 Alright, so we know that the expression is not constant in simple harmonic motion, right? So we cannot use those equations of motion that we have learned in the kinematics because those equations of motion v is equal to u plus a t, s equal to u 2 plus half a t square and the third one, all of them were derived by assuming expression is a constant, right? So we can never use those equations in the case of any oscillation, okay? And we are learning a special kind of oscillation, which is simple harmonic motion, right? So we'll start the class with a couple of examples on what we did last class. And before I do that, before I take up the examples, we'll list down the equations of motion. So if you type in what are the equations of motions you have, all of you, what, which equation of motion you remember, type in, okay, Rubau, we cannot use that. That equation is valid only for the constant acceleration, alright, okay? So the first equation is this, a is equal to minus of omega square x, alright? Then the next one, what else? Anything else? Mudit, this is your first class, first session you were joining, Mudit? Yes, sir. So you are aware of simple harmonic motion? Anything? No, sir. No? So you are joining in the middle of a chapter, okay? Yes, sir. So it may be half an hour, 15 minutes to get in the groove, right? Now, since you are in middle, it might be a little difficult, but then you can make an effort. It will be not so difficult also. Yes. Alright, so, fine. So we have another one, Siddhas is mentioning this, v is equal to omega root over a square minus x square, right? So first equation, first equation of motion is between acceleration and x. What is x, everyone? What is x? Displacement, no. What is x? Anyone? It is the distance from mean position, okay? It is a distance from mean position. And when you measure the distance, one side can be positive and other side will be negative. You can do that. Displacement, you know, requires two points. You have to say from where. So this is from the mean position. So v is equal to omega root over a square minus x square. And by the way, do you know what is the name of omega? Omega is what? Write down, omega is angular frequency. Now any other equation, other than these two, okay, Aditi has written 1x is equal to A sin omega t plus phi. Now tell me what is A, what is it? Capital A is amplitude, correct? And how you measure it physically, what it is? How will you identify the amplitude when an oscillation is going on? Physically what it is? Maximum distance from the mean position, okay? Maximum distance from the mean position. What is the condition for a mean position? How you know where is the mean position? Condition for it is what? Acceleration is 0, right? You can see A is equal to omega square x. At mean position, x is 0. So at mean position, at mean position, acceleration is 0. So force will be what? Force is also 0, okay? In which direction along the SHM direction, right? So we have seen all of this last class, I am quickly summarizing. Any other equation? First one is between acceleration and displacement. Second one is between velocity and displacement. What is x versus t? Next, you can differentiate x to get velocity. So v is equal to omega A cos of omega t plus phi, okay? And you can differentiate it again, you will get A is equal to minus omega square A sine of omega t plus phi. So how many equations of motions are there? One, two, three, four, five. You already have five equations of motion for SHM. Do not use those three, all right? So let's take a few questions on these equations. How we can use these equations to analyze the simple harmonic motion. And by the way, what is phi before I proceed with a question, phi is what? Phi is what? We have derived it last class, tan inverse x by A. Do you remember this? Do you remember it? Look at your notes, all right, fine. So first of all, if equation of motion is this. x equals to A sine omega t plus phi. You need to tell me what is the time period of the oscillation, time period of oscillation is how much, how will you calculate it? Find out, wait, sorry, it was sine inverse, it is corrected, it was sine inverse, okay? That is what we derived last class. Time period of oscillation, all of you find out, yeah, yeah, sine inverse. Anyone close, no one, okay, all of you. So if capital T is the time of oscillation or time period, okay, then whatever was happening, and it means whatever was happening at T equals to T, same will happen at T equals to T plus capital T. Yes or no? Oscillation should repeat after every fixed interval time, and that is what the time period is. So we need to find out what is this capital T. All of you agree to this? Everyone type in, right? So if X is equal to A sine of omega t plus phi, after T plus capital T, it should be the same. This should be equal, also equal to omega t plus capital T plus phi, both should be equal because after capital T it will come back to the same thing. This is equal to A sine omega t plus phi plus omega capital T. This should be equal to A sine omega t plus phi. So what should be the value of omega capital T? So that both are equal. One value is omega into capital T should be equal to zero. What is the other value for which these two are equal? So what should be the value of omega into capital T should be equal to, everyone, you know, right, how is