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[Assistant] Is equal to e to the ix plus e to the minus ix over 2. You can check that. And sine x.


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[Assistant] Is equal to e to the ix, minus e to the minus ix, over 2i.


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[Assistant] Okay, so what does this mean? It means the trigonometric functions.


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[Assistant] You don't need them. You've got exponential functions. You can manufacture trigonometric functions out of them, provided


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[Assistant] You're not afraid to go to exponents with complex number i in them.


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[Assistant] And all the identities you know about science and cosines will follow from this.


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[Assistant] For example, if we take cosine square plus sine square.


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[Assistant] You're supposed to get one, but you can square the right hand side, you can square the left hand side, and you will get one.


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[Assistant] To get the one, you better remember the following. e to the ix times e to the minus ix is what?


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[Assistant] You know how you combine exponents. You raise it to one power times the same thing raised to another power is e raised to i x minus x, which is e raised to 0, which is 1.


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[Assistant] When you raise the number to a power, you multiply it by the number times a different power, the product is the number to the sum of the two powers.


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[Assistant] Right, powers combined. 2 cubed times 2 to the 4th is 2 to the 6th.


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[Assistant] That's true for 2, it's true for e, it's true for everything. Exponents add when you multiply them. That's why e to the ix and e to the minus ix combine to give you 1. Once you know that, you can prove this.


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[Assistant] So I introduced to you the number i by saying it's the square root of minus 1, and complex numbers entered our life, even though we didn't go looking for them. You can write down equations with real numbers.


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[Assistant] With no intention of invoking anything fancy, like this.


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[Assistant] And you find there is no solution to this equation, even though everything there is completely real.


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[Assistant] So you can say well x squared is minus one and you can manufacture


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[Assistant] And number i with the property i squared is minus 1, then of course you can have x equal to plus or minus i as your answer. So complex numbers are always trying to solve quadratic equations.


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[Assistant] So let me write you a slightly more interesting quadratic equation, x square plus x plus 1 equal to 0.


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[Assistant] So there's no funny business, all real numbers. We want to solve this equation. So we go back to our good old sandbox days when we knew what the answer for this was.


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[Assistant] And we already have a problem because we don't know what to do with square root of minus 3. So we will write it as square root of minus 3.


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[Assistant] We'll write it as square root of minus 1 times square root of 3.


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[Assistant] The square root of minus 1 we will call i, and we will say this equation has two roots.


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[Assistant] These are solutions formally to this equation in the following sense. Take this x that I'm giving you.


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[Assistant] Put that into that equation, square the x and add the x to it and add the 1, it will in fact satisfy the equation.


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[Assistant] So people realize that if you enlarge numbers to include complex numbers, then you can solve.


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[Assistant] Any polynomial equation with that many number of roots. If it's quadratic, it'll have two roots. If it's cubic, it'll have three roots. They may not all be real. They may involve the complex number i.


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[Assistant] The whole thing is a single complex number. Don't think of it as a sum of two numbers. It's a single complex number.


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[Assistant] Instead of minus, you'd get something like 1 plus or minus root 3 over 2. Take the positive root. You don't think of it as two numbers. This is a single number.


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[Assistant] Okay, so that's what we are going to do now. We're going to generalize this particular case and introduce now a complex number z.


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[Assistant] And we're going to write every complex number z as a part that has no i in it and a part that has an i in it.


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[Assistant] The example I had here, x is equal to in this example,


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[Assistant] And we're going to visualize that complex number as a point in the xy-plane.


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[Assistant] You just measure x horizontally and y vertically, put a dot there, that's your complex number z.


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[Assistant] Then we say a new complex number called Z star, and it's defined to be X minus IY. It's called the complex conjugate.


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[Assistant] Complex conjugate of Z is obtained by changing the sign of I.


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[Assistant] So if you like, if Z is sitting here, Z star is sitting there, reflected on the X axis.


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[Assistant] Okay, so, how does one work with complex numbers?


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[Assistant] You add them, you subtract them, you multiply them, and you divide them. If you know how to do that, you can do with complex numbers everything you did with real numbers.


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[Assistant] If you draw pictures, then if Z1 is that number, and Z2 is that number,


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[Assistant] Then Z1 plus Z2 really is like adding vectors. This number is Z1 plus Z2.


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[Assistant] It's exactly like adding vectors, but you don't think of


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[Assistant] This plus that as two disjoint numbers, but as forming a single entity. In a complex number z, x is called the real part.


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[Assistant] If you want to get the real part of a complex number Z, you add to it as complex conjugate and divide by 2.


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[Assistant] Because you are adding x plus i y to x minus i y, you get 2x and you divide by 2, you get that.


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[Assistant] The y, which is called the imaginary part of z, you take z, minus z star, divide by 2r.


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[Assistant] It's like saying, find the x component of a vector, it's whatever multiplies i, y component of the vector is whatever multiplies j. Similarly, a complex number can be broken down into the part which is real and the part that multiplies i.


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[Assistant] To extract the real part, add to it the conjugate and divide by 2. To extract the imaginary part, subtract the conjugate and divide by 2i.


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[Assistant] Again, if you have seen this, you don't need this. If you haven't seen it, it may look a little fast.


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[Assistant] That's why I posted on the notes, uh, on the website, chapter 5 from this math book, which has got all this stuff. So you can go home and you can read it.


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[Assistant] Okay, the next question is, what is z1 times z2?


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[Assistant] Well, just open out the brackets, and remember, I squared is minus 1. That's all you've got to do. So, that gives me x1, x2. Then let me multiply the Iy1 times the Iy2, and that gives me minus y1, y2.


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[Assistant] Because i squared is minus 1, plus i times x1y2 plus x2y1. This how we multiply two complex numbers.


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[Assistant] There's something very nice that happens when you multiply a number z by its complex conjugate.


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[Assistant] What happens when you multiply z by its conjugate? You are saying take x plus iy, multiply by x minus iy.


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[Assistant] Now, A plus B times A minus B is A squared minus B squared. But remember, B is Iy. So, when you take the square of that, you should keep track of all the signs, and you will find it's x squared plus y squared.


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[Assistant] As this, and this thing is called the modulus.


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[Assistant] You should know, whenever you take a complex number, multiply by its conjugate.


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[Assistant] Because I'm going to use that now to do the last of the manipulations, which is, what is z1 divided by z2?


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[Assistant] How do you divide these crazy numbers? So on the top, I write x1 plus iy1. The bottom, I write x2 plus iy2.


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[Assistant] If I only had x2, I know how to divide. It's just a number. Divide that by x2 and that by x2. But I got the sum of these two numbers in the bottom, and you can ask, how do you do the division?


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[Assistant] The trick always when you run into this problem is to multiply the top and bottom by the complex conjugate of the bottom.


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[Assistant] Multiply by x2 minus y2, top and bottom.


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[Assistant] Then something nice happens to the denominator, because the denominator then becomes x2 squared plus y2 squared, which is an ordinary real number. Nothing complex about that.


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[Assistant] Now, don't worry about all the details. All it means is, if you know how to multiply to complex numbers, you can also divide by a complex number. That's the key point.


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[Assistant] Why? Because if you've got a complex number denominator and you don't like it, multiply top and bottom by the complex conjugate of this guy, the denominator turns into a purely real number. So this whole thing could be 36, for example.


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[Assistant] Well, we know how to divide the numerator by 36, right? x1 x2 plus y1 y2 is some number, maybe 9, then you divide by 36, one-fourth.


