequency, it's trivial, right?
If the ω_1 =ω_2,
ψ = 2 A cos ωt; where ω is the common
frequency. That just means they reinforce
each other. At every instant you get double
what you got before. But if the frequencies are not
equal, then initially they are in step.
This cosine is 1, and that cosine is 1 and the
amplitude is 2A. As time goes by
ω_1t and ω_2t start
differing, so the cosines are not in step anymore.
And what can happen after awhile is the two cosines are
offset by half a cycle or the angle between them is π.
Then in fact, they will cancel. Then, if you wait long enough,
again, they will be in step. One way to think about it is
imagine two runners going around in a track.
If they have slightly different speeds they start out together,
they go around and one starts lagging,
behind the other, and they will be on different
parts of the circle, and a time will come when
they're on opposite parts of the circle.
When they're on opposite parts of the circle,
the angle or difference between them is π,
and that's when we say they cancel.
If you wait long enough they'll again line up,
but with one difference, one guy would have done one
more revolution than the other; then again they will be in step
and out of step. That's what you expect the
formula to give you. So, you got to go back and use
the trigonometric formula. And the answer is,
2A cos (ω_1 -
ω_2,) over 2 times t,
times cos (ω_1 + ω_2), over 2
times t. This says cos A - cos B
is 2 cos A - B over 2, cos A - B over 2.
Alright, so we are interested in the case where
ω_1 is very nearly equal to
ω_2 so two frequencies are very close.
A typical situation--You have a piano tuner, the piano tuner
hits the fork, it vibrates at some frequency,
maybe 440 Hertz, and then your piano is slightly
off, maybe it's 439. What sound do you hear when you
put them together? Well, let's look at this.
If ω_1 - ω_2 is a very
small number, this cosine here is changing
very, very slowly; this cosine is oscillating at
roughly the--it's oscillating at the average of the two
frequencies. So, if one was 401,
and one was 399, this is oscillating at 400,
but here ω_1 - ω_2 is 2,
divided by 2 is 1. So, it's oscillating much
slower. So, what will you get if you
plot this? Think of the whole thing as an
amplitude that's slowly varying with time.
So, we have done it the other day when I wrote e to the
minus γ, t over 2 times some
cos ωt; this is an envelope that
controls the amplitude. That day it was falling
monotonically, but here it'll oscillate very
slowly. So, let me plot this cosine
here. This is just the amplitude,
but with that amplitude, this guy is oscillating like
crazy. So, what I do mean to convey is
that you'll have oscillations like this.
So, it'll be loud, silent, loud,
silent, loud, silent.
So, you will hear another beat, this is called a beat,
in which you will hear a hum in which together they add and
cancel, add and cancel. And we want to calculate the
beat frequency. Now, the first guess is to say
the frequency of the beat is (ω_1 -
ω_2) over2, because that's what seems to
appear here. But actually,
the correct answer is the beat frequency is simply
ω_1 - ω_2.
And I will tell you why. At t = 0,
when the cosine had a value 1, you got an amplitude 2A.
Wait till ω_1 - ω_2 over 2 is not
equal to 2π, but equal to π,
I mean times t. If this whole combination is
equal to π, this number becomes a negative
number of ψ's 2A, but that just means it's here.
But if you multiply that huge negative number by the cosine,
it is still oscillating very wildly here, as much as it is
here. In other words,
5 cos ωt and -5 cos ωt have the same amplitude.
The minus doesn't change the fact that it's having large
excursions from plus 5 to minus 5.
So in other words, I'm telling you to compare some
answer like this, 5 cos 400t,
-5 cos 400t. Plot this and plot this;
you'll find if 400t is very rapid oscillations,
this has an amplitude as big as this one.
And the ear is just picking up the amplitude;
the ear is not picking up anything else.
The ear is just picking up the loudness of the sound,
so it will seem to be loud now and loud there.
But once you've gone from here to here, you only had to change
the argument of this whole thing by π and not by
2π. That's why the beat frequency
then happens to be simply the difference in frequencies.
Or I ask you to do another thing, if you're not happy with
this. Go home and do the following
thing. One is vibrating as
ω_1t, the other is vibrating as
ω_2t. Find out when the difference
between the phase angles is equal to π;
that's when this guy and that guy will cancel each other;
that's when you come here. Then, you ask yourself when
they'll come back to where they are, 2π.