the sine curve? Sine curve is this. This is the sine curve, right? This is the sine curve, all right? So you are, let's say you're talking about displacement X. This is X and this is T, all right? So same thing will happen at this point and then at that point, okay? Now you will say that, no, no, there is a point, you know, there is this point, same thing is happening here also, okay? You can say that it is not 2 pi, but then you will take this point, then you have to go away a distance of 2 pi only. This distance you have to take, okay? If this is the angle, okay? This is your angle you are plotting. Then after every 2 pi angle, sine function will repeat itself, okay? This is zero, this is pi, this is 2 pi, this is 3 pi, okay? So when omega T becomes 2 pi, same thing will repeat for every point. I don't, I'm not concerned with only one of the points. All the points, whatever is the value of small t and phi, same thing should happen and that will happen when omega T is equal to 2 pi. So time period of the oscillation is 2 pi by omega. This is a time period and from here, frequency of oscillation, which is 1 by time period is omega by 2 pi. All of you understand this? If you understand, right? So if you know what is omega, you can find out the time period, 2 pi by omega and omega by 2 pi is the frequency, okay? Now, take few more questions. So like I told you last class, this chapter is divided into two parts. One part deals with the kinematics of simple harmonic motion. Other part deals with the laws of simple harmonic motion. So right now we are focusing on the kinematics part of it, okay? So let's say this is the mean position, okay? Amplitude is A, fine? So at mean position, let's say velocity is U. Velocity is U. You need to find out the distance from the mean position, how much it will be for point which has the velocity equals to half of the mean positions velocity. All of you do it. These are very direct type of questions. Little bit of carefulness if you show, you will get the answer. Equations of motions I have already written. No one? Which equation of motion will you use here? Which one will you use here? All of you tell me. These are the equations of motion. Which one will you use? I am asking you that where is the location at which the velocity becomes half of the mean position? At which the velocity becomes half of whatever is its velocity at the mean position? Which equation will you use? Tell me. Second one, there is no doubt about it, right? Second one only, isn't it? This one will use, alright? So let me write that equation since you have identified it. P should be equal to omega root over a square minus x square. Can you tell me what is the velocity at the mean position with respect to this formula? Using this formula, can you tell me at mean position what will be the velocity? Everyone, at mean position, correct, at mean position x equal to 0. So v is equal to omega a, v is equal to omega a. So we need to find out a point which has velocity, how much? Which has a velocity of omega a by 2, okay? Half of the mean position velocity. So just substitute v as omega a by 2 and you will get the answer. Can you tell me quickly what is the answer? Omega is gone. You could have watched the video, recorded video of the previous class. So after the class, ping, push power me, I'll forward you the previous class recording, okay? Okay. So x square is equal to 3a square by 4. So x is root 3 by 2 times a. Is this the correct answer? I mean, can there be more than one point? Everyone? Is this the only point possible? Can I say plus minus? Other direction also? No. If you talk about at this point, there will be one velocity going this way and other velocity going that way. At every point, you have both the directions of velocity while going away and while coming back, okay? But the point is there will be two such points. x equal to root 3a by 2 and x equal to minus root 3a by 2. Because the SHM is symmetrical about the mean position. Whatever that whatever happens on the one side of SHM, same thing happened on the other side of the SHM also, okay? Okay. So I hope it was a straightforward thing. So this one, you don't have to solve anything. Just describe the motion a, b, c, d, these are the forces acting on a mass. So can you tell me what happens if this kind of force acts on a mass? Anyone? You don't have to find anything. When this kind of force 3x plus 3, will it do SHM? Will it do SHM? If I apply that kind of force, what will happen? This force if I apply, let us visualize, you know, there is a mass like this, okay? It is kept at let's say x equal to 0. At x equal to 0, what is the force acting on this block? How much is the force? 3 Newton? 3 Newton, so 3 Newton is acting on it. What will happen to this block? This blocks, this block will move forward, right? This block moves forward because of the force. Now after x will become, let's say here x was 0. After it moves forward, x will become equal to 1. So now what is the force? What is the force now? 6 Newton? So do you guys see that as it is getting displaced on the right hand side, the force is getting increased more and more, right? So will it perform SHM? Will it ever come back? Will it ever come back? No, right? So no SHM, okay? For A. Now similarly, I want you to analyze B also. Everyone, do it. Is B SHM B? Is this SHM equation? Will it do SHM? No. At x equal to 0, what is the force? What is the force at x equal to 0? Minus 3. So the force will act on this side. Force acts on this side, okay? Then x becomes, let's say minus 1. What is the force now? When x equal to minus 1. Now what is the force? 