Derivation of the 1D Wave Equation

hello everyone and welcome to another video today I would like to talk about one of the most famous partial differential equations around the one-dimensional wave equation the focus of the video today is actually just on deriving the equation here we're going to leave the solution of this equation to another video and while this video here is designed to stand on its own I do want to mention that this is actually part of our multi video series discussing partial differential equations I'll leave a link in the description below to all of the other relevant videos in this series so with that being said why don't we jump in and talk about the 1d wave equation now this shows up in a lot of different engineering and physics and science applications and the picture that goes along with it or at least one of the applications where I think is the easiest to physically understand where this equation comes from and what it describes here is to consider the case of a of a string here so let's just draw ourselves a string here maybe we'll draw it like such I don't know maybe it looks something like like that here doesn't have to be sinusoidal here it could be some odd deflection here but maybe let's just draw you know an x axis or something like this here like this and what I want to talk about here is let's talk about the deflection here away from the horizontal position so you can imagine that maybe this is like a guitar string or something like that where you it would vibrate up and down sort of in this plane here right so maybe you know let's put this and here at zero here and maybe we'll say this thing has a length of L here and what I would again like to say here is we want to measure what is the deflection let's call this just you here now what makes this interesting is if you think about this long enough the deflection of the string here you it's actually a function of many different and it well not many in this case two independent variables namely this thing is going to vibrate up and down here so there's some time aspect here to it right and the other thing that's interesting is every position here of this string here has some x-coordinate associated with it as well here right so in this what we're talking about here is I would like to find this function you here which is a function of time and space X here so in this case with our guitar string or a vibrating string example U of T X this might represent the deflection of the string in the vertical direction however like we said earlier this vertical direction here this is at a given time T and given spatial location X right in other words the horizontal position here so that's what we're looking for here um to set the stage even further to help visualize this now might be a great time actually to pause here and look at our other video that discusses standing waves then we are gonna actually do I do a little demonstration showing how this rope might deflect it and some of the interesting behavior and modes that come out of it here so what we want to do now is derive the equation of motion which governs how does this string vibrate up and down here in this scenario so the way we want to do this is let's approach this and look at a small infinitesimal little gap here a chunk of this this string here this chord here let's pick two points let's call this maybe P and Q here so P is located at a distance let's call it X and we're going to consider a this point Q to be a distance Delta X away so this location here is obviously X plus Delta X right and I think it might behoove us to actually zoom in on this scenario or this little small chunk here and draw it a little bit larger so let's do that over here let's redraw the picture here but only focus on the points P and Q here so what I'm going to do is again we'll draw this in an exaggerated fashion let's have P here let's say you've got the string doing something like this maybe and then you got Q over here okay again we said P is located at a spatial location X and Q is located at a spatial location X plus Delta X here okay so that's the physical setup let's make a couple of simplifying assumptions for this so in terms of assumptions here we're going to make a couple of them one here let's assume that the mass of this string is uniformly distributed across the the entire length of the string here so let's just write this down and say how about that the mass per unit length you know this is in units of kilograms per meter here is constant so there aren't any lumps or knots or super dense sections of the string it's basically the same density across and again let's make a note here that we're gonna be using units of here of row here right and row is going to be in this this sorry we're gonna be using a variable of Road - didn't note the linear density here in units of kilograms per meter here right so if you buy yourself a one meter long rope here it's and the density here is 0.5 kilograms per meter that whole rope is going to weigh half a kilogram but that half kilogram of that mass is evenly distributed over the entire length of the string here right so that's our first assumption let's go ahead and write down a second assumption here all right sorry I'm back my laptop ran out of batteries here so let's come back here to our assumptions here the second assumption here that we'd like to make here is that we want to assume here that the string is perfectly elastic here so let's write that down so string is perfectly elastic and offers no resistance to bending to bending right so in other words