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[Assistant] This could be i times 18. It would be 36. You get i over 2. So this number could have been 1 fourth plus i over 2. So dividing by complex numbers is not a problem.


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[Assistant] Why did I do all that stuff today about e to the ix? You're going to see that now.


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[Assistant] The rationale for going through that math is the following. So let's take this complex number.


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[Assistant] And let's introduce, just like we would for an ordinary vector, this angle θ and that length r.


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[Assistant] Whenever you have a vector, you can talk about the Cartesian components x and y, or you can talk about the length of the vector and the angle it makes. Then it's clear that x is equal to r cosine theta.


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[Assistant] And y is called r sine theta, where theta is now an angle associated with a complex number.


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[Assistant] Which is x plus iy is equal to r times cosine θ plus i sine θ, which you can now write as r times e to the i θ.


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[Assistant] That's why we did all that work to tell you that a complex number can be written in this form.


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[Assistant] You can either write it as x plus iy or as r e to the i theta. They both talk about the same number.


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[Assistant] One talks about how much real part and how much imaginary part it has. Other way of writing it talks about how long the vector is and at what angle it is located. It contains the same information.


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[Assistant] So that r is equal to, if the inverse of this formula is that r is square root of x squared plus y squared, and theta is tan inverse y over x. In other words, tan theta is y over x. This is y, and this is x.


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[Assistant] Now the advantage of writing a complex number in polar form is the following. Suppose I give you two complex numbers. z1 is r1 e to the iθ1.


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[Assistant] And z2 is r2e to the iθ2. The product z1z2 is very easy to calculate in this form.


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[Assistant] Because r1, r2 multiply to give you this, and the exponentials combine.


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[Assistant] It's a lot easier to multiply them in this form than when I multiply them somewhere here. See, if you write it in this Cartesian form, it's a big mess. In polar form it's very easy. The rule is, if you want to multiply two numbers,


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[Assistant] Multiply the lengths to get the length of the new number, and add the angles to get the angle of the product. So what I'm telling you is that if z1 looks like this guy, with some angle theta 1, and z2 is that guy with an angle theta 2,


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[Assistant] The number z1, z2 has a length equal to product of these two lengths and an angle equal to sum of these two angles, so it will look like that.


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[Assistant] To go from this to that, you multiply the lengths and you add the angles.


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[Assistant] So, the polar form is very well suited for multiplying. And it's even better suited for dividing. If I say, give me z1 over z2.


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[Assistant] Well, you can all do that in your head. It's r1 e to the iθ1 divided by r2 e to the iθ2.


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[Assistant] The modulus of the new number is just r1 over r2. And how about this? e to the iθ1 divided by e to the iθ2.


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[Assistant] Dividing by e to the iθ is the same as multiplying by e to the minus iθ. So this is really e to the iθ1 minus θ2.


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[Assistant] So I'm assuming you guys can figure this part out. 1 over e to the iθ is the same as e to the minus iθ.


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[Assistant] Well, if you doubt me, multiply, cross multiply, and all I'm telling you is that 1 equals e to the iθ times e to the minus iθ, which you know is true when you combine the angles.


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[Assistant] When you got ertiai theta downstairs, you can take it upstairs with reversed angle. And that's a useful trick. I hope you will remember that useful trick.


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[Assistant] So that's one part of this thing that I want you to carry in your head, because it's very, very important. When you take a complex number,


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[Assistant] It's got a length and it's got a direction. When you multiply a second by a second complex number, you are able to do two things at the same time. You are able to rescale it and you are able to rotate it.


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[Assistant] You rescale by the length of the second factor and you rotate by the angle carried by the second factor.


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[Assistant] The fact that two operations are done in one shot is the reason complex numbers play an incredibly important role in physics and certainly in engineering and mathematical physics.


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[Assistant] Okay, I'm going to now change gears, but I want to stop a little bit and answer any questions any of you have. So how many people have seen all of this before?


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[Assistant] So if you saw something, you saw the whole thing. But look, that's not such a big number. So I am cognizant of the fact that not many of you have seen it. But you will have to go and learn this today.


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[Assistant] I'm going to post the problem set for this sometime in the day so you can start practicing.


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[Assistant] Don't wait. Promise has nothing to do with next Wednesday. It has a lot to do with today.


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[Assistant] When I assign a problem today, I imagine you're going breathlessly to your room, not able to wait, jumping into the problem set and working it out. That's the only way you're going to find out.


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[Assistant] What these things mean if you've never seen them. Seen them before in high school, taken a math course, and you've got lots of practice then.


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[Assistant] I'm not talking to you, but I'm going to use this at some point. So you should understand complex numbers, and I'm telling you there are very few branches of any science where complex numbers will not be used.


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[Assistant] You may not believe this now, but that is true. They make, if an electrical engineer, absolute must. If you solve any kind of differential equation, which can occur in biology, in chemistry, that's a must. So, it's very, very important.


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[Assistant] Alright, so now I'm going to do, I'm going back to physics now. Going back from mathematics to physics, and the physics of this


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[Assistant] Topic for today has to do with something very different from the past. So forget all relativity. Now we are going back to


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[Assistant] Newtonian days, kinetic energy is one half mv squared. It's a little difficult to go back and forget what you learned.


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[Assistant] On the other hand, for some of you it may not be so hard, if you didn't learn anything, then you are that much ahead of the other guys. Okay, but so remember now, the truth is the relativistic theory, we're going to go back,


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[Assistant] So what we're going to study now is what's called small oscillations, or simple harmonic motion.


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[Assistant] It's a ubiquitous fact that if you took any mechanical system, which is in a state of equilibrium, and you give it a little kick, it vibrates. You know, if you got a pillar, ceiling to float, you hit it with a hammer, it vibrates.


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[Assistant] If you take a gong and hit it, it also vibrates. If you've got a rod hanging from the ceiling through a pivot,


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[Assistant] The standard example everyone likes to use is if you got a particle in a bowl.


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[Assistant] It's very happy sitting in the bottom. If you push it up a little bit, it'll go back and forth.


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[Assistant] And the example that we are going to consider is the following.


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[Assistant] If you have a mass, M, connected to a spring, and the spring is not stretched or contracted, it's very happy to be there. That's what I mean by equilibrium. Equilibrium means the body has all the forces on it, adding up to zero, it has no desire to move. The question is, if you give it a little kick, what will happen?


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[Assistant] Well, there are two situations you can have. You can have a situation where a particle is on top of the hill.


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[Assistant] That's called unstable equilibrium, because if you give that a kick, it's gonna come down and never return to you.


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[Assistant] That's equilibrium, but unstable equilibrium. I'm talking about stable equilibrium. That's because there are restoring forces.


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[Assistant] If you stray away from the equilibrium, the force is bringing you back. And in the case of the mass and spring system, the force, MD to MA,


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[Assistant] So we want to understand the behavior of such a problem. We want to solve this problem.


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[Assistant] So how do you solve this problem? Well I did it for you long back, when I gave you a typical paradigm for how you apply Newton's laws, I gave you this example. So I'm going to go through it somewhat fast.


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[Assistant] Our job is to find the function x that satisfies this equation, and we would like to write it as follows. d2x over dt square is equal to minus ω square x, but ω square is the shorthand for, ω is the shorthand for square root of k over m.


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[Assistant] By the way, I'm using big X and small x like crazy, so it's just small x. I don't mean anything significant between this and this. They're all the same. If you want, you can take this. Pardon me?


