 Hi, I'm Zor. Welcome to Unizor Education. Let's solve a couple of problems related to damping oscillations. In this case, we will talk about viscose environment. Now, this lecture is part of the course called Physics for Teens presented on Unizor.com. I suggest you to watch this lecture and all other lectures from the website because it's a course, which means there is a logical sequence between lectures. There is a menu, which basically brings you from the whole topic, Physics for Teens, to certain parts. Now, right now, we are talking about waves. This is the part of this course. And in particular, we are talking about mechanical oscillations. And this is the problem related to oscillations in the viscose environment. Now, this website contains also the prerequisite course, Mass for Teens. You cannot study physics without knowing mass, especially calculus. So, I suggest you to refer to Mass for Teens part of this course, this website, if there is anything missing in your calculus knowledge. Anyway, so, we will talk about oscillations in the viscose environment. Now, what's important about viscose environment, which is different, by the way, from the previous topic, which I was discussing, damping oscillations when there is a friction involved. The friction introduces the constant force of resistance to oscillations. The viscose environment actually is different, because it resists the movement depending on the speed of movement. If the object moves slowly in the viscose environment, let's say, you are moving in the water. And whenever you are moving slowly, there is no problem. Basically, water doesn't really resist your movement. But if you try to move faster, the faster you move, the stronger will be the force of the water to resist your movement. So, when we are talking about damping oscillations, which are occurring in viscose environment, we have to, in addition to some kind of a spring or whatever else, is initiating the oscillations. We have to consider the resistance force of the environment, which is, well, generally speaking, proportional to speed. So, the lecture, which preceded this particular problem, they were actually dealing with an equation of motion where there are forces of the spring, which involves the oscillation, obviously, and the force of resistance, which is proportional to speed. Now, the mathematics of this are, well, not very complex but involved. So, this particular problem is not about deriving something like an equation of motion, which basically I was doing during the previous lectures. This is about, okay, let's say that the motion is given to you and the motion is damping oscillations and in such a way that it actually is a result of viscose movement. And now I would like to determine certain characteristics of the movement, like where it goes back to equilibrium, where it receives the deviates to a maximum distance from the equilibrium, etc. So, this is this particular lecture about given the equation of motion determine certain characteristics of the movement. Okay. So, here is the equation of motion. X is deviation from the equilibrium point. T is time, obviously. And I have it as 4 times e to the power minus t over 2 times cosine of 5t minus p over 4. I just took it from the head. I mean, there is no significance in any kind of numbers. What is significant? Significant is this multiplier because as t is increasing, this multiplier is decreasing. So, if t is equal to 0, it's 1. And as t goes to infinity, this goes to 0. Staying positive. So, basically, this is the graph of e to the power of t over 2. By the way, the graph of the whole function is presented in the notes for this lecture. Now, this lecture, as any other has notes, it's on the Unisor.com website. And the graph is really important. It helps to solve the problem. Now, I will just draw it here. But whenever you will review the problem, it's in the notes. Now, cosine is obviously a periodic function. So, what happens here is the following. Without this multiplier, cosine would look like this. But with this multiplier, as we are moving down the time, we are decreasing the deviation from the equilibrium. So, the whole graph would look like this. So, if this is my e to the power of minus t over 2, the graph would be something like this. It goes from both ways. It diminishes from both pluses and minuses. That's how it will go. So, every extremum, maximum or minimum, will be multiplied by some positive number, which is getting smaller and smaller and smaller. So, that's why you will have this type of graph. So, this is a guiding line for maximum values. And this is the real graph. My first problem is, based on this equation of motion, I would like to find out where we are returning to the equilibrium points. Not where, when. Where on the time axis. So, when, at what time moments, we will return to equilibrium. So, if we start from some point, we stretch the spring. Let's say the whole thing is in water. So, we stretch the spring and let it go. So, it starts oscillating. And that's how it will go through the equilibrium point. And then back to another extreme, where it will be smaller and then another extreme. Squeeze, this is a stretch. This is squeeze, stretch, squeeze, etc. Okay, so how can I determine the points of equilibrium? Well, it's just an equation, basically. X of t is equal to zero. You have to solve it. Well, how to solve it? Very easily. Now, if this is equal to zero, this multiplier never equals to zero. So, it's basically cosine equals to zero. Now, when cosine is equal to zero, five t minus pi over four equals to zero. Well, the cosine equals to zero when angle is equal to pi over two, plus n times pi. Right? If you forgot trigonometry, review it, please. But that's the... Okay, so, and this is five t minus p over four. So, whenever my argument is equal to this, where n is basically any integer number. Now, from this, we can find out five t is equal to... What is it? Three pi over four, plus... Now, let me do this also over four. Four n pi over four. So, I will... This is a common denominator. And now, I can take pi out of the parenthesis, and I will have this. Now, I will get rid of this, and I put here 20. This is the answer. So, now, what kind of n? Now, I said it's any integer. Well, actually, not exactly, because if n is negative, then the whole thing would be negative, which is not good, because this is a time. So, t is time, so it's supposed to be positive. So, n is any integer static from zero. So, n is equal to zero, one, two, et cetera. All non-negative integer numbers. So, this is the answer to the problem of when exactly we are returning to equilibrium. Now, what's important here? What's important is that if n increases by one, no matter what it is, the whole thing will increase by four pi over 22, which is pi over five. So, this is my point. The distance between these is pi over five. Constant distance. So, no matter how far we deviate from the equilibrium, we will