0. 0, right? Right now it is 3 Newton this way. Now the force is 0. And then it further displaces to x equal to minus 2. Because acceleration is 0, velocity is not 0. Velocity is not 0, acceleration is 0. So let's say you go to x equal to minus 2. What is the force now? At x equal to minus 2, 3 Newton. So you'll get a force. Force will change its direction or not? Force will change its direction. Earlier it was minus 3, now it is plus 3. So 3 Newton on the other side. So you can see that it is, when it is going from here to there, it was decelerating, okay? So now the object will come back, alright? And when it goes this way, x will become positive. Force will become negative. So it will get pushed this way. When it goes this way, it get pushed that way. And when it goes that way, it get pushed this way. So every time it is getting pushed towards the mean position. And that is why here some sort of oscillation is going on. All of you are able to understand this. Everyone is able to understand, okay? So can you tell me where is the mean position? What is the value of x where the mean position is? Mean position is what? x is equal to what? x is equal to 0, that is the mean position, okay? So minus 3x minus 3, this is the force. x is equal to 0, so force will also be 0. So at x equal to minus 1 meters. This is the mean position, alright? This x and the x in the equation of motion, they are different, alright? This is x coordinate. Remember that this is x coordinate. And that x which you have in this equation, that x is a distance from the mean position. At x equal to minus 1, this value of x is 0. I hope you are able to understand this. They both are x only, but this is the distance from mean position, this is the coordinate, alright? So there is a simple way also to identify whether it is an SHM or not, you have to just see whether the force x coordinate, they are of opposite sign, okay? So if f is equal to minus cx plus k, where c and k are constant, then it will always be an SHM, okay? If f is minus cx plus k. So we have discussed it last class, right? That if expression is of opposite sign of x, it will be an SHM. If expression is opposite sign of x, the force is also opposite sign of x because force is mass and expression, okay? So now can you tell me is c SHM or not? You don't need to analyze it now. Is c SHM, everyone? If it is SHM, what is the mean position? It is an SHM because coefficient of x is negative and coefficient of force is positive, opposite sign. SHM mean position at x equal to 1, where 3x plus 3 is 0. What about d? Is d SHM? No, d is not a SHM, okay? Do this, do this. What happened? Where are you guys going? Area and punch, where are you going? All of you do this. Equation of SHM is this. X is equal to a sin omega t plus phi. Basically, I want you to find out what is the value of phi for this? Phi is how much? It is at a by 2. When t is equal to 0, it is at a by 2. Everyone, what is phi? If you are getting phi by 6 or 30 degrees, that is not correct. See, you need to understand one thing that phi is equal to 0 when? When it starts from here and go towards it, right? Phi is 0, right? When it reaches here, what is the value of phi when it is at a? Sin inverse x by a is what? When you go to a, at a, what is a phase? At a, what is sin inverse x by a? Phi by 2 at x equal to a, alright? So, this position is how? This position is that particle goes to a and then it is trying to come back. So, it has to be greater than pi by 2, right? So, if I tell you that particle is going like this, then your answer is correct. But it has gone like this and it is returning back. Are you getting it? So, it is not pi by 6. So, now tell me what it is. Don't treat it like a mathematics, okay? It is physics also. Do you all understand why it is not pi by 6? Everyone understood why it is not pi by 6? Okay, so tell me what it is? Phi belongs to, please remember, 0 to 2 pi. Phi cannot be anything but this. I don't know whether, I mean, you guys should come with whatever we have done last class. It is sad that you guys don't even look at what we have done, okay? We have discussed all of this. 0 to 2 pi, the angle will change. All right, so it has to be greater than pi by 2, all right? This is 0 angle. This is pi by 2 angle, okay? And then at this location, we need to find out what is the value of the phase 5. So, let us do one thing. At t equal to 0, x is equal to what? x equal to a, see this equation should be valid no matter what. So, a by 2, x is a by 2, this is equal to a sin of phi, okay? So, sin of phi is equal to half. So, according to this, the value of phi should be equal to what? Pi by 6? Can it have any other value? Can it have any other value? Can it be pi minus pi by 6 also? Can it? Do you guys know that sin of 30 and sin of 150, both are equal? Do you guys know this? So, when I say sin of phi is half, there are multiple angles for which sin of phi is equal to half. There are multiple angles, right? And if you talk about angles only between 0 and 2 pi, there are two angles. One is pi by 6, other is 5 pi by 6, okay? All of you understand