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[Assistant] This is k, and this guy is k. And the other equations we use to follow the lecture today, these are all the same.


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[Assistant] Okay, so don't say what happened there when he moved from one to the other. Alright, so what did I say we should do?


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[Assistant] You can make it a word problem if you say, I'm looking for a function.


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[Assistant] Which when I take two derivatives, looks like minus itself, except for this number.


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[Assistant] And we just saw that today. Trigonometric functions are the property that you take two derivatives, they return to minus themselves. So you can take a guess that x looks like


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[Assistant] If you took this function of time, it in fact obeys this equation.


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[Assistant] Provided the ω that you put in is the ω that's in the equation.


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[Assistant] First time you get minus omega sine omega t, second time you get minus omega squared cos omega t, which means it's minus omega squared times your x itself. And a is whatever you like.


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[Assistant] Then I said, let's plot this guy. When you plot this guy, it looks like this. And A is the amplitude.


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[Assistant] Well, omega is related to the time or the frequency of oscillations as follows.


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[Assistant] If I start with t equal to 0, where the x is equal to a, how long do I have to wait till I come back to a? I think everybody should know.


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[Assistant] That, I have to wait a time capital T so that omega times t is 2. Because that's when the cosine returns to 1.


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[Assistant] That means the time that you have to wait is 2π over ω.


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[Assistant] You can also write it as 2π times frequency, where frequency is what you and I would call frequency.


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[Assistant] How many oscillations it does per second. That's the inverse of the time period.


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[Assistant] So, if you pull a mass and you let it go, it oscillates with the frequency.


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[Assistant] Which is connected to the force constant and the mass.


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[Assistant] If the spring is very stiff and k is very large, frequency is very high. If the mass is very big, the motion is very sluggish and f is diminished. So all this stuff comes out of the equation.


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[Assistant] One very remarkable part of the equation is that you can pick any A you like. Think about what that means.


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[Assistant] What is the meaning of A? A is the amount by which you pulled it when you let it go.


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[Assistant] You are told, whether you pull the spring by 1 inch or by 10 inches,


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[Assistant] The time it takes to finish a full back and forth motion is independent of A.


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[Assistant] This frequency here is independent of A. Yes?


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[Assistant] So the remarkable property of simple harmonic motion is that the amplitude does not determine the time period.


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[Assistant] If you pull it by two inches, compared to one inch, it's got a long way to go.


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[Assistant] But because you pulled it by two inches, the spring is going to be that much more tense, and it's going to exert a bigger force, so that it'll go faster for most of the time.


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[Assistant] That's very clear, but the fact that it goes faster in exactly the right way to complete the trip at the same time is a rather miraculous property of the fact that this equation has this particular form.


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[Assistant] If you tamper with this, if you add to this a little bit like 1 over 100 times x cubed, then all these bets are off. It's got to be that equation for that result to be true.


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[Assistant] Okay, so this is simple harmonic motion. Then you can do the following variant of this solution, which I will write down now.


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[Assistant] But it's the same physics, it's the same equation. And that really comes from the fact that we had one more latitude here of adding a certain number of phi, which is called the phase to the oscillator.


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[Assistant] Whatever you pick for phi, it'll still work. And whatever you pick for capital A, it'll still work. So whenever you have an oscillator, namely a mass and spring system,


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[Assistant] And you want to know what this x is going to be at all times.


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[Assistant] You need to somehow determine the amplitude and the phase. Once you know that, you can calculate x at all future times.


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[Assistant] So let me give you an example. Suppose an oscillator has x equal to 5 and velocity equal to 0 at t equal to 0.


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[Assistant] What does that mean? I pulled it by five, and I let it go.


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[Assistant] I tell you the spring constant k, and I tell you the mass, and I say what's the future x? You've got to come back to this equation, and you've got to say 5


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[Assistant] So that tells me 0 is minus ωA sine Φ, ω is not 0, A is not 0, therefore sine Φ is 0, that means Φ is 0, you get rid of that, so the subsequent motion is x equal to 5 cos ωt. This is a problem where we did not need Φ.


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[Assistant] But it could have been that when you join the experiment, you are somewhere here, then the mass is actually moving, then I would say at t equal to 0, velocity was some other number, maybe 6. Then you've got to go back and write two equations, 1 for x,


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[Assistant] Take the derivative, which looks like this, and instead of putting 0, you put 6. You've got two unknowns, a and phi. You've got to juggle them around, and you've got to solve for a and phi given these two numbers.


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[Assistant] Simple exercise to solve for them, and when you do, you completely fit the problem.


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[Assistant] In other words, simply to be told a mass is connected to a spring of known mass and force constant is not enough to determine the future of the spring.


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[Assistant] For example, if nobody pulled the mass, it's just going to sit there forever. If you pulled it by 10, it's going to go back and forth from 10 to minus 10. So you have to be told further information to nail down these two numbers. They're called free parameters in the solution.


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[Assistant] On the other hand, no matter how much you pulled it and how you released it, this ω and therefore the frequency are not variable. Yes?


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[Assistant] If you started at rest, indeed it will be zero, because the velocity at t equal to zero is minus omega a sine phi. If you want to kill that, the only way is phi equal to zero. You could put phi equal to pi if you like, but if you put zero that will do.


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[Assistant] This is the elementary parts of this thing, and I already told you the following things. If x sub t is equal to a cos omega t,


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[Assistant] Now let's put phi equal to zero. In other words, we'll agree that since there's only one oscillator, it is perverse to set your clock to zero at any time other than when the oscillator is at one extreme point.


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[Assistant] If there are two oscillators oscillating out of step, it's impossible to make phi 0 for both of them. Because if you can sync your clock when one of them is at the maximum, that may not be when the other one is at the maximum. But with one oscillator, to choose phi other than 0 is perverse.


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[Assistant] It's like saying I'm doing a projectile problem, I'll pick the place where I launch the projectile to be x equal to y equal to 0. You can pick some other crazy origin, but it doesn't help. So we'll assume phi is 0 in this problem.


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[Assistant] If that is x, what is velocity at time t? Take the derivative of this, you get minus omega a sine omega t.


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[Assistant] That means the velocity is also oscillating sinusoidally, but the amplitude for oscillation is omega times a.


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[Assistant] So if a is the range of variation for x, omega times a is the range of variation for velocity. Velocity will go all the way from plus omega a to minus omega a. The acceleration, which is one more derivative, is minus omega squared a cos omega t, which is really minus omega squared times x itself.


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[Assistant] So the amplitude for oscillation for the acceleration is omega squared times a.


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[Assistant] So I think you should understand if omega is very large.


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[Assistant] Then the velocity will have very big excursions, and the acceleration will have even bigger excursions, because the range of this A really means if you plot x, it's going to look like this, going from A to minus A.


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[Assistant] This says if you plot the velocity, it's going to look like this.


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[Assistant] The range will go from ωA to minus ωA. If you plot the acceleration, it will go from ω square A to minus ω square A.


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[Assistant] Last thing you want to verify is the law of conservation of energy.


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[Assistant] I think I mentioned this to you when I studied energy, but let me repeat one more time. If you take in this problem,


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[Assistant] 1 half kx squared is going to involve a squared cosine squared omega t. 1 half mb squared is going to involve something, something, something sine squared omega t. And by magic, sine squared and cosine squared will have the same coefficient, so you can use the identity to set them equal to 1, and get an answer that does not depend on time.


