Free Physics NEET Video Lectures for Oscillations

00:>00:01 [Music] hello I am going to give you a few lectures on oscillations and waves and what this kind of motion oscillatory motion and wave motion what kind of motion is this we are going to look into that and how we describe it mathematically and where does it occur so oscillatory motion comes a class of motion called periodic motion so let us understand what does periodic motion mean periodic motion is something is a motion that repeats itself so let us understand what does that mean if a particle is going

00:02:15 around in a circle so let's say it started from here it goes around in a circle and then keeps repeating it and every time it goes around it does the same motion then the motion is periodic let us see if I want to calculate the X component or the Y component of this motion what does it look like so if the particle is going around in a circle and let us say it covers a distance theta T in time T right starting from here on the x axis then X T would be a projection of this on the x axis this will be X T and this would be

00:03:21 if the radius of the circle is our R cosine of theta P and similarly the Y component of this motion YT Y of T is going to be R sine of theta T and suppose this completes an entire circle in time capital T right so it comes back in time capital T then you can see X starts from some distance R initial distance if starts from this point is R then it changes in a certain manner and when it reaches here at the upper point which I'm doing number two this was number one started from number one reaches number two then X starts

00:04:27 becoming negative becomes minus R at number three goes up becomes zero again at number four number three number four and could do something like this back to number one and if this motion repeats itself over time that means theta T is going to be exactly the same after it has completed one cycle then the motion is going to be periodic so if I were to plot it carefully let me show it here here is X and y axis this is a particle going around the radius is R I'm plotting X T I plot it in a certain way

00:05:18 it could be any general function however it repeats itself exactly in the same manner then it is a periodic motion with time period being from here to the point it reaches point one again so let me show this this was point one this was point two this was point three maximum this is 0.4 and this is 0.5 which is again point 1 and the motion repeats itself if theta is in general changing very arbitrarily then the motion would not be periodic however if it repeats itself after a certain time it is periodic it could so happen that

00:06:00 the particle goes around and after it completes two circles once and twice then the motion repeats itself then the periodicity or the time period would be twice as large but the main point is that if the motion repeats itself if a motion repeats itself after time T it has to be exactly the same then the motion is periodic with time period t other samples of this would be examples of periodic motion Earth's rotation on its axis right and this motion repeats after T Tauri equals 24 hours other example

00:07:29 would be moon going around the earth and this motion repeats itself after roughly 29 days so this is a periodic motion with time 20 time period 29 days or the earth going around the Sun with time period roughly equals 365 days or a year and the motion repeats itself after that much time so these are some examples of periodic motion so what I have defined for you is time period which is T whatever units we use a related quantity is going to be frequency and what frequency means is in per unit time how

00:08:43 many times the motion repeats itself so frequencies usually is denoted by F and this is 1 over T in time T the motion repeats itself so obviously in per unit time it's 1 over T times the motion takes place right so this is a frequency third we also define something called the angular frequency meaning of it would be clear after some time and this is two pi times F usually denoted by Omega and this is two PI over T all right so this is the angular frequency time could be given as seconds then

00:09:37 frequency is going to be per second sometimes also written as H said or what it means is Hertz if time is in hours then frequency is per hour and time is in days then frequency is going to be per day so we talked about I'll go back to now a motion of a particle in a circle and I'll now specialize to motion where the speed of the particle is constant sometimes also called uniform what it means is that when this particle is going around in a circle its speed here is re it's the same all the time and

00:11:05 therefore if the circle radius is R the time taken for the particle to go around is going to be 2 PI R divided by V and this is the time period you can see it for yourself if the speed is uniform the motion is going to repeat itself after the particle reaches its initial point it's going to repeat itself so time it takes to go around once is going to be the time period and that is 2 pi R over V the frequency F of motion is going to be 1 over T which is V over 2 pi R V over R I'm going to call Omega

00:11:55 which is angular frequency divided by 2 pi now you understand the meaning of angular frequency it takes time T to go around the angle 2 pi alright so angular speed is nothing but 2 pi over T and that's the same as angular frequency and which is nothing but 2 pi and F so we have all these things are related as you can see now why this motion is interesting is because if I look at this motion a particle going around in a circle of radius R with uniform speed V then the angle it covers in time T the angle is