Ask for when the difference in phase is 2π and you will
find that that occurs at the frequency, which is what I told
you. So whether you can,
if you found this factor of 2 confusing, you've got to
understand the argument I gave you.
That in order for this thing to sound just as loud as it did at
t = 0, it's not necessary that this
number come to 2A, it's good enough if it comes to
minus 2A; because minus 2A times
the rapid cosine, which is oscillating on top of
the minus 2, will also look like plus 2;
you cannot tell the difference. So, beat frequencies are used
to tune instruments. So, what the piano tuner does
is to strike the fork, and strike that thing on the
piano and listen to the two, if they're not quite matched,
you'll hear a slow hum that goes up and down in intensity.
You keep on fiddling with it till it goes away.
That's when you match the two frequencies.
So, beat frequencies have a lot of interesting applications in
physics. In fact, when you go to a radio
station, you know, this station is what,
960 megahertz or something. No way your ears can hear that.
That's not a frequency of the sound;
that's a frequency of the signal, called a carrier wave.
Then, you modulate with that, with something,
and you produce a beat. It's the beats that we are able
to hear. Anyway, this is something
fairly simple, a trigonometric problem.
Now, I will do a slightly more complicated problem,
where I'm going to add two waves.
And we'll do a lot more of this in the next term but right now
you're supposed to have first pass at this topic;
this is called a double slit experiment.
You have a source of waves here. The source is so far away that
you can treat these to be just parallel lines,
and you hit a double slip, namely, an impenetrable barrier
with two holes in it. If you want think of it as
water waves, and you've erected a wall on your estate,
where there are two holes. So, you know what'll happen,
right? These two holes will themselves
start generating their own waves.
And you want to stand on a line here, go up and down and ask
what's the water doing where I am.
So, first take a point exactly midway between these two points.
What are you getting? You are getting a signal from
here, and you're getting a signal from here.
One signal travels that way; the other travels that way
[pointing at diagram on board]. Do you understand that because
you're symmetrically located, the two signals will be in step
when they come to you? Because they were on step when
this front hit this double hole, and started making its own
waves. So, these waves are emitted in
step, they march along and come to you without,
with the same time delay. So, when a crest from this
reaches you, a crest from that will reach you,
trough reaches, trough reaches.
So, you'll just double the answer.
So here you will get, ψ will become
2ψ, or A will become 2A.
If you have an amplitude A from this and A
from this, you will get 2A here.
So, the water will be very choppy there.
If you're living on a little boat, the boat will rock up and
down with amplitude 2A. If you shut off this hole,
the amplitude will go back to just that from one hole,
it'll be A; 2 gives you 2A.
That's what makes a lot of common sense.
But pick another point here, I will tell you how to pick
that point, so that these two lengths are not equal.
So, let's call that length L_1 and length
L_2. Suppose L_1 -
L_2 is half a wavelength.
As you know between every two crests there is a trough where
the function is minus the amplitude, at the crest is equal
to plus the amplitude. So, if you're half a wavelength
apart, when a crest from here appears, a trough from here will
appear. When a trough from here
appears, a crest from here will appear.
In fact in every instant, whatever comes here,
minus of that will come here. Let's understand why.
The way the wave travels is 2πx over λ,
where x is, say, measured from here.
For the other guy, it is 2π times x
prime over λ, or x prime at that
length. If that length is longer than
this, by λ over 2, if this is x plus
λ over 2, look what you're getting extra:
2π times λ over 2 divided by λ,
is really cosine of π, plus what the other person got.
But cosine of π added to π added to anything
changes the cosine to minus itself.
So, the way to think about it is when you're half a wavelength
behind coming from here, whatever this one does,
this does minus of that. Because shifting by half a
wavelength is adding a π to the cosine;
adding a π to the cosine changes it to minus its
value at every instant. Not only does crest come with
trough and trough comes with crest, every possible
displacement comes with minus of that from here,
and you get nothing here. So, this is called an
interference minimum. Or we say, here the
interference is constructive, and here we say the
interference is destructive.
How will you find the point of destructive interference?
It's a matter of simple geometry.
You measure that length, and you measure that length;
take the difference and set it equal to λ over 2.
That's the first destructive interference.
That's a very interesting point to sit in.
Look what it's telling you. It says if you sit here with
your boat, the boat won't move at all, even though there are
two holes letting waves in at that point.
That's the principle behind noise cancellation.
You can actually cancel one sound with another sound if you
bring the second sound off by half a cycle.