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[Assistant] That's the beauty. Even though position and velocity are constantly changing, this combination, if you crank it out, will not depend on point.


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[Assistant] And what can it possibly be? If it doesn't depend on time, I can calculate it whenever I like.


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[Assistant] So let me go to the instant when the mass has reached one extremity.


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[Assistant] And the sub-order swings back. At that instant, it has no velocity. It only has an x which is equal to the amplitude of motion. So that's the energy of an oscillator, 1 half a square.


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[Assistant] When x is, when b is equal to 0, x will be plus or minus a. At any other intermediate point, you can find the velocity if you give me the x. We did that too.


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[Assistant] So this is all I want to say about the linear oscillator.


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[Assistant] Well let me mention to you another kind of oscillation, which is very interesting. Suppose you suspend a rod from the ceiling, like this.


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[Assistant] It's saying that if you give it a twist, then it'll go twisting back and forth. Like that.


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[Assistant] That's also simple harmonic motion, except what's varying with time is not the linear coordinate, but the angle θ by which the rod has been twisted.


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[Assistant] What's the equation in this case? The equation here is I d2θ over dt square. You guys remember now that's the analog of MA. That's going to be equal to the torque.


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[Assistant] Now what will happen in the problems we consider is that theta is a small number.


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[Assistant] And the torque will be such that it brings you back to theta equal to zero. So this number will be approximately equal to minus some number kappa times theta.


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[Assistant] For small oscillations, since τ vanishes at θ equal to 0, it will look like some number times θ. If you like, it's the Taylor series for τ for small angles.


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[Assistant] If you knew this kappa, you are done because mathematically this equation is the same as md2x over dt squared is minus kx. Because the omega for this guy will be square root of kappa over i.


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[Assistant] You can steal the whole answer, because mathematically, mathematicians don't care if you're talking about theta or if you're talking about x. If you compare this equation to this equation, md2x over dt square, equal to minus kx. If the ω there was the root of k over m, the ω here will be the root of kappa over i.


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[Assistant] Give me an example where tau looks like kappa times theta. So I'm going to give you that example.


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[Assistant] So that example is going to be the following, a simple pendulum.


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[Assistant] Take a pendulum like this. It's got a mass m. It's got some length l. It's hanging from the ceiling. This is a rigid rod. Not a string, but a rigid, massless rod. And the mass m is at the bottom.


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[Assistant] So it's happy to hang like this, but suppose you give it a kick, so you push it over there to an angle θ, which is whatever you like.


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[Assistant] The angle between the force, and you're trying to do rotations around this point, this angle here is the same theta.


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[Assistant] So the equation of motion for the pendulum really is the following. I d2θ over dt squared is equal to minus mgl sinθ.


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[Assistant] Now we cannot do business with this equation. This equation is a famous equation, but you and I don't know how to solve it with what we know.


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[Assistant] But we have learned, even earlier today, look at the formula for sine x. Sine x begins with x minus x cubed over 3 factorial and so on. So if θ is small, then you may write it as mgLθ.


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[Assistant] So for small angles, sine theta is theta, and the restoring torque in fact is linear in the coordinate, the angular coordinate. So by comparison now, we know that this fellow is our kappa.


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[Assistant] So kappa for the pendulum is mgl. So kappa may not be


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[Assistant] Universal in every problem. In each problem you've got to find kappa. You find that by moving the system off equilibrium.


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[Assistant] Finding the restoring torque, and if you're doing it right, restoring torque will always look like something, something, something times theta. Everything multiplying theta is our kappa. So let's now calculate omega for the simple pendulum.


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[Assistant] It's equal to moment of inertia of this mass.


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[Assistant] Around this point, the rod is massless. For a single point mass, we know the moment of inertia is just m times L squared.


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[Assistant] So you find ω is equal to square root of g over L, and that is 2π over t. So you get back the famous formula you learned, t is 2π square root of L over g.


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[Assistant] This is the origin of the formula you learned in school. It comes from small oscillations of a pendulum.


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[Assistant] So think about this problem. If you pull the pendulum by an angle,


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[Assistant] So large that sine theta is no longer approximated by theta, but you need the theta cubed over sixth term. It can happen for large enough theta. Then the equation to solve is no longer this. And what that means is, if you took a real pendulum,


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[Assistant] The time it takes to finish an oscillation, in fact, will depend on the amplitude.


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[Assistant] Only for small oscillations, when sine theta can be approximated by theta, you will find whether you do that oscillation or that oscillation, the time is the same. But if you go too far, the other terms in sine theta will kick in and the result is no longer true.


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[Assistant] The period of the pendulum, in fact, is not just a function of L and G, but also a function of the amplitude. Only for small amplitudes, you get this marvelous result.


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[Assistant] There's one problem I want you to think about. I may do it myself, or I may want, I think you should try to read up on it, and I will try to explain it next time, which is the following


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[Assistant] I will give the problem to you and I will do it for you next time, but I want you to think about it. Here is a hoop, like a hula hoop. It's hanging on the wall with a nail. And it's very happy to be where it is. Now I come and give it a little push. It will start oscillating.


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[Assistant] It'll look like that, and sometime later it'll look like that. It's going back and forth.


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[Assistant] So you move it a little bit, and you find the restoring park. Then you are done. But you gotta do two things right to get the right answer.


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[Assistant] You've got to find the moment of inertia of a circular loop, not around the center, but around a point on the circumference. So you guys know the magic word you've got to use, parallel axis theorem.


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[Assistant] And to find the restoring torque, here is the part that blows everybody's mind.


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[Assistant] We said when the loop goes to that position, where does gravity act?


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[Assistant] Gravity is acting at the center of mass. Center of mass is somewhere here, in the middle of the loop, even though there is no matter there.


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[Assistant] So center of mass of a body need not lie on the body. The ring is a perfect example.


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[Assistant] So if you remember, the center of mass of the loop is in the center of the loop, and put the mg there, you will get the right torque and everything will just follow.


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[Assistant] By the way, my offer is always open. If there's something about the class...


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[Assistant] Uh, you want a change, you find this hard, this, that, you can always write to me. Okay, you can. I want your feedback, because I just don't know, uh, which parts are easy, which parts are hard, which are slow, which are fast. Or if you haven't had enough math and you need some help, so send me some email, and I will try my best to answer that.


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[Assistant] And say good morning to all the baby fishies. Okay, here is the fish tank. I don't really want to show you that. It's not really feeding time yet. But over here, we have this guy. This is a cichlid.


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[Assistant] They flipped it over by themselves and laid all their eggs there. And so they take turns fanning it. And here comes the other partner. They take shifts. They take turns watching the babies. Good morning fishies. Hello, how are you? Hey guys, how's it going over here? We have this monster. This is something that you would see in your dreams.


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[Assistant] Like a scary movie, this creepy thing. We spent a long time cleaning this place, and it's finally looking really nice. Yesterday was like an off day, well the past few days have been off days for me because you guys, I've, I've had so many backed up of the 10k challenge.


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[Assistant] No, simple wedding. We were just talking last night, we're like, I'm so afraid to kiss in front of my parents. He's like, I don't wanna kiss you in front of them either. I'm like, we're gonna have to, right?


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[Assistant] the beans the melons


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[Assistant] Ah, it's good to get our old car back again.


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[Assistant] And I think our old car likes us too! Don't you?


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[Assistant] So it is very important that Daddy Pig knows where his glasses are.