00:12:52 going to be measured in radians as you can see earlier I said it's 2 pi where it goes around once in full circle this theta theta is going to be equal to this distance here is going to be V times T this radius is R so theta is going to be V times T the arc length divided by R we can you can see this Omega times T which is a well-known relationship and therefore XT the X component here is going to be equal to our cosine of Omega T YT is going to be R sine of Omega T so in this case if I plot the motion so let me show this to

00:13:47 you again I'm considering the case where the particle is going around in a circle right of radius R the speed is V and this angle theta is omega T where Omega is V over R X as a function of time is R cosine of Omega T and if I were to plot xt versus T at T equal to zero the displacement is r it comes down as a cosine curve goes to zero that is let me show this again by positions position 1 position 2 is going to be here position 1 is here and then X becomes negative and at position 3 its maximum negative

00:14:57 so it goes like this this is position 3 and after this the X starts reducing again becomes less negative it comes here at position 4 becomes 0 again and then goes up and back to 5 or 1 so this entire time that it takes to come back is time period t and this is a cosine curve so it's X T equals R cosine of Omega T if I were to plot y T right if I were to plot YT this is going to be R sine of Omega T and if I were to plot it at this point position 1 y 0 this is position 1 and then it goes up reaches the maximum

00:16:12 y equal to R at position 2 comes down and then goes up again negative and then goes up again this is position 3 position for position 5 and then it repeats so this is also periodic with the same time period this point is supposed to be right here this is this is position 5 they're supposed to be same same point T ok so to beat itself and this is a very specific kind of motion so what is happening here is if a particle is going around in a circle with uniform speed then XT is equal to R cosine of Omega T

00:17:06 YT is equal to R sine of Omega T moves around with exactly one frequency right this is called simple harmonic motion all right so this this is motion which is simple harmonic it contains a cosine Omega T or sine Omega t time dependence and this is going to be the focus of our study in these lectures but before that I just want to give you some more periodic motions and how to represent them on the displacement versus time graph motion and their representation on displacement versus time graph okay so

00:18:31 let's see look at that number one let me take a particle which can move along the x axis so let's say this is the x axis from 0 to L and then there are these some hard walls at these positions so that this particle goes from here with the uniform speed V goes here hits this wall and immediately it turns back so you can see it's going to be going back and forth and repeating its motion so this is a periodic motion let us see how does X versus T graph look for this so if I were to plot xt versus T and let's

00:19:09 say T equal to zero it was at this point on the left hand X equal to zero then x increases goes to uniformly goes to a value L so this is L and as soon as it reaches here it starts coming back when it starts coming back with the same speed without losing energy it goes down exit decreases goes to zero right and then the motion again repeats itself you see again that motion is repeating itself it's exactly the same triangle that keeps coming back again and this time from here to here is going to be

00:19:59 the time period what is T equal to the total distance traveled is 2l divided by V that's going to be the time period let's take the second example a ball is released from height H comes down and without losing energy it bounces back so it goes up to height H again and then it comes down and goes up and this motion keeps on repeating this is also a periodic motion because exactly the same motion is taking place after certain time and if I were to plot the height of the ball YT versus the time T it starts

00:20:51 at a height H comes down and you know from your equations that the height is going to be like this because I'm going to have Y P equals H minus one half GT square so it looks like this now it hits y equal to zero and it starts moving up is exactly the same speed goes up goes to height H and then again it starts coming down so you see motion repeats after this much time this is the time period how long did the ball take to come down well we know that y equal to 0 implies T equals to H over G square root but that

00:21:50 is not the time period because that is the time they took to reach this point here we are making a big block so total time period t is going to be twice as much which is 2 square root of 2 H over G that is the time period whenever the motion is periodic all the related quantities are also periodic so what we have to run in the examples I have shown you we have shown the displacement of a particle XT which is periodic that means this is changing I'm also giving you all sorts of words changing

00:22:36 periodically with time period t so in the first example let me just go back again we took a particle which was going back and forth between two rigid walls and XT plotted against time look like X was increasing when all the way up to L came back X was increasing came back I want to show you now how other quantities are going to look like so suppose I were to plot its velocity versus time right well I'm saying it velocity because I'm going to take negative and positive both now as the particle moved up to this point capital

