18. Simple Harmonic Motion (cont.) and Introduction to Waves

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 Title 18. Simple Harmonic Motion (cont.) and Introduction to Waves Title of Series Fundamentals of Physics I Part Number 18 Number of Parts 24 Author License CC Attribution - NonCommercial - ShareAlike 3.0 Unported:You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this license. Identifiers 10.5446/46683 (DOI) Publisher Release Date 2006 Language English

 Subject Area Physics Abstract This lecture continues the topic of harmonic motions. Problems are introduced and solved to explore various aspects of oscillation. The second half of the lecture is an introduction to the nature and behavior of waves. Both longitudinal and transverse waves are defined and explained. 00:00 - Chapter 1. Free Vibration: Oscillation Under F=0 08:20 - Chapter 2. Initial Conditions at Start of Oscillation 18:52 - Chapter 3. Solution to Harmonic Equation under Driving Force 30:01 - Chapter 4. Properties of the Oscillating Function, Resonance 39:23 - Chapter 5. Complete Solution = Complimentary + Particular Solutions 43:40 - Chapter 6. Introduction to Longitudinal and Transverse Waves 52:55 - Chapter 7. Derive Wave Equation as Differential Equation 01:04:40 - Chapter 8. Solution to Wave Equation

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Okay, so what problem were we trying to solve? We were trying to solve the problem of masses, coupled to springs, and acted upon by various forces. And the equation that we need is the following [writes equation on the board]. So, this is the most general problem we looked at. This is F 0. I might have just called it F last time. It doesn't matter; it's a matter of this symbols, okay? Alright. Let us first divide everything by m. Divide everything by m, then it becomes x double dot, plus ?x dot, plus ? 0^(2)x, equals F 0 over m, cos ?t. I'm using the fact that ? 0 was defined as the root of k over n. We used to call it ?, but now I'm calling it ? 0, and you should understand why I'm doing that, because there are now two ?s in the problem. There is, the ? of the driving force, so you should imagine an oscillatory force grabbing this mass and moving it back and forth. ? 0 is the natural frequency of the mass. If you pull the spring and let it go, that's the frequency that it will vibrate. There are really the natural frequency that is fixed by k and m and ? is for you to decide. You can decide to apply to it any force you like of any frequency. Professor Ramamurti Shankar: So first, take the easy case in which the right-hand side is 0. No driving force. So, I'm saying let's consider a simple case, F 0, or F, the applied force vanishes. That's the problem I spent a considerable amount of time on yesterday. You can say, "Well, if there's no driving force, why does anything move?" That's actually correct. If you took a mass in a spring and didn't do anything to it, the solution of this equation is x = 0, and that is fine. But it also has interesting non-trivial solutions because you can pull the mass and then let it go. After you let it go, there are no forces on the system except the force of friction and the force of the spring, and that is this problem. Now, how do you solve this problem? I told you the way you solve this problem is you make a guess: x of t is some number, A times e to the ?t. See if it'll work. And you find it will work and you can do this in your head. I certainly don't want to repeat all of that. The reason it's going to work is that every time you take derivative of x, you bring down an ?, and still keep e to the ?t. So, when you put that guess into this equation, you're going to get ?^(2) plus ??, plus ? 0 square times A, e to the ?t equal to 0. And if this is going to vanish, it's not because of A and it's not because of the ?t, it's because this guy in brackets is going to vanish. So, when that's 0, you all recognize this to be a quadratic equation. You go back to your days in the sand box [earlier in the course or your life] when you all knew what the answer to this is; the answer to this is minus ? over two, plus or minus ? over two square, minus ? 0 square. These are the roots. Now here, you have to look at--So anyway, what does this mean? This means that yes, you can find a solution to this equation that looks like A to the ?t; A can be any number you like. No restriction is placed on A. But ? cannot be any number you like. ? has to be the solution to this quadratic equation and there are two solutions. So, you can call them ? plus and ? minus. That's when I branched off into two different cases. Case one, you can see look, many of the quadratic equations, you know the way it's going to fall out is depending on whether the stuff inside the bracket is positive or negative. So, let's take the easier case when it's positive, that means ? over two is bigger than ? 0n. ?, you remember, is the force of friction so we want it to be not only non-zero, but bigger than this amount controlled by the natural frequency. In that case, let's not go into details, but you can all see you put whatever value you have for ?, the square root of this is some number. The whole thing is going to be two real choices you get. One will be minus ? over 2, plus whatever is inside the square root. The other will be minus ? over 2, minus what you have in the square root. The only thing I want you to notice is that both these roots are negative. This guy's obviously negative because it's a minus ? over 2 minus something else, that's definitely negative. You have a little [inaudible] worried [inaudible] here, that this is minus ? over two and this is plus something. But the plus something is smaller than the minus something, because what's inside the square root is ? over 2 square. If you took the square root of that you'll get exactly ? over 2. But you're taking away from it, this number; therefore, this square root is smaller than this number, so the overall thing will still come out negative. So, why am I so eager that it should come out negative? Why am I eager that the root should be negative? Yes? Student: [inaudible] Professor Ramamurti Shankar: No, even if I had positive, if I took two derivatives, it would work out. Yes? Student: Because if alpha is greater than one, then the e won't converge to cosine. Professor Ramamurti Shankar: Won't converge to what? Student: The cosine function? Professor Ramamurti Shankar: This is not a