So that if you had only one hole in the wall that you'd
built between your little estate and the beach,
and the waves from here were rocking the boat,
miraculously putting a second hole at the right place can
actually calm the waters where you are.
And the minute somebody blocks one of these holes,
the water will get choppy again where you are.
It's the property of waves that two waves can lead to nothing,
but one wave cannot because two waves have a chance of working
against each other and canceling.
This is something you've got to get very straight.
Now, if I go further up here, until I reach a point where the
difference in those two lengths is the full wavelength,
then I'm back to being in step. Of course, a full wavelength
means that these are not doing the same length,
but this is differing from that by a whole wavelength.
That means when the 13^(th) wave from here arrives,
the 14^(th) or the 12^(th) from here will arrive.
But we don't care, as long as a crest comes with a
crest; we're happy where we are.
So things get--start with constructive,
destructive, constructive,
destructive; it oscillates both above and
below the midpoint. Yes?
Student: How is the amplitude of the wave,
once it goes through that amplitude [inaudible]
Professor Ramamurti Shankar: It could be,
it need not be the same. It all depends on the aperture
and so on. But what is clear by symmetry
is that these two have the same amplitude.
In practice, we believe the amplitude will
also be the same, because amplitude has to be
continuous when you go through a slit.
The water cannot be jumping up by two inches before the slit,
and suddenly on the other side jump to different lengths.
Amplitude is always continuous. So in fact, this amplitude and
that amplitude will be equal. You can also see from the wave
equation if the ψ was not continuous;
dψ/dx will be infinity. There's no room for any
infinities in the equations. ψ will be continuous;
in fact--Yes? Student: [inaudible]
Professor Ramamurti Shankar: Here?
Student: Yes. Professor Ramamurti
Shankar: You're not able to read this equation,
because it's very badly written.
Let me tell you what I tried to do.
Cos 2πx over λ;
let's pick a certain time, so that I don't worry about
ωt. In fact, let me write it as
L_1. The other one is cos
2π, L_2/λ.
I'm adding two waves. Suppose L_2 =
L_1 plus half a wavelength.
Then, this one becomes cos 2π
L_1/λ + π. Right?
L_2 = L_1 + λ/2.
This is the L_1.
If you put a λ/2, times 2π over
λ, you get a π. Then, you've got to understand
that whenever you add a π to a cosine,
you get minus the cosine, without the π.
There you can see this guy is producing a signal,
which is exactly the minus of that guy.
I did this at t = 0, but you can put an ωt
to both. But nothing changes the fact
that when you subtract the same ωt from the 2,
it doesn't change anything. So, I didn't worry about the
time. It's the spatial mismatch
that's the problem here.
Okay, now sometimes people write this formula for
destructive and constructive interference as follows.
There's an approximate formula that's very handy.
The correct way to do this of course is simply measure that
distance and that distance and take the difference,
okay? In other words,
if L_1 - L_2 = mλ,
where m is an integer, 1,2 et cetera,
or plus minus 1, plus minus 2,
it is constructive. If it's a half integer,
which we like to denote as follows: m + ½
λ, again m = 0 plus minus 1,
et cetera is destructive. The way of writing--m is
usually the standard symbol people use for integers;
m and n are common symbols for integers.
So, if the path difference is an integer multiple of
λ, it's harmless; if it is a half-integer
multiple of λ, it's harmful,
meaning destructive. The whole approximation has to
do with what is L_1 - L_2.
One is to actually measure these lengths exactly,
and using Pythagoras' Theorem, given the difference between
the slits, given the distance to the screen.
If that distance is d, one can do the math and
calculate it. But here's what people do.
They say imagine this screen is very, very far away.
Very far away so that the rays leaving this are actually
parallel because they meet at infinity, and that's where your
screen is. It's not really at infinity but
it's far enough for this purpose.
Then, you can tell very easily that the extra length traveled
by this wave is just this part here.
Do you see that? Because this part cancels out.
So, drop a perpendicular here, those two are equal in length
going out in infinity; that's the extra section.
How long is the extra section? With that as d,
distance between the slits, if that angle is θ
here, then we say d sin
θ is = L_1 - L_2.
And that's what you want to be either mλ or (m +
½)λ, for constructive and
destructive. But there's a way to relate
this θ here to this direction.
They are the same angle. You may want to go look at your
book where the graphics are better than mine.