00:23:40 T by 2 it was moving with positive velocity and the velocity was certain value V as soon as it reached this point the other wall it got hit and started moving the other way so velocity became negative and then it came here this was the velocity from this point onwards up to this point and then the velocity became positive again up to this point and then it became negative again so it is red color up to this point so velocity of this particle going back and forth between two walls is like this just I'm showing the dotted line

00:24:51 it's not well-defined then again like this here then the dotted line there like this dotted line like this and you can see that after time period t the velocity is repeating itself so velocity is also a periodic function of time so if XT is periodic so is the velocity and this is given as DX by DT slope of X with respect to time that is also periodic what I've shown you in this example again particle going back and forth between x equals 0 and x equals L I've shown you X curve which I'll quickly

00:25:49 plot now like this I've shown you the velocity curve or let me put here X T versus T I've shown you a velocity versus time curve which looks like this and so what about the acceleration let's plot the acceleration also you see when the particle was moving with uniform speed the acceleration is zero and then suddenly it has a huge negative acceleration so that the velocity becomes negative is zero again then a positive acceleration and zero again negative acceleration zero again and so on positive acceleration zero

00:27:02 again negative acceleration and zero again what you can see is this acceleration is also repeating itself from here it becomes negative positive and then zero again this is the time period in real life when this ball hits it's going to be squeezed a bit this is for a short time there's going to be some force which we are showing that short time to be roughly zero but in real life is going to be slightly different the actual curve may look something like this zero then it hits the wall then it hits the wall that's a

00:27:38 huge acceleration for some time and huge acceleration for some time the actual curve may look something like this nonetheless the point is that even acceleration is repeating so XT is periodic velocity VT is periodic with the same period and so is the acceleration I'll let you think what this integral if I were to take an integral from time T up to this time T equal to 0 to say this time of the acceleration 8002 let's say this is T 1 T 1 DT what it would be I'll give you the answer the

00:28:39 answer would be minus 2 V you figure it out why it is so it's basically related to that the acceleration a all right a T let me write it on this side 80 keep writing on the right-hand side this is XT the next point is VT there's nothing but DX by DT and the acceleration 80 is DV by DT which is d 2 X by DT square and this is what this is what leads to this integral let's look at the next example that I did and that was a ball dropped form height H and bouncing without losing any energy and in this case when I plotted X

00:30:03 T versus T or height Y T versus T it looked like this where this up to this point is the time period let us now plot as velocity also VT versus T and you know from your daily our earlier exercises in this case it started with zero velocity and out here by the time it reaches this point let me show it by black the velocity is negative so this velocity goes like this and keeps on increasing up to this point then the ball bounces and s velocity just changes direction and therefore it becomes positive the same magnitude and then it

00:31:05 again it starts flowing as it moves up velocity decreases reaches zero when the ball which is the highest point where it stops momentarily and then it comes down again and then the motion repeats itself and so on this is the velocity curve so you can see that velocity goes up to this point and then changes it is like this and the time period time over which it repeats itself is from here to here what about the acceleration again if I were to plot acceleration is minus G up to this point it minus D

00:32:07 throughout minus G but at this point the body changes velocity so it goes up a huge acceleration in the positive direction and then minus G again all the way up to this point changes again all the way up to this point changes again and this is the time period you see motion is repeating itself so it doesn't matter how I write the time period I can write it between these two points these two points and so on in real life again the acceleration would look something like this it'll go like here goes up

00:32:45 comes down goes up comes down this will be real life acceleration so Peaks are not that sharp they are spread a bit and you have if I were to take the integral of time let's say t1 to t2 just like earlier integral t1 to t2 a T DT will be equal to 2 V 0 where V 0 is the velocity when it hits the ground which is going to be 2 square root of 2gh and all this follows from that this is of course the first term is YT the second term is VT which is dy t over DT and the third term acceleration ay T is the wheat

00:33:55 by DT which is same as d2y over DT square so if you integrate this equation you get this answer the third motion which I said will be interested in what are their lectures that follow is the simple harmonic motion which we said is nothing but if I take a particle going around in a circle of radius R then XT which is equal to R cosine of Omega T right this XT is called simple harmonic motion which contains the term cosine Omega T or equivalently I could also call its component YT which is R sine of