the picture that goes along with this raise we're looking for you know a string like we looked at in that standing waves dem tration where you know it's it's very easy to deflect here it's got no ability to resist a moment in bending here so so we're not dealing with something like you know a string of plastic pens here which which you know if you try to bend this thing it doesn't want to really vibrate very well because there's got some it's some resistance to bending it won't credit yeah nope well as you can see why that's that's not gonna work for our PDE here so let's just go with assumption number two here string is perfectly elastic and has no resistance to bending here how about number three here let's say that the gravitational force on the string is negligible so let's say gravity or gravitational force is negligible and that's pretty reasonable if you think about this long enough here because we're going to have a string which is stretched nice and tight here and the tension in this string is actually going to be the dominating force and the gravity acting on this little bit of string here really should not influence it if you have a problem with this assumption here maybe just think about taking this string and instead of having it oscillating vertically just turn in ninety degrees so it's oscillating in a plane that's perpendicular to gravity so again gravity here won't play much of a role here okay um how about an assumption number four here let's assume that the string performs small or none small / negligible transverse deflections so in other words this is in the X direction so in another way to think about this is really the only motion that's happening with this string is that it's going up and down sure this tiny little bit here as it moves up and down it might go left and right a tiny little bit but that's going to be negligible compared to the other motions here the string so again assumption number four is I don't think it's going to be breaking anything too badly so number five here let's make our last assumption here and that we can get down to this here let's assume that the slope of this string is small at all times here so in other words the vertical deflections are small enough here let's write this down so vertical deflections okay in the U Direction are small at all times and therefore slopes are small so in other words coming back to our our little rope here right if you stretch this thing nice and tight and if this thing is vibrating up and down in the string in the plane right it really is not vibrating a lot up and down so we're not going to consider cases like this right where you've got this huge giant Bend here and now the slopes are definitely not small along this string here so instead we're gonna case it look at cases where yeah this is fairly fairly tight and it moves up and down here but the slope at any given location here is small okay so great with that I think that's a reasonable set of assumptions here I think we can go ahead and take this and stretch start trying to turn this here into a partial differential equation here so again maybe what we should do here is let's erase this half of the world we'll leave our picture here up because we're going to be adding to that in a second okay so let's write this down here so let's go ahead and talk about the derivation of the 1d wave equation okay so the picture that goes along with this here so let's go ahead and add some forces to this picture here so again we said this is that small little infinitesimal chunk here of string here and let's assume that maybe on the left side there's a tension T 1 on the string and it's gonna act in the direction of the string here at location P so maybe what we can do is we can draw that like this let's call this tension T 1 similarly on the right side of the string we're going to have a tension T 2 which acts in its direction here aligned with the string let's call this thing T 2 right perfect and what we can do now here is let's go ahead and also look at the angle that this force makes with the with the horizontal so we see let's call this alpha this angle alpha and then similarly over here on the other side let's call this angle beta okay all right so if we do that here what we can do he is let's go over here and write down what are some of the consequences so namely let's look at how about consequence number four here so if consequence in order for assumption four to be true we see that the horizontal components of these forces have to equal each other right so let's write this down for assumption a three two hold right we need the horizontal components of force to equal to be equal so let's just write down what are the horizontal components of the force so on the left hand side over here we see that the horizontal component it's what it's t1 cosine alpha right so I have the left side forces t1 cosine alpha right and that better equal the horizontal force on the right side which we see is two times t2 cosine beta all right boy sorry I'm not having good luck with the technical issues today so now the camera ran out of batteries halfway through that so I went through and then realized something was wrong but I think I've backed up and got us back to this exact same location so like we were talking about earlier we said that in order for assumption a 3 2 hold mainly that the thing doesn't move in the transverse direction the horizontal components had to equal each other here right now remember one of our other assumptions I believe a5 which unfortunately I erase but hopefully you have in your