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Lecture 19
 Waves
Overview
Waves are discussed in further detail. Basic properties of the waves such as velocity, energy, intensity, and frequency are discussed through a variety of examples. The second half of the lecture deals specifically with superposition of waves. Constructive and destructive interferences are defined and discussed.
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htmlFundamentals of Physics IPHYS 200  Lecture 19  WavesChapter 1. General Solution of Wave Equation [00:00:00]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 lefthand 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. Chapter 2. Spatial and Temporal Periodicity: Frequency, Period [00:08:51]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 λsized 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. Chapter 3. Wave Energy and Power Transmitted [00:17:39]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. Chapter 4. Doppler Effect [00:30:02]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′, 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′ 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, subobserver [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. Chapter 5. Superposition of Waves [00:38:58]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_{1}t. That’s the second source of sound, which is A cos ω_{2}t. 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_{1}t + cos ω_{2}t. 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_{1}t and ω_{2}t 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_{1}t, the other is vibrating as ω_{2}t. 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. Chapter 6. Constructive and Destructive Interference, Double Slit Experiment [00:48:57]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′ over λ, or x ′ 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 halfinteger 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. Chapter 7. Modes of Vibration: Application to Musical Instruments [01:01:30]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 oneshot 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 antinode. 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 antinodes 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 sidetoside; this is the extent of sidetoside 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 antinode here, an antinode 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. [end of transcript] Back to Top 
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