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[Assistant] Without his glasses on, Daddy Pig cannot read his newspaper.


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[Assistant] Pepper and George are looking for Daddy's glasses. Pepper looks under the newspaper, but Daddy Pig's glasses are not there.


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[Assistant] Peppa looks under the pillows, but Daddy Pig's glasses are not there.


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[Assistant] George looks in daddy's slippers, but the glasses are not there either.


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[Assistant] Pepper and George cannot find Daddy Pig's glasses anywhere.


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[Assistant] I've been looking at why individualism, I mean, individual differences are such an important area of management studies.


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[Assistant] They're all fundamentally a group of people working together.


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[Assistant] But it's what these individuals contribute to their places of work that makes you realise how important they are.


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[Assistant] Of course they bring different ideas, but it's also their attitudes and their experiences of learning. Diversity is important in these areas too.


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[Assistant] So why do people behave so differently from one another at work?


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[Assistant] There are lots of reasons, but research has shown that a lot of it comes down to personality.


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[Assistant] And the other factor is gender. It's a well-known fact that men and women do lots of things in different ways, and the workplace is no different.


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[Assistant] Did you look at the effects of this variation on companies? Yes I did.


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[Assistant] On the positive side, exposure to such diversity helps encourage creativity which is generally an asset to a company.


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[Assistant] But unfortunately, individual differences are also the root of conflict between staff, and they can lead to difficulties for management, which can sometimes be serious.


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[Assistant] Well, managing people means you not only have an understanding of your employees.


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[Assistant] In fact, for some organizations, creativity and individuality may be the last thing they want to see during working hours. Very true.


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[Assistant] Yes, but managing people isn't as easy as it looks. For example,


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[Assistant] Change in the workplace can be quite tricky, especially if there's a need to increase profit.


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[Assistant] And at times like these, managers may have to give priority to profit rather than individual staff needs. Hmm. Yes, and that creates difficult situations for people. Yes.


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[Assistant] They should be thinking not only about the strain on their staff, but take time out to think of themselves. Absolutely.


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[Assistant] Thanks to all of you for coming along today to hear about how the Robotic Float Project is helping with ocean research.


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[Assistant] Well, first of all, we'll look at what a robotic float does and its use. So let's start with the device itself.


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[Assistant] It looks a bit like a cigar, and it's about one and a half meters long.


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[Assistant] More importantly, it's full of equipment that's designed to collect data.


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[Assistant] So it can help us in building up a profile of different factors which work together within the world's oceans.


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[Assistant] That's quite true, but this project is a really good example of international cooperation.


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[Assistant] Over the last five years, scientists from 13 countries have been taking part in the project and launching floats in their area of ocean control.


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[Assistant] And next year this number will rise to 14 when Indonesia joins the project. That's impressive. But let's move on to how floats work.


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[Assistant] The operational cycle goes like this. Each of the floats is dropped in the ocean from a boat at a set point and activated from a satellite.


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[Assistant] Then the float immediately sinks about 2000 meters. That's two whole kilometers down in the water. It stays at this depth for about 10 days and is carried around by the currents which operate in the ocean at this level.


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[Assistant] During this time it's possible for it to cover quite large distances, but the average is 50 kilometres.


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[Assistant] Well, at this stage, nothing. But as it rises to the surface, it collects all sorts of data. Most importantly, variations in salinity, that's salt levels, and the changes in temperature, a bit like underwater weather balloons.


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[Assistant] Then, when it gets back to the surface, all the data it's collected is beamed up to the satellite.


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[Assistant] After about five hours on the surface, the float automatically sinks, beginning the whole process again.


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[Assistant] Well the information is transferred direct to onshore meteorological stations like our one in Hobart and within four hours the findings can be on computers and they can be mapped and analyzed.


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[Assistant] Some of the data has already helped in completing projects. For example, our understanding of the underlying causes of El Nino events is being confirmed by float data.


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[Assistant] Another way we're using float data is to help us to understand the mechanics of climate change, like global warming and ozone depletion. That's part of an ongoing variability study, but the results are still a long way off.


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[Assistant] However, this is not the case with our ocean weather forecasting.


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[Assistant] Because we know from the floats what the prevailing weather conditions will be in certain parts of the ocean, we can advise the Navy on search and rescue missions. That's happening right now, and many yachtsmen owe their lives to the success of this project. In addition, the float data can help us to look at the biological implications of ocean processes.


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[Assistant] Yes, and advising governments on fisheries legislation.


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[Assistant] We're well on the way to completing a project on this. We hope it will help to bring about more sustainable fishing practices. We'll be seeing the results of that quite soon.


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[Assistant] Yes it does. It's also a powerful agricultural tool.


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[Assistant] If we were aware of what the weather would be like, say, uh, next year, we could make sure that the farmers planted appropriate grain varieties to produce the best yield from the available rainfall.


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[Assistant] I agree that this concept is still a long way in the future, but it will come eventually, and the float data will have made a contribution.


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[Assistant] So uhm, I really love the way the camera doesn't pan or move, it just sits. And can you say something about that? The framing?


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[Assistant] Motivated you to, uhm, wish Ted Kaczynski happy birthday.


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[Assistant] She finally went in and asked to see a dress that was in the window.


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[Assistant] And decided to punish the assistant next day. She returned to the shop the following morning dressed in a fur coat.


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[Assistant] With a handbag in one hand and a long umbrella in the other.


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[Assistant] After seeking out the rude assistant, she asked for the same dress.


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[Assistant] 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.


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[Assistant] Then you pull it and you release it, it's going to be doing something. This height.


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[Assistant] From the normal position at the point x is the actual variable psi of x and t.


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[Assistant] So at every point x along the string, there is a displacement of the string in the transverse direction. That's psi of x and t.


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[Assistant] I showed you that this psi obeys the following equation.


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[Assistant] divided by mass per unit length, I'm going to change my notation into force.


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[Assistant] Because t is also used for time periods, 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 squared is f over mu. F is the tension on the string.


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[Assistant] So, how did we get this? Let me remind you how I got this. Okay?


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[Assistant] This is very easy to understand. The meaning of this equation, I claim, is just F equals MA.


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[Assistant] You realize that this is the A of the string, moving up and up. This is the acceleration.


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[Assistant] 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.


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[Assistant] Mu times dx is the mass, this is mass times acceleration. The left hand side is the force on it.


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[Assistant] Why is this the force? I tried to explain to you that if you took a piece of that string,


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[Assistant] Then the tension is pulling like that there and like this here.


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[Assistant] And the two don't cancel because if you resolve this force in the vertical horizontal part, that in the vertical horizontal part, you can see this is going to be bigger than that.


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[Assistant] So the slope itself is changing, and this is the rate of change of the slope.


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[Assistant] 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 with a positive curvature.


[User] Please transcribe this text

[Assistant] Negative curvature, the force is down. String has positive curvature, that probably means


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[Assistant] You're 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.


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[Assistant] One of you asked me a very good question in after class, which is,


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[Assistant] The question was, you took care of the vertical forces as T sine theta plus delta theta here.


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[Assistant] That's also not quite balanced, the angles are not equal.


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[Assistant] But the point was sine theta was approximated by theta for small angles, but cosine theta, you guys remember, begins as 1 minus


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[Assistant] Theta squared over 2 and so on, but we are not keeping track of things more than the first part of theta in the calculation. So cosine theta is just 1.