notes here is that the slopes were all small so namely alpha and beta were small everywhere here right so if alpha and beta are small here cosine of something small is basically 1 right so we see that what this ends up with his the t1 and t2 are basically equal to the same tension here there's no there's one tension through the string here and what I can write here is this is just equal to T here right so everything's are the same this this cosine of something small and cosine of something small really is not going to matter here so we end up with this relationship here so let's box this up here because we're gonna use this in a second here so this is also the same basically saying that the horizontal components are all equal here all right okay so how about the vertical direction here so hopefully there's some motion in the vertical direction right that's the whole point of having this string move here so let's talk about how about thinking about this string chunk here as like a very very small mass which is moving in the vertical direction well here on earth apparently I think this guy Newton came up with this second law here which tells you how things move around here on earth in an inertial frame isn't that something like some of the forces in the vertical Direction are is equal to the mass of this thing times the acceleration in the vertical direction right okay so all we have to do is start adding this up so first of all let's go and get this the forces in the vertical direction so looking at our picture here we see the vertical forces are basically what it's t2 sine beta minus t1 sine alpha right so we end up with here t2 sine beta right minus t1 sine alpha right is equal to the mass times the acceleration in the vertical direction now acceleration I think you'll remember from high school physics acceleration is the second derivative of position here right well what's the position of this little mass of string that's you hear right so in other words I could say that the acceleration in the vertical direction it's the second partial of U with respect to time right here is acceleration in the vertical direction here right great okay how about oh sorry maybe what we should do is we should have figured out mass here what the mass here another way to describe the total mass of that little chunk here right its Rho times the did instance of the string here right which we said was Delta X right the string is Delta X long in this little chunk here so Rho times Delta X okay great I think we're all consistent now alright next step here is let's go ahead and divide both sides by T here so let's divide this thing by T divide this by T here and maybe we'll rewrite it over there give me a second sorry all right so now if we divide this by T here what we can do here is let's rewrite this whole system here as I'll come over here maybe now is the this is a better spot here so left-hand side I could rewrite this thing as t2 sine beta minus t1 sine alpha this is divided by T here this is divided by T is equal to Rho Delta x over T times partial u squared with respect to T squared okay great now let's make a note here so if you recall previously here the thing that we boxed up earlier T here was just cosine want our t1 cosine alpha t2 cosine beta and T that's all the same thing so tell you what how about for this T let's write this as t2 cosine beta and for this T let's write this as t1 cosine alpha in other words this left side could be let's do this let's do this one first here so now this here this T I want to replace this with t2 cosine beta here right minus t1 sine alpha and then for this T I want to replace it with t1 cosine alpha so this is t1 cosine alpha right is equal to same thing on the left hand side let's not deal with that yet dot same thing ok now this is great the t's cancel here and if you remember what sine over cosine isn't that just tangent here so I can write this whole thing here as tangent beta all right - tangent alpha is equal to this thing over here again let's make a dot dot dot right it's the same thing carried all the way down now let's look a little bit more closely at this term tangent beta and tangent alpha let's let's for it look a tangent beta first if you look at this thing long enough the picture over here you see where is tangent beta tangent is if you remember tangent is what it's opposite over adjacent I think so it's opposite like say like this height here here's your here's your opposite if you want to think about it and this kind of shooting from the hip method here right here's the adjacent here so actually tangent of beta is the same thing as the slope of this thing right here right so that's basically giving you the slope of this of this red arrow here in other words it's giving you the slope of the string here at X plus Delta X right so maybe let's be very explicit here and write this down here so we see that tangent beta right this is slope of string at X plus Delta X right what's another way to describe the slope of the string here right isn't that the same thing as the partial of U with respect to X here at X plus Delta X that's like another way to explain it here right and by that exact same logic you can do the same analysis for tangent alpha so tangent alpha over here opposite over adjacent or rise over run it's basically saying tangent alpha is the slope of the string here at location X here so let's write that down here so tangent alpha is basically partial U with respect to X at X right that's the other way you could look at this okay great so replacing that into our expression let's