 Hi, I'm Zor. Welcome to Unisor Education. We will solve a couple of very simple problems related to mechanical oscillations with damping effect of the friction. Now, this lecture is part of the course called Physics for Teens, presented on Unisor.com. So this is the website where I recommend you to watch this and all other lectures because it presents a course which means it's a logical connection between different lectures. There is a menu which drives you in a logical sequence from one lecture to another, from one topic to another. Every lecture on this website has a very detailed notes which basically can serve as a textbook. The site is completely free. There are no advertisements, no strings attached at all. You don't even have to sign in if you don't want to. Also the same website contains Mass for Teens course which basically is a prerequisite for physics. You do have to know mass. I'm using, especially the calculus, I'm using differentials and integrals practically in most of the lectures. Physics is all about calculus, quite frankly. Okay, so damping effect of the friction. Okay, consider the hand falling problem. So you have a table. You have a spring attached to the wall and attached to the object. Now, let's characterize the object with mass which is equal to 0.3 kilograms. I'm using the C units of measurements. Now, the spring has elasticity equal to 2 Newton by a meter, which means if you stretch it by a meter you will get 2 Newton's force back to... So any extension by certain amount of certain distance will cause the spring force multiplied by 2. The distance multiplied by 2. So if you stretch 1 meter it will be 2 Newton's. If you stretch half a meter it will be 1 Newton. If you stretch 2 meters it will be 4 Newton's. So that's what elasticity coefficient is. And we also need friction which is 0.1. So this is a ratio between... Okay, if you have certain force of the weight you have to multiply by this friction coefficient to get the force of the friction which prevents the movement of the body. Sometimes we differentiate between static friction when the object actually is sitting and it needs an initial effort to move it. From dynamic friction whenever it's already moving the resistance to the movement. And usually static is greater than dynamic. It's more difficult to start movement from some kind of a standing position than to continue movement. But let's just for simplicity consider that these static and dynamic are the same. Okay, so what's my problem? My problem is whenever I'm stretching the spring to a certain distance I will have the force of the spring which will pull it back to initial equilibrium position. Initially let's assume it's equilibrium so the spring is not stretched, not squeezed. But now at the same time when I stretch it to a certain position not only the spring will attempt to pull it back also the friction will prevent it to move, right? So if I...now this force of friction is always the same. It's weight times coefficient. So the friction force f friction is equal to mu times m times g. m is a mass, g is acceleration of the free fall. So m times g is the weight and times mu gives me the force. The force is always constant and it always prevents the movement. Now the spring whenever I'm stretching it to certain distance if you multiply this distance by the coefficient of efficiency will give it also the force. So my problem is what's the maximum distance I should move this particular body so the spring will have enough force to initiate the movement. Because if I will just move it just very very little my force of the friction will be exactly the same. My spring pulling force pulling back to the equilibrium will be very small because this distance I'm stretching very small. But the greater I stretch the greater the force of the spring. So I can actually manipulate with a position. So the spring will have different force, the force will be stronger the further I will move it from the equilibrium. But the friction is exactly the same. Looking for a specific position, maximum position I can stretch without spring being able to move it back. So again if I'm stretching too little spring will not be strong enough to pull back because the force of the friction will prevent it. So when exactly will be that maximum distance I can move where my spring will not be able to pull it back. Well obviously my force of the spring depending on x if x is a distance I'm stretching then the force of the spring which obviously depends on the x would be equal to k times x. That's the meaning of the Hooke's law. The distance I'm stretching times coefficient of elasticity that's the force. Now this force will be very small when x is small but the bigger is x the bigger is force. So I'm looking for the maximum when the force will be equal to the force of the friction. So basically I have to resolve this equation from which x is equal to mu mg divided by k which is equal to 0.3 times mu is 0.1 times mass is 0.3 times g is 9.8 divided by 2 which is equal to 270 to 94 to 94000 so it's 0.294 divided by 2 which is equal to 0.147 which is we're talking about meters right equals 147 millimeters. So under these conditions I have to move my object no more than 147 millimeters in order to spring not being able to move the body back. If I will stretch it more than 147 millimeters then the spring will have sufficient force to move it back. How much back? It's a different question but that would be actually a second problem. Okay that's the answer to the first problem. Now the second problem it has the same setup. Okay now I'm looking for another distance I have to stretch. Now if I will move it further than 147 let me just write 147 millimeters it's by the way it's called critical distance and there is a special letter lambda which I'm using so that's the critical distance. Before that critical distance my movement will not initiate the pulling back starting from just a little greater than 147 my spring will be sufficiently strong to start movement. How much farther? Well not much because as soon as my spring it will start moving back to the original it will cross 147 millimeters boundary but then after that the strength of the spring will not be sufficient because we are shortening spring so the pulling back strength will not be sufficient to move it much farther so it will move just a little bit beyond this point and stop. So my point is how far I have to really move it outside so the spring will initiate movement but then it will be strong enough all the way until the body will reach the equilibrium point and stop. That's what's interesting it's stop. What it means? It means that the energy which has been accumulated by moving my object to the right potential energy