But I'm telling you the same old trick we used from incline
planes. If you got two lines,
this line and this line, the angle between them is
θ, then the perpendicular of the
two lines have the same angle between them.
Perpendicular to this line is this one, and perpendicular to
the screen is the horizontal one.
So, θ is also that angle.
Therefore, the way you will do any of these problems is,
suppose someone says find out when I get the first destructive
interference, given that length to the
screen, and someone says, "How far should I move?"
What you do is, you say d is the
distance between the two slits, sin θ equal to say
λ over 2, for destructive.
Solve for θ. Because the d is known,
sin θ is not known, you want it to be λ
over 2; λ will be given to you.
Once you get the angle, you're basically telling the
person to go at that angle, and that's where the intensity
will be a minimum. But then, it's trigonometry to
understand that d over L is just tan θ.
So, if you knew θ, you can find D over
L. So, D over L is
tan θ.
So, the problems you can get are either λ is given to
you and θ has to be found,
or θ is given to you, λ has to be found.
Somebody can say in some experiment with waves I find the
first minimum at that angle, what's the wavelength?
The distance between the slits is so and so.
But the basic idea is really to find the path lengths;
take the difference, set it equal to either integer
multiple or half integer multiple of the wavelength.
Okay, last of the interference problems has to do with all
these musical instruments, and it goes like this.
Take a string. Now, there's a wall here and
you start shaking it, and ask what happens.
So, let's say I shake it just once, and it sent a pulse like
this, moving to the right. Well, this pulse,
by the time it gets to this place where the string is
anchored and not free to move, it cannot vibrate here.
So, how is that going to happen? Well, some forces will be
exerted on the string by the wall to make sure that the
signal the wall sends out, along with the signal you send
out, add up to zero at the point of contact,
because no vibration there is allowed.
And the result of that is that when a signal goes out like
this, it comes out reflected in direction and inverted in its
sign.
For those of you who wanted more satisfactory proof of this,
let me say the following. Suppose there is no wall.
I saw you send a signal, and I manufacture a mirror
image of that to the right, traveling towards this point.
The two are coming near each other, and when they meet,
by design, I've arranged it so that this one and this one,
when they overlap, are exactly opposite of each
other, so that they cancel each other at this point.
Then, they go through each other, and this will come out
here, that'll go to the other side.
But throughout this encounter -- here's the punch line --
throughout the encounter this point was not asked to move,
because the two waves cancelled.
That means even if I put my finger there,
it won't make a difference to this problem because that point
is not planning to move anyway. It also means that if you put a
barrier here that doesn't allow the string to move;
it doesn't change the outcome. That's the reason we convince
ourselves that the incoming pulse this way comes out that
way. Okay, now what I want to study
is not a single one-shot pulse, but a pulse that looks like
this, A cos kx - ωt.
A continuous signal goes in, how does it come out?
It comes out as [-A] cos ωt + kx.
So, I've reversed the amplitude and I've reversed the direction
of motion. So, this is a result of an
incoming wave, and the one that's sent back by
the wall; together they coexist.
But that goes in and that comes out, and together they form what
I will find in this region. I keep sending stuff to the
wall, the wall keeps sending that back to me,
and both exist. Well, you've got to do another
trigonometric identity, to show you that this is really
2A sin kx, sin ωt.
I don't want to go through the trigonometric identity.
But what I want you to notice then, is that the resulting
physics looks like this. Here is the wall,
at a given point x, the amplitude is 2A sin
kx, and it oscillates with a frequency ωt.
And if you draw it sin kx, it's going to look like
this.
So, whenever kx = π, if kx = π where that
distance is λ/2, there is no vibration.
There is no vibration. This is called a node;
in between is an anti-node. This kind of thing will be set
up all the way back. Now you see what that means.
So, this point is not going to vibrate.
If I put another barrier here, it won't make any difference,
because that point is--these two waves add up to zero
identically at that point. If you put your finger there it
won't matter. So, in addition to making no
difference if you hold the string there,
it also doesn't matter if you hold it here.
So, what that means is that if you turn the problem around and
say, look, I give you the following problem,
I give you a string which is clamped at these two ends,
what are the possible vibrations of the string?
They have to look like the following.
Your string length has to be equal to that or that or any of
these multiples. That means what can happen in
your string is either that, or it can do that,
or something with more and more lobes in it.