go ahead and erase some of this let's see where we end up now okay so we said tangent beta here what we could say is we can write this thing as partial u with the respect to X at X plus Delta X right okay - partial U with respect to X evaluated at X here right so again I don't mean to belabor the point here but I want to make sure we're on the same page here right this is the same thing as tangent beta this is the same thing as tangent alpha here okay this is the left hand side here is now equal to let's maybe now's maybe a good time to rewrite this thing here it's Rho Delta x over T times partial U with respect to T squared right okay let's divide through by delta x move that to the other side so I can write this whole thing as maybe let's write this as brackets here is 1 over Delta X here and then here we got partial U with respect to X evaluated X plus Delta X minus partial U with respect to X at location X here is equal to Rho over T partial U with respect to T squared right ok let's look at this again here let's look at this term in brackets here right and look at it in conjunction with the picture over here what is this asking this term here this here is the slope over here at Point Q this here is the slope at Point P the difference here between these two is basically how much did the slope change here right so if you if you want to write this down in some English interpretation this is how much did slope in other words partial U with respect to X here right again we've got to be careful when we say slope I mean physical slope the the deflection with respect to changing the position how much did the slope change between Point P and Q right you can think of that almost like a rise here right because if you look at this there's this term over here Delta X what is Delta X delta-x is literally like the run of how far is is point P and point Q different from each other right so if you think about this long enough here if we take the extension here of Delta X let's take look and think about this as Delta X gets really really small here right what is this whole left hand side the left hand side is asking how much did the slope change as exchanged right in other words it's asking about the rate of change of the slope or this whole left side is basically what it's the second derivative of U with respect to X right if we take the limit as Delta X gets small hair right so this whole left side boils down to you tell you what let's come over here and erase this and write it down with a little bit more clear board space cuz I think we're getting really close here all right so the left hand side becomes basically partial u with respect to x squared here right so again let's let's make sure we're all on the same page here right the term in brackets is how much did the slope change and the denominator is how far in the X direction did you move here so again yeah that seems reasonable that is the second derivative of U with respect to actually the second derivative of position with respect to X here right or we can think of it as the first that never might I don't want ya I'm belaboring the point here right okay so great we got the left hand side and now the right hand side it's just Rho over T times partial U with respect to T squared great all right and you know what rearranging terms what usually happens here is people want to solve for the derivative of the respect to time is how you'll typically see this equation presented so let's just let's just move these constants to the other side and solve for partial U with respect to T squared so I got partial U with respect to T squared is just equal to tension over linear density here times partial u with respect to x squared and typically what you'll see here is you know this is just some mishmash of constants it's tension over density however in this case and in most cases for the 1d wave equation right this is this is a constant number right it doesn't make any sense to have like a negative tension here and doesn't that make any sense to have a negative density here so usually to denote that this coefficient has to be positive what you'll typically see people right here is this is partial U with respect to T squared is equal to C squared partial u with respect to x squared so again this is the more common representation here of the 1d wave equation here so this is 1d wave equation okay and all this C squared constant is the noting here is it's trying to tell you that this coefficient is usually positive I lost something like tension over over density here right so here we are this is our 1d wave equation here and this is the governing partial differential equation and if you look at this thing long enough you see that what it's a second order partial differential equation it's linear it's homogeneous here so there's a lot of tools that we can go forward with trying to solve this here however that's the topic of a of another video here so I think we've gone ahead and successfully derived the 1d wave equation here so if you enjoyed the video please subscribe to the channel because we're gonna have another video coming up which discusses actually how do you go about solving this here and talking about other engineering topics as well so with that being said I hope to catch you at a future video bye

Channel: Christopher Lum

Views: 40,576

Rating: 4.9252338 out of 5

Keywords: 1D wave equation derivation, wave equation derivation, derive 1D wave equation, partial differential equation, wave equation PDE

Id: IAut5Y-Ns7g

Length: 26min 16sec (1576 seconds)

Published: Sun Nov 11 2018