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[Assistant] In this approximation and tan theta and sine theta are both equal to theta.


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[Assistant] 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 theta. But the vertical forces are proportional to the first part of theta, and that's what we get. So that's the origin of the wave equation.


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[Assistant] Have a little more time, I can tell you how to get the solution to this equation. I don't have that much time.


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[Assistant] So what we were going to do was to be content with writing down a solution.


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[Assistant] Then convincing you it's a solution analyzing its properties. So solution I wrote down looked like A cosine kx


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[Assistant] Minus omega t, 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. You see that?


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[Assistant] So, if you put this here, two derivatives will pull out a k-square. Two derivatives here will pull out an ω-square.


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[Assistant] If you want the two sides to match, you will want the ω square side divided by v square to equal the k square. That means ω equal to k times v.


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[Assistant] That's the only condition. If that is satisfied, this, if this omega is k times v, you have a solution.


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[Assistant] No, I tell you it's real and positive, and two miles long, yes.


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[Assistant] Not only that, the whole calculation assumed it was a small displacement problem, right? That's why sine theta and tan theta are all equal to theta and cos theta is roughly 1. So, once you make an approximation,


[User] Please transcribe this text

[Assistant] And you get an answer, you should not blindly apply the answer to circumstances that are not valid.


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[Assistant] 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 now. Alright, so A is anything. So let's put in the result I've learned.


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[Assistant] But it turns out any function of x minus vt actually satisfies the wave equation. This is a very interesting exercise for you calculus buffs. Take f unknown function, whatever you like.


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[Assistant] 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 like.


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[Assistant] 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


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[Assistant] Take any function you like at a given time t equal to zero.


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[Assistant] It has a certain shape. Our cosine has a certain shape, but take any shape you like. A t equal to zero. Let us say it is peaked here. If you wait a little bit...


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[Assistant] If you'd like to be peaked at x equal to 6, for example, if you wait a little bit, this term minus vt, as subtracting it from it, you've 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.


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[Assistant] In other words, if Vt becomes non-zero, x should become non-zero by the same amount if you want to write the crest of that wave. That means the signal is moving to the right.


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[Assistant] So it turns out, you can have x minus bt or x plus bt. They're both solutions of 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.


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[Assistant] Let's ask the following question. What is k and what is omega? So let's take this function psi of x and t.


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[Assistant] 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.


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[Assistant] Every point on the string has a variable psi at that point.


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[Assistant] I look at this function and I say let me at least understand this function at t equal to zero. At t equal to zero,


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[Assistant] Drop that term. In other words, this is a snapshot of the function. What does it look like?


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[Assistant] If I take a picture of t equal to 0, then I vary space. Well, that's cosine kx, and you can all plot cosine kx.


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[Assistant] This is going to start out at A, then it's going to oscillate because that's what the cosine does.


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[Assistant] Now, this distance from one peak to the next peak, we call the wavelength.


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[Assistant] It's not 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.


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[Assistant] What is the relation of k to λ? Now you gotta ask yourself. When x was zero, this angle was zero.


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[Assistant] I've got to increase x till it's come back to what's essentially zero. For a cosine, you're back to essentially zero if this angle is 2 pi.


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[Assistant] Can also be 4π or 6π, but the first time you're back to 0 is when it's 2π.


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[Assistant] Therefore, the length lambda up there has a property that k times lambda is 2 pi, because if I move to the right by an amount lambda,


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[Assistant] The angle and the cosine are changed by two points. So this number called k,


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[Assistant] Now you can say, well, you pick time t equal to zero, maybe if you pick time t equal to one second.


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[Assistant] You would have gotten a different picture. Think about it. If I put t equal to one second, as long as it's a fixed number, that's some fixed angle phi. That'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 lambda.


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[Assistant] 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


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[Assistant] With a dynamic picture by saying, if you know very time, imagine this sliding by.


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[Assistant] 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.


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[Assistant] What does it look like to a person sitting at a certain location as a function of time?


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[Assistant] 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 waves are going to go past you.


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[Assistant] So you will bob up and down, and that's what I want to understand. So I pick some location, I pick x equal to 0 for convenience, then psi,


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[Assistant] At x equal to zero as a function of time, looks like A cosine omega t.


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[Assistant] I didn't miss the minus sign. For a cosine, it doesn't matter if you change the angle to minus the angle.


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[Assistant] 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


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[Assistant] This is a picture at one point where the water or the string goes up and down.


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[Assistant] So we like to call this, from one maximum to the next maximum, the time period. That's the time over which the wave repeats itself.


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[Assistant] Well, it follows then that omega times the time period is 2 pi. Because when I put little t equal to time period, I've gone from this maximum to that maximum.


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[Assistant] K and omega, which we put into the wave, are related to the time period and the wavelength through this reciprocal formula.


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[Assistant] 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...


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[Assistant] Add to little t an amount of time big t, nothing happens because 2π's don't matter. So this makes it very clear the periodicity in space and time. But this is more compact because we don't want to carry 2π's and λ's in this complicated form. We give a new name to 2π over λ and 2π over t.


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[Assistant] What about the velocity of the wave? Remember, it satisfied this condition. It's ω over k.


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[Assistant] But now let's write this in terms of this new guys. 2π over t divided by 2π over λ is λ over t.


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[Assistant] So the velocity of this, this wave is called a sine wave or a cosine wave or a plane wave. You gotta understand.


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[Assistant] Waves in water and waves in string always have a velocity. If you plug the string and let it go, a little blip will travel from left to right.


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[Assistant] But not every wave has a wavelength or a time period. This is a special function that has periodicity in x and t.


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[Assistant] In other words, if you go to a lake and drop a rock,


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[Assistant] It'll set out some waves and after a while they'll spread out. It's not going to happen over and over and over again.


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[Assistant] But if you want a wave to be present for all time, for all space, you should keep on agitating the water.


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[Assistant] 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.


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[Assistant] 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.


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[Assistant] For that simple situation, the velocity is connected to the wavelength and time period in this fashion, or you may also like to write it as lambda times the frequency.


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[Assistant] 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 the λ over t is in fact simply the velocity. Here's another way to understand this.


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[Assistant] Suppose I have a string that goes on to infinity and I'm catching it at this end and I start


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[Assistant] Wiggling the string. When I start wiggling this, these pulses will travel. Let us say they come up to this point.


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[Assistant] In one more second, I would have manufactured a few more of these pulses, and the wave would look like that. How many pulses would I make per second? F. How long is each one?


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[Assistant] Okay, so now I consider another aspect of these waves, and that's the energy in the wave.


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[Assistant] First of all, you've got to understand a vibrating string has some energy compared to a string which is not vibrating.


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[Assistant] Okay, you have an intuitive feeling. The vibrations are a cause of energy.


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[Assistant] I want to calculate now the energy in a string vibrating according to this formula here.


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[Assistant] If it's an infinitely long string, the energy in it is infinite.


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[Assistant] 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.


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[Assistant] 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


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[Assistant] Far to the right, we don't care how far. Take a portion of the string.


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[Assistant] You 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


[User] Please transcribe this text

[Assistant] It's just undergoing simple harmonic motion. That's pretty obvious from this.


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[Assistant] So we just need to write down the energy. So how do you write the energy? The energy


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[Assistant] Is going to be either kinetic or potential, or the sum of the two.


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[Assistant] Now you can find the energy whenever you like because it's a constant. Let's find it when the velocity is a maximum.