this? So, one of them is correct answer here. How will you find out which one is correct? The answer to that is we need to write velocity also, differentiate x, you will get v is equal to a omega cos of omega t plus phi, okay? Now, put t equal to 0, v is equal to a omega cos of phi. This velocity, this is the velocity at t equal to 0. The velocity should be positive or negative. v should be greater than 0 or less than 0, everyone. All of you, this v should be less than 0 or greater than 0 in this scenario, in this scenario at t equal to 0. Less than 0. So, I have two values. You need to substitute these values and check which one will give you negative value of velocity. So, it is obviously that 5 pi by 6, when you put here, we will become negative. So, if you just look at x, there will be two values of angle while going and while coming back. But you have to look at the velocity also and its direction to find out which one of these two is the correct answer. So, phi is equal to 5 pi by 6, okay? This is the way mathematically you solve it, all right? Physically, you can solve it a lot easier. How? Can you tell me this red line, this one starting from here and this coming back? What is the total angle? Going to the amplitude and then coming back to the main position. Total angle is how much? Pi, 180 degree, 180 degree, right? From here to here, what is the angle? This angle, 30. So, subtract 30 from 180 because it is going towards the main position, all right? So, your equation of SHM is A sin of omega t plus 5 pi by 6, okay? Fine, let's proceed. If we are doing simple questions only, very simple questions these are, just that you are doing it for the first time, that's the reason why you may find it difficult to do this. Amplitude is how much? What it is? It is obviously 5 meters, 5 is the amplitude, okay? Time period, how do you find time period? You just did 2 pi by omega, omega is what? What is omega in this equation? X equal to A sin omega t plus pi is the equation of SHM, right? You compare it with equation of SHM, coefficient of the time t is omega, which is pi. So, t is 2 seconds, maximum speed is what? Vmax, how to get the maximum velocity? Maximum velocity happens where in SHM, at which location? Main position. At main position, what is the velocity? Omega A, what is omega? Pi, what is A? Phi, what is the answer for maximum velocity? Vmax is omega into A, what it is? 5 is A and pi is omega, 5 pi meter per second, okay? Now, do the next part. Velocity at t equal to 1. How you do this? Next part. Velocity at t is equal to 1, yes. Do it. How Anusha, how it is like that? Using fourth equation. For velocity, do not memorize things. It will, there is x and we know derivative of x is velocity, right? Derivative of x is V is equal to 5 pi cos of pi t plus pi by 3. Now, put t equal to 1 in this. So, 5 pi cos of t is 1, so pi plus pi by 3. How much it is? Cos of pi plus pi by 3, isn't it minus of cos pi by 3? Everyone, this is equal to minus of 5 pi by 2 meter per second. All of you understand? Everyone understood? Type in. Is there any other way you can get the velocity? Any other way you could have get the velocity? Tell me. How else you can get the velocity? How do you get V from x? Differentiate dx by dt. V is dx by dt. Okay. Yes, another way is find x at t equal to 1. Then use V is equal to omega under root a square minus x square. You will get the same answer. If you are getting a different answer, there is a silly error somewhere. Do this. Anyone got equation of SHM? Find the value of pi. Will pi be less than pi or more than pi? What do you think? More than pi, right? If it starts from here, goes there pi by 2, then comes back, becomes pi and then it has travelled further this side. So pi plus whatever extra it has travelled this much. Okay. Or I got something. Others? Pi is equal to sine inverse x by a. So sine inverse of minus of root 3 by 2. So how much it is? Minus root 3 by 2. What is the value? Pi plus pi by 3. 4 pi by 3, right? All of you understood? Type in, is it clear? So 5 is 4 pi by 3. So we have got equation of SHM. So this is your first part. Second part. Do this. Assume time period is capital T. In terms of capital T, you have to find out. Let's say it is given. How will you proceed? What you will do? From its root 3 by 2 position, it has to go to the left extreme. From here, it has to go there. Take x equal to minus a. Very good. X is equal to minus a. This is equal to a sine of omega t plus 4 pi by 3. So from here, sine of omega t plus 4 pi by 3. So from here, sine of omega t plus 4 pi by 3 is equal to minus 1. What will I write? Omega t plus 4 pi by 3 as then, what is the angle at this point? When it reach here, sine inverse of minus 1 is 3 pi by 2. Omega t plus 4 pi by 3 is equal to 3 pi by 2. So what will be t equals to from here? Omega t is equal to x6 common 9 minus 8. So it will become pi by 6. t is equal to pi by 6 omega. Omega is equal to 2 pi by 2 pi by omega. Sorry, 2 pi by t is omega. So when you substitute that, you will get t by 12. This much time it will take. Can you go through it and let me know? Is it clear? Everyone? Is it clear? Clear? Parvati, is it clear? Gurman, is it clear? Yes, sir. Okay. After the class, you get in touch. We'll send you some recordings. Alright, do the C part. Find the time taken by the particle to reach the main position. What do you put X as