of the spring will be spent completely on the way from that position back to the equilibrium and then this object will stop which means there is no kinetic energy and no potential energy. No potential energy because it's equilibrium point and the spring has no potential energy in equilibrium point and kinetic energy will be zero because I'm saying that the condition of this particular problem is the body should stop, the object should stop at that point. So it's all about energy. Again, as soon as we stretch the spring it accumulates the potential energy. As soon as we let it go potential energy will start converting into kinetic energy of the movement so potential energy is diminishing because the spring will be shorter and shorter from the squeezed position but kinetic energy will still be something whatever the kinetic energy is but energy will be spent to friction because friction will have to do some work. This is a constant force of friction so on the way from that stretched position to the neutral position will always be working against the movement so energy will be spent. So the idea is to find out what's the potential energy whenever I'm stretching to a certain distance and then that same distance should be used to calculate the work which friction force will perform and just equate them and that would be my equation. Okay, so how can I calculate the potential energy of the spring when it's being stretched? Well, I did address it in the lecture but let me just talk about this again. Let's consider this is your spring and you have stretched this particular body to a distance x and then x plus dx. So dx is infinitesimal infinitesimal increment of the distance. Now whenever my object is at position x the force of the spring which depends obviously on this position is equal to coefficient of elasticity comes the distance increment so this is zero and this is x. So my force obviously depends on the distance x. Now as always in physics and mathematics we are assuming that on this infinitesimal increment of the distance my force is the same and so my work which this force should actually perform on this distance should be differential of w of x is equal to f spring of x times dx. So the force times the distance, that's the work. So the force I assume is the same on this particular infinitesimal distance which is equal to this and then force I multiply by the distance to give my work and this is only differential of the work an infinitesimal increment of the work which spring is doing whenever I'm moving an object from x to x plus dx. How should they do how should they calculate my work if I would like to stretch it on the distance a? Well I have to integrate it from zero to a which is basically a summation of this infinitesimal number of infinitesimally small pieces of distance and infinitesimally small increments of the work. So what is this? This is integral from zero to a f is this, this is a very simple integral. Now this is a definite integral to get it using the Newton's wave needs formula I have to have an indefinite integral which is kx squared divided by 2 because the derivative of this is kx k is a multiplied derivative of x squared is 2x divided by 2 would be x so this should be substituted from zero to a so I substitute the first one a minus the second one but the second one is zero so it would just total work which depends on a would be equal to k a squared divided by 2 and that's all I need right now from the spring. Now this is the work which spring is doing well against the force of spring by stretching the spring so we have stretched the spring by distance a and that's the potential energy which we are giving to this particular object because we are spending our energy to overcome the resistance of the spring so we are basically transferring our energy working by stretching the spring and we are actually giving it to the body this object now stretched by the distance a has this amount of potential energy okay now whenever I let it go and the spring will start pulling back my condition is that this same energy should be completely spent on the way back to the position of zero to equilibrium position now this energy is spent to a friction friction has always the same constant this is the constant force so to find out the work which friction is doing I have to just multiply this constant force by the distance well distance is a so my work which friction course performs is mu Mg times distance which is a so this is the work which friction performs on the way back to equilibrium and it should be exactly equal so the kinetic energy would be zero at the end of the point no speed and the potential energy is zero because it's equilibrium point so it should be equal to the potential energy accumulated by the object when I stretch the spring and that's the equation and A is an unknown variable so we'll just solve this equation which is very simple A can be cancelled out so A is equal to 2 mu Mg divided by k which is equal to well I didn't write it down but this lambda was equal to mu Mg divided by k so it's twice as much very interesting, right? so the critical distance is this which is 147 and the distance which is supposed to be such that the object will move back exactly to the equilibrium point where it started is twice as much so let me just calculate it so it's 2 times mu which is 0.1 times mass times acceleration divided by 2 which is equal to so it's 3 times 0.294 meters 294 millimeters which is twice as much as this one so what's interesting is whenever I am stretching beyond the critical point and I did actually talk about this in the lecture my spring will start pulling back and it will stop on a point which is symmetrical to critical point so this is lambda, critical point so this is the beginning this is 0 and this is A so if I am stretching it twice the critical distance it will go back symmetrically relative to this if I stretch it greater it will go beyond the zero point so it will start oscillating now if it's within from lambda to twice lambda it will just go back maximum to a zero if I stretch it less than to lambda it will stop symmetrically if I stretch it greater than to lambda it will go here and then it will start oscillating because on this side it will be repeating the same thing relative to the critical position on this side well, basically that's it these are two problems I do suggest you to read the same two problems on the website and Unisor.com you have to go to physics for teens course this is the part which is called waves and the topic is mechanical waves now in the mechanical waves topic you have lots of lectures and this is the problem number two within the mechanical oscillation topic that's it, thank you very much and good luck