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[Assistant] Like for a harmonic oscillator, it's like finding the velocity when the mass is just swinging past the normal position x equal to zero.


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[Assistant] Mu dx is the mass of the segment I'm looking at. What's the maximum velocity?


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[Assistant] You guys remember from harmonic oscillators, what's the maximum velocity that a harmonic oscillator has, whose motion is given by this?


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[Assistant] Uh, omega. This is getting a Yale education, so that's the difference between $40,000 and $10,000 is that this is not W, but omega.


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[Assistant] By the way, I could not get any of my kids to say omega. They keep saying w. I'm trying to have more success with this class than with my own offspring.


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[Assistant] The Greek alphabet is essential to communication. So, this is Aω as the maximum velocity, and I've got to square that.


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[Assistant] Now this is the energy of a strict segment of length dx, so what we normally define is a quantity called little u, and that is the energy per unit length.


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[Assistant] That means you just divide by dx, and you get 1 half mu a square omega square.


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[Assistant] It is just the kinetic energy of a segment of unit length.


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[Assistant] It's the, it's the total energy, but I chose to compute it when it's all kinetic.


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[Assistant] Mu is the mass per unit length. You understand? You take the string, you find its mass, and you divide by length. That's so much mass per centimeter or per meter. That's mu.


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[Assistant] That's what, if you go back to the derivation of the wave equation, we were thinking in terms of mu.


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[Assistant] But let's say it has 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.


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[Assistant] 1 half mu a square omega square times the velocity.


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[Assistant] That is the power I have to give in to produce this wave continuously. Do you guys follow that? If I wait one second.


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[Assistant] A segment of length V has now started moving in the front end. The wave is being produced by me and sent out into the medium. The distance it travels in one second is V.


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[Assistant] 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.


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[Assistant] A special property of life in one dimension is that this wave goes undiminished in amplitude.


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[Assistant] It can go 10 miles or 100 miles from me. Amplitude is still the A that I produced here.


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[Assistant] That's because all the energy goes along this line. That's sort of atypical. What typically happens is,


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[Assistant] When you live in, when you take into account vibrations of say sound, in two or three dimensions,


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[Assistant] If you have a tower on top of which you put a speaker that's making noise,


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[Assistant] The energy radiates out in concentric circles.


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[Assistant] And the power at the source is now spread over bigger and bigger spheres.


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[Assistant] So in three dimensions, we have the notion of intensity, which is the power per unit area.


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[Assistant] What I'm saying is, if you've got a speaker that is sending out sound waves, I take a one meter by one meter window and hold in front of me and ask how much power crossed this in one second. That's


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[Assistant] Intensity. You don't have the notion in one dimension, because you cannot hold a window in one dimension. Okay? Three dimensions, you take a little square, one by one meter, and you define intensity to be power per unit area.


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[Assistant] Of course, it doesn't have to be one meter big. You can take a tiny square, provided divided by the area of the tiny square. That is this.


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[Assistant] 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


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[Assistant] 4 by r squared, where r is the radius of the sphere.


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[Assistant] And you are a distance R from the center. And the power is going out uniformly in all directions. In your share,


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[Assistant] For the whole sphere was p, the share for the unit area on the sphere is p divided by 4 pi r squared.


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[Assistant] And that's measured in watts per meter squared.


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[Assistant] This one? Uh, what else do you think it should have? T for time period? Why?


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[Assistant] It's energy per length, but if I wait one second, the wave has gone a distance b times one second.


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[Assistant] And so the extra energy I give in every second has to pay for V more meters.


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[Assistant] Yeah, 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.


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[Assistant] Okay, so this is called the intensity. And for sound, we have another way to measure intensity. We define something called β.


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[Assistant] Which is 10 times the logarithm of the intensity divided by a standard intensity.


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[Assistant] The standard intensity has a value 10 to the minus 12 watts per meter squared.


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[Assistant] So you measure the loudness of a sound in decibels.


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[Assistant] You must have heard that everywhere you look at your, even your amplifier will have so many dB's in it. So what's the dB talking about?


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[Assistant] The DB says, your speaker's making some, uh, 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.


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[Assistant] The power coming to me was say 10 to the minus 11 watts per meter squared. Then I over I naught will be 10 to the minus 11 divided by 10 to the minus 12, which will be 10 to the 1. And log of 10 to the 1 is just 1, and beta will be 10 dB.


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[Assistant] 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 squared.


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[Assistant] Beta 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


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[Assistant] Thousand times bigger, the decibel goes from 10 to 40.


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[Assistant] 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.


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[Assistant] It grows logarithmically. So every time I add an extra 0 to the power, you go from 10 to 100, 100 to 1000, decibel goes up just by from 10 to 20 to 30 to 40.


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[Assistant] And I believe people use decibels because the human ear is able to take a whole range of intensity and hear it.


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[Assistant] 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, but the intensity ranges over several powers of ten.


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[Assistant] 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, you deal with the logarithm of the number, then it's more manageable. So you can imagine trivial problems involving finding beta or given beta, finding the intensity and so on.


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[Assistant] I'm finished with the basic properties of waves. Decibels is a property just for sound. You don't apply that to light.


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[Assistant] And the whole standard intensity, I0. Intensity is well defined for everything, even electromagnetic waves of intensity.


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[Assistant] Okay, now for another totally different property of wave propagation, which I'm going to again apply to sound, is the familiar Doppler effect.


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[Assistant] 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,


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[Assistant] 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.


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[Assistant] So to proceed with this, I want to remind you this relation λ times f is v.


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[Assistant] 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.


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[Assistant] Now suppose the source is moving to the right at a certain speed u. u is the speed of the source.


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[Assistant] 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 the forward direction like this.


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[Assistant] So the waves will get squashed because the spacing would 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.


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[Assistant] 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.


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[Assistant] T is 1 over n. So, what is the new frequency that I will hear?


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[Assistant] The new frequency I will heal with the velocity divided by the new wavelength.


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[Assistant] 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.


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[Assistant] 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. The medium is air. It can travel only at that speed.


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[Assistant] So you don't have a new velocity when the thing begins to move. 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.


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[Assistant] So we multiply top and bottom by F. So we write it as F times something, which is V divided by λF minus U, and we fiddle with that and write it as V divided by V minus U, because λF is V.


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[Assistant] 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 got this, the rest of it is just algebra.


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[Assistant] The point is to realize that the waves get crunched because when you emit a crest,


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[Assistant] And you rush towards that crest and emit the second crest, you reduce 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.


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[Assistant] The frequency is the normal frequency divided by 1 minus something.


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[Assistant] So 1 minus something is less than 1, so the frequency will go up. If you want to see what happens here,


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[Assistant] 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 one plus something. So we'll write both solutions as minus or plus.


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[Assistant] In one case the frequency will be increased, other case the frequency will be decreased.


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[Assistant] 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, use the plus sign. But you should know intuitively at least what sign to pick.


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[Assistant] You gotta know if the source is coming towards you, the frequency will go up. 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.


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[Assistant] Ah, relativistic speeds, a lot of things happen. That's correct. When you handle light, you've 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


[User] Please transcribe this text

[Assistant] 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 will slow down. So relativistic problems are more complicated. Yes?


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[Assistant] Okay, sonic boom comes in, in the following thing. f' is f divided by 1 minus u over b. Let me embellish the formula by writing u sub s, where s means source.


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[Assistant] 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.