These are the allowed modes of vibration of the string,
fixed at two ends. What I want to extract from
this is what sound will you hear, what frequency will you
hear. So, you have--this is the
family of problems that you guys are expected to do.
So, you have to think as follows.
That is the length. How is the length related to
the wavelength, that's all you got to figure
out. Is it clear to you guys that
this is half a wavelength? If you've got that, you're home.
Because you're saying half a wavelength is equal to L,
or the wavelength is equal to 2L;
the frequency is the velocity of sound divided by the
wavelength, which is the velocity divided by 2L.
That's the frequency you will hear from this guy.
When it vibrates like this, I think it's clear to you that
it's gone through one full cycle here, this is λ = L.
You do the whole thing you will find frequency is now
v/L, which is two times the lowest frequency,
which we denote by f_0,
which means the fundamental frequency.
So, the bottom line is very simple.
A string clamped at two ends, and has length L,
can only vibrate in either the fundamental frequency or an
integer multiple of the fundamental frequency.
And when it does that, the two ends of the string will
be nodes, and you can have any number of anti-nodes in between,
provided you start with a node and end with a node.
Now, this is also valid in the following problem.
Take a tube, clamped [correction:
should have said blocked] at both ends,
and there are some longitudinal sound waves going back and
forth. This is transverse waves on a
string; this is sound waves going back
and forth. You realize the sound waves
going back and forth cannot go back and forth at this wall or
at this wall. Therefore, the motion of the
air molecules has to have a node here and it has to have a node
here. So, the possible modes of
vibration for sound also look exactly like this.
The frequencies will be the same for vibrations on a tube,
closed at two ends. The only subtlety,
which I don't want you to forget--;Here,
the picture really talks about the string moving perpendicular
to the length. This does not indicate here
that the sound, the air molecules are going up
and down. This is the amplitude for
longitudinal vibration. If this number is three inches,
it means the molecules here are moving back and forth by three
inches. So, we just out of habit plot
everything on the y axis; that doesn't mean the motion is
in the y direction. For a longitudinal wave the
motion is side-to-side; this is the extent of
side-to-side motion. So, the only problem where you
may get stumped is the following one.
I will tell you how to do that and then we are done.
Take a tube, like this. It's closed here and it's open
here. You know, take a Coke bottle or
something and try to blow on it; you hear a certain sound.
Why is it picking that one frequency?
That's what I want to understand.
Now, since I'm making the noise here, here is where the
amplitude will be the biggest, and of course there's nothing
doing at this end, because that's a wall which
cannot go back and forth. So, what pattern can I draw?
Take a minute and think about it.
I want a maximum vibration here, nothing here.
The lowest one I can draw looks like that.
The next one I can draw is constrained to have a maximum
here and then a minimum at that end.
Now, what is the relation of the length of the tube to the
wavelength? I think you can see the length
of the tube is now a quarter of a wavelength,
it's not from here to the top. Coming down is the other
quarter and then the other half is missing.
So, what's the frequency? It's equal to v/λ;
it's going to be 4v/L. I'm getting the frequency to be
velocity divided by wavelength, and I bet you will get this.
Oh, did I get it wrong? Oh yes, I'm sorry,
it's v/4L. 4L is λ.
Now, what's the answer here? What is the relation of the
length to wavelength? The length happens to be a half
a cycle plus another quarter, is 3/4^(th) of λ.
And the frequency, if you do this,
will be 3v/4L. And this is all I want you to
notice. That in this peculiar problem
where one end is open and one end is closed,
the frequencies are odd multiples of the fundamental
frequency, 3 times that,
5 times that, 7 times that.
And in exams, you can be asked a variety of
questions, either this one or some other exam you take some
time later in life. You should be able to do
the--Let me do one practice thing.
Here's a tube open at both ends. Draw me the lowest frequency
you can draw. I want an anti-node here,
an anti-node here, but between them I got to have
a node. So, it'll look like that.
Then, you can draw more complicated patterns,
provided the two ends look like this.
And I think you can see here, that this is half a wavelength.
So, L = λ over 2. And again, you will find here
all frequencies are integer multiples of the fundamental.
So, the story is this, as long as the two ends are the
same, both open or both closed, all frequencies are integer
multiples of the fundamental one.
If you got a funny situation, one end is open and one end is
closed, you'll get odd multiples of the fundamental.
But I think you might have to go do some reading in the book.
I don't have a better way to explain this,
and this is somewhat easier part of the course,
and I think you should try to be on top of this particular
topic.