when it reaches the main position? Just like you did the B part, same way. You have to do the C part also. The value of X will be what? 0. Of course it is 0. So 0 is equal to A sin omega t plus 4 pi by 3. Now my question is this will give you sin of omega t plus 4 pi by 3 is equal to 0. So omega t plus 4 by 3, 4 pi by 3 will be equal to what? What is the angle at 0 when it comes back after one round? Okay. So from here, when it comes back pi, then it takes another round from there and then comes back. So the angles, sin is equal to 0 when you do. You have to take it as 2 pi. Otherwise you'll get t is negative, which you don't want. In fact, you get multiple values of sin for which this is 0. So this is 0 for 2 and pi. Okay. But then since first time it is coming back, we are doing it equal to 2 pi. Alright. So you'll get t is equal to 2 pi minus 4 pi by 3 divided by omega. This is how you do it. Right? So I hope you're getting idea about how to use the kinematics equations for the SHM. Alright. What is the doubt? Quickly type in, directly type in your doubt. Don't tell me that, don't make statements. Type your doubts. How it is 2 pi. Sin of 0 is what? Tell me. Sin of some angle is 0. So that angle should be equal to what? It can be 0. It can be pi. It can be 2 pi. It can be 3 pi. It can be 4 pi. It can be many angles. Okay. So 0 we have rejected. If you keep pi, you'll get t is negative because 4 pi by 3 is more than pi. So next value you have to put 2 pi. Time cannot be negative. It is time taken. Okay. Maybe 1 or 2 more questions. Just one more question. Then we will start something new. Answer for A part is what? Answer for part A. Yes. What is the reason? Force and coefficient of X. They are in opposite sign. Force is positive coefficient. X has a negative. So see it. Next. B. Anyone got the answer for B? Gourman got it. Aditi, Anusha. Equilibrium position. You know the name itself suggests it is equilibrium. So net force should be 0. Right. So when you put this 0, you will get the value of X to be equal to 4. So X equal to 4 is the equilibrium position. Fine. C part. Before you answer the C part, I want you to tell me what is the amplitude for this? Draw a diagram. You are solving a physics question. Don't do it without a diagram. Main position is X equal to 4. Where is the extreme position? Can you identify? Can you tell me where is the extreme position? Extreme position. Now read the question carefully. It is released at rest from X equal to 6. So suppose it is here at X equal to 6 meters. Which direction the force is acting on it? Everyone tell me which direction it is? Left or right? Force is negative or positive? At X equal to 6 meters, force is negative or positive? Negative. The force is acting this way. So there is a point where the object is at rest and it has started moving towards the main position. So can I say that this is an extreme position? Can I say that? Extreme position is X equal to 6. All of you agree? What is the definition of extreme position? Particle should be at rest and then after its velocity becomes zero, it should move towards the mean position. Are you getting it? All of you understand? X equal to 6 is the mean position. Sorry, extreme position X equal to 6. So now tell me what is the amplitude here? Amplitude is how much? Correct. Amplitude is 2 meters. Now proceed with the C part. Write the equation of the motion. The equation of motion is X equal to A sin omega t plus phi. You need to tell me three things in this equation. A, omega and phi. X equal to 6 is an extreme position because it is at rest and it has started moving towards the mean position. That is what happens in the extreme position, isn't it? Extreme position particle is at rest and it starts moving towards the mean position. That is what it is. A, you have already found out two. Can you tell me phi? What is phi? It starts from the mean position on the right hand side. So phi is equal to what? What is the angle there? What is the angle on the right extreme? Over here. Angle is zero here. Over here, what is the angle? Phi by 2. So phi is phi by 2. A is 2. What is omega? How will you get omega? Tell me. Sine of omega t plus phi by 2 is equal to cos of omega t. I am asking you what is omega? Anybody got the answer for omega? Only shortage. Force A is 8 minus 2x. According to Newton's second law, this is equal to mass times acceleration. So acceleration is equal to minus 2 by m times x plus 8 by m. M is what? M is 2. So A is equal to minus of x plus 4. Now looking at this, can you tell me what is the value of omega? Your mean position is at x equal to 4. So you can write this as A is equal to minus of x minus 4. So your x minus 4 is your distance from the mean position. Let's say capital X. Now you can compare it with equation of SHM. A is equal to minus omega square x. So from here, omega is equal to 1. So omega is 1. So you have phi, A and omega. Clear to everyone? Type it out. See this x is different from this x which is in the equation. In the equation, the x is distance from the mean position. Distance from the mean position is x minus 4 because mean position itself is at x equal to 4. Mean position is not at x equal to 0. All of you type in, is it clear? Then we are done with this kinematics of SHM. So we are done with the kinematics of SHM. Now let us move towards the larger portion of the chapter.