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[Assistant] So that's when, imagine the waves getting crying still, 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,


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[Assistant] The way the waves emerge from you is so that you leave behind a trail. There's nothing in front of you.


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[Assistant] 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 will hit you before the sound waves hit you.


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[Assistant] Okay, so sound is not the absolute limit, so there are possibilities of, I mean, every time you go in a concert, well, not anymore, but when you did,


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[Assistant] You would be crossing this thing pretty soon after takeoff, because you go several times faster than the speed of sound.


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[Assistant] Okay, the last Doppler thing I want to do is the opposite one where


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[Assistant] The siren is emitting nice spherical waves, but you are rushing to meet the siren. You are going with the speed u sub-observer. Then what happens?


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[Assistant] 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 will see them at the normal frequency. But now, you are, these waves are spaced at distance λ apart.


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[Assistant] But your velocity relative to sound is now u0 plus v.


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[Assistant] Your relative velocity. Sound is moving to the right, you are moving to the left. So with respect to the stress, you are zipping at a speed u0 plus v.


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[Assistant] And the crests are spaced a distance lambda apart now. There's nothing wrong with the spacing between them. They're just like shown here. You're just rushing to meet them now from your side. So how many will you intercept in a second is this. This is the relative speed and that's lambda.


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[Assistant] So when u moves towards the source, the correction factor appears in the numerator. When the source moves towards u, correction factor appears in the denominator.


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[Assistant] You gotta use common sense to see is it plus, is it 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.


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[Assistant] 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.


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[Assistant] It'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.


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[Assistant] One essential property of waves. So the whole time I'm going to do, this is one important property of waves that has to do with interference of waves.


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[Assistant] Look at the wave equation I wrote down. It's a linear equation. You see that? Don't be fooled by this d2ψ. That doesn't make it quadratic. The first power of ψ on both sides. Then, do the following exercise in your mind. ψ1 is a solution, ψ2 is a solution.


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[Assistant] Add the two equations, you can check that psi 1 plus psi 2 is also a solution.


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[Assistant] 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.


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[Assistant] Then I turn you off and I turn on another speaker here. Let's say the speaker makes a sound.


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[Assistant] If the 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.


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[Assistant] So if one cause produces one effect, the second cause produces the second effect, then both are turned on, the wave they produce will simply be the sum. You can simply add them, because that respects the, satisfies the wave equation.


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[Assistant] So all I'm going to do from now till end is take a variety of situations where I'm going to be adding two waves. So the simplest problem is the following. Pick a certain spot. You just sit there and listen to two waves.


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[Assistant] 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 ψ1 equal to a cos ω1t.


[User] Please transcribe this text

[Assistant] That's the second source of sound, which is A cos ω2t.


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[Assistant] I've chosen the two to have the same amplitude, but not necessarily the same frequency.


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[Assistant] Now you can do the trigonometric identities and figure out what had happened, but you have to think a little bit about what's going to happen. Think about these two waves. If they had the same frequency,


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[Assistant] It's trivial, right? If ω1 is equal to ω2, ψ is equal to 2a cos ωt, but ω is a common frequency.


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[Assistant] That just means they reinforce each other at every instant. You get double what you got before. But if the frequencies are not equal,


[User] Please transcribe this text

[Assistant] Then, initially they are in step. This cosine is 1, and that cosine is 1, and the amplitude is 2a. As time goes by, ω1t and ω2t start differing.


[User] Please transcribe this text

[Assistant] So the cosines are not in step anymore, and what can happen after a while is, the two cosines are offset by half a cycle, or the angle between them is pi. Then in fact they will cancel. Then if you wait long enough, again they will be in step.


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[Assistant] 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, one starts lagging.


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[Assistant] Behind the other and they will be on different parts of the circle and the time will come when they're on opposite parts of the circle.


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[Assistant] When the opposite part of the circle, the angular difference between them is pi, and that's when we say they cancel. If you wait long enough, they'll again line up.


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[Assistant] But with one difference, one guy would have done one more revolution than the other, then again they would be in step and out of step. That's what you expect the formula to give you. So you've got to go back and use the trigonometric formula, and the answer is 2A cosine


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[Assistant] This says cos A minus cos B is 2 cos A plus B over 2 cos A minus B over 2.


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[Assistant] Alright, so we are interested in the case where ω1 is very nearly equal to ω2.


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[Assistant] Typical situation, you have a piano tuner, piano tuner hits the fork, it vibrates some frequency, maybe 440 Hz, and then your piano slightly off, maybe at 439. What sound do you hear when you put them together? Well, let's look at this. If ω1-ω2 is a very small number,


[User] Please transcribe this text

[Assistant] This cosine here is changing very, very slowly.


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[Assistant] This cosine is oscillating at roughly the, it's oscillating at the average of the two frequencies. Say 1 was 401.


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[Assistant] And 1 was 399. This is oscillating at 400, but here, ω1-ω2 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.


[User] Please transcribe this text

[Assistant] We have done it the other day when I wrote e to the minus gamma t over 2 times some cosine omega t. 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.


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[Assistant] 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 will have oscillations like this.


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[Assistant] So, it'll be loud, silent, loud, silent, loud, silent. So you will hear another


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[Assistant] Beat, this is called the 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.


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[Assistant] It's ω1 minus ω2 over 2, because that's what seems to appear here.


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[Assistant] But actually the correct answer is the beat frequency is simply omega 1 minus omega 2, and I will tell you why.


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[Assistant] At t equal to zero, when the cosine had a value one, you got an amplitude 2a.


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[Assistant] Wait till omega 1 minus omega 2 over 2.


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[Assistant] It's not equal to 2π, but equal to π. I mean, times t.


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[Assistant] If this whole combination is equal to π, this number becomes a negative number of size 2a.


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[Assistant] 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.


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[Assistant] In other words, 5 cosine omega t and minus 5 cosine omega t have the same amplitude.


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[Assistant] The minus doesn't change the fact that it's having large excursions from plus to minus 5.


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[Assistant] So in other words, I'm telling you to compare some answer like this. Cos 400T minus 5 cos 400T.


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[Assistant] Plot this and plot this, you'll find if 400T is very rapid oscillations, this has an amplitude as big as this one.


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[Assistant] And the ear is just picking up the amplitude. The ear is not picking up anything else.


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[Assistant] Here it's just picking up the loudness of the sound, so it will seem to be loud now and loud there.


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[Assistant] But once you got from here to here, you only had to change the argument of this whole thing by π and not by 2π.


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[Assistant] That's why the beat frequency then happens to be simply the difference in frequencies.


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[Assistant] Or I ask you to do another thing if you're not happy with this. Go home and do the following thing.


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[Assistant] One is vibrating at omega 1t, other is vibrating at omega 2t.


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[Assistant] Find out when the difference between the phase angles is equal to pi. Because that's when this guy and that guy will cancel each other. That's when you come here.


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[Assistant] Then you ask yourself when they will 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.


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[Assistant] So whether you, you can, if you found this factor of too confusing, you've got to understand the argument I gave you.


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[Assistant] That in order for this thing to sound just as loud as it did at t equal to 0, it's not necessary that this number come to 2a. It's good enough if it comes to minus 2a.


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[Assistant] If they are not quite masked, 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.


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[Assistant] So beat frequencies have a lot of interesting applications in physics.


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[Assistant] 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 the frequency of the sound. That's the frequency of the signal called the carrier wave. Then you modulate with that, with something, and you produce some beats. It's the beats that we are able to hear.