always return to equilibrium with exactly the same intervals. That's what makes calling this particular movement like almost periodical. Now, a true periodical movement is when there is a complete repetition. So, if amplitude would not decrease, it would be a pure periodic movement. With amplitude decreasing all the time, I can't really say it in the pure sense of the word that this is a periodic. It's almost periodic because it's really returning to the same position, which is equilibrium, after the same interval of time, constant interval of time. So, as time goes by, I'm returning to equilibrium with exactly the same intervals. Now, you understand that as soon as our amplitude is decreasing the distance which the object is supposed to cover from one extreme to another, getting smaller. But the speed probably is also getting smaller and that's why the timing of returning is exactly the same. So, from this extreme to this extreme, the time will be the same thing as from this extreme to this extreme. Otherwise, we would not have equality of the timing. Okay, so that's my first problem. Now, my second problem is where exactly we are reaching the maximum of the absolute value of deviation. So, this is our points where my deviation goes to maximum. So, these are local extremum of the function. And the question is how can I find the local extremum? Well, let's go back to mathematics and we know that for any function good function, smooth function, differentiable function, if you would like to find points where it actually achieves its maximum value. Well, maximum value is where the horizontal tangential line exists, right? If it's tangential and if it's horizontal, it's local, in this case, maximum, in this case, minimum, extremum. Okay, so, when is it occurring? Well, you know from the math, from the calculus, that to find the local extremum you just have to find the derivative of this function. Now, derivative is a tangent of the angle of the tangential line with x-axis, or t-axis in this particular case. So, whenever the first derivative is equal to zero, tangent of this angle is zero, that's the local maximum or local minimum. So, all we have to do is to solve equation first derivative is equal to zero. So, the previous problem was solving the equation x of t is equal to zero, to find these points, these points. Now, this, if we want to find this point, and this point, and this point, then we have to find the solution to this equation. Okay, so, let's just take the derivative of this thing and again, I hope you remember how to take the derivative. So, it's product of two function and you know that if you have a product of two function, it's the first function derivative times the second plus, this is u, this is v. Okay? So, this is the rule for having a derivative of the product. So, this is the product of this and this. Well, four is just a constant multiplier, so it's applied to everything. Alright, so, four times, so it's the first function, derivative of the second function. Okay, e to the minus t over two. Now, derivative of the cosine is minus sine. So, I have to put minus here, minus one times sine of whatever it is. But, this is the function of function. The first function is 5g. The second function is cosine. Whenever you have these combined functions, you have to go into the inner function derivative and the inner function is 5t minus 5 4, minus 5 4. Derivative of this is just a plain function, so that's 5. Okay, plus. Now, we have to take derivative of this and the function of this. Now, derivative of e to the power is the same e to the power and then we have an inner function which is minus one two derivative. e to the minus t over two. And then the function, the second function remains as it is. So, that's my derivative and it's equal to zero. Now, I can obviously get rid of four. I can get rid of e to the power because it's never equal to zero. So, what do I have? Well, I have minus this, times five, so it's minus five, sine of 5t minus p over four, minus one half, cosine of the same thing, minus p over four. Okay. And this is supposed to be equal to zero. Okay. Now, it's minus and minus and it's zero, so I can make it plus and plus. It's easier. I don't like fractions, so I will multiply by two and I will have ten here. So, if I will put cosine to one place with a minus sign and then divide every, so it's ten sine of 5t minus p over four equals minus cosine of 5t minus p over four. If I divide by this, I will have ten tangent of 5t minus p over four. Right? Sine divided by cosine is a tangent. It's equal to minus one. Or tangent is equal to zero point one. I divide by ten. From which 5t minus p over four is equal to arc ten of minus zero point one. Now, tangent has a periodicity of pi, so I have to add pi times n, where n is any integer. Now, from here I can just get the t is equal to, so pi over four goes to there, so it's arc tangents of minus zero point one plus pi over four plus pi n, and this is all divided by five. Right? Divided by five, I will put this coefficient 0.2. Now, let me just simplify this a little bit. So it's four divided by four, so four is a common denominator, and pi can go outside with one plus, and I don't have this. So that's what it is. Or, you know what, that's probably easier if I will just open this, and this is one-fifth, so I will put twenty here. That's easier. So that's the answer. And, well, I do put a couple of values of this into the notes calculated for n is equal to zero if n is equal to one and n is equal to two, and I'm getting these points. And now, as n increasing, as you see, again, we are increasing by, if n is increasing by one, my whole thing, my time is increasing by four pi over twenty, which is pi over five, which is exactly the same as... So the distance between local extremes is also pi over five. So that's what makes, again, it's almost periodic. My final problem is what are the extreme values, maximum, minimum, maximum, minimum, deviation to stretching or squeezing, stretching or squeezing. Now, how can I find them? Well, since I know this point and I know the equation, all I have to do is my first extreme is equal to x of t zero, where t zero is n is equal to zero, whatever it is, then a one is equal to x of t one, etc. So, and again, I did have some calculations. I calculated the values of this, this, this, and this. And obviously, it is diminishing. This is a plus something, this is a minus, but absolute value is smaller. This is, again, plus value is smaller and this is minus value is smaller, etc. So it goes exponentially diminishing amplitude. So these are the problems which are related to this cause environment and oscillations in this environment. I do suggest you to read the notes for this lecture. They have a much better picture and then there are some numerical calculations and you can actually see that the numbers are decreasing and these numbers are exactly the same. That's it. Thank you very much and good luck.