00:34:45 Omega T which is also called the simple harmonic motion if I were to plot xt versus T it looks like a cosine curve with this maximum displacement being R or minus R the corresponding velocity VT versus T is initially slope is almost zero so it's going to be zero however as time increases as the particle goes around like this is going into the negative x-direction so velocity increases and it reaches a maximum here you can see at this point to its maximum then it starts decreasing and becomes

00:35:40 zero again at this point the velocity looks like this and then it reaches a maximum positive here so let me show the corresponding point this is point one point two point three point four this is 0.1 0.2 0.3 0.4 and then zero again at point one or point five and then it starts repeating itself VP here is DX D over DT which is nothing but minus R sine of Omega T there is the Omega here minus Omega R sine of Omega T how about the acceleration acceleration at point one is you can see the change is very large so it's going to be

00:36:39 negative and then it becomes zero at point two so one two becomes positive large at this point largest at this point and then goes down like this okay and this becomes the time period this is the time period acceleration 80 is given as DV over DT which comes out to be minus Omega square R cosine of Omega T which is nothing but minus Omega square X T okay so acceleration goes up comes down sir is exactly negative of this it's negative multiplied by Omega square R of course because this largest value

00:37:26 is going to be Omega square R so it's multiplied by sorry Omega square times X T and change of sign that's the acceleration so let's see what it means in simple harmonic motion XT has been given as R cosine of Omega T the corresponding speed or velocity is nothing but DX DT which is minus Omega R sine of Omega T and the acceleration a T which is nothing but DV over DT which is same thing as d2 X over DT square there is nothing but minus Omega square X I could also have written this whole thing

00:38:28 as the motion along the y axis so I could have written YT is equal to R sine of Omega T corresponding velocity VT is dy over DT which would be Omega R cosine of Omega T interesting thing is that the acceleration which is DV over DT which is nothing but d 2y over dt square is this 2 comes out to be minus Omega square Y that is it's again minus Omega square times the displacement or in general I can write a motion XT given as a constant a cosine of Omega T plus some other constant B sine of Omega T then

00:39:23 the velocity T would be equal to DX over DT which is minus Omega ay sine of Omega T plus Omega times B cosine of Omega T and the acceleration a T which is DV over DP which will be minus Omega square a cosine of Omega t minus Omega square B sine of Omega T which is again minus Omega square X T so what you see is that whether I take motion to be purely cosine purely sine or a combination of this the acceleration comes out to be minus Omega square X T and that is a typical sign of simple harmonic motion

00:40:20 in other words if I have an equation X double dot which is nothing but D 2 X by DT square if this is equal to minus some constant C times X where C is a positive number then I'm inverting the whole argument now I came from X derived the acceleration now I'm going to go backwards then X dot T is going to be of the form that was given and XT is going to be if I integrate this of the form a cosine of root C T plus B sine of root C T I can write X dot T as some coefficient a 1 cosine of root C T plus

00:41:15 B 1 sine of root C T so earlier I showed you that if a particle is moving around in a circle right uniformly that means its angular speed or V is constant then the motion comes out to be its X or Y component motion X component of the motion y component of the motion comes out to be a combination of either pure cosine Omega T or sine Omega T or in general when I combine the two it's a combination of the two the important thing is in this motion is that the acceleration comes out to be exactly minus Omega square times the

00:41:57 displacement let us understand that so when a particle is moving around in a circle you it is at this point here the only acceleration it has is centripetal acceleration which you know is V squared over R or Omega square R that's the magnitude and if I take its components its X component is going to be this in this direction if this is Omega T this is Omega T you can see this is nothing but minus Omega square R cosine of Omega T which is nothing but minus Omega square X similarly the Y component of

00:42:40 the acceleration is going to be minus Omega square R sine of Omega T which is minus Omega square Y T so the acceleration in this case the x and y component which show the simple harmonic motion the corresponding acceleration is nothing but minus Omega square times that displacement and that is simple harmonic motion so simple harmonic motion is nothing but a very specific case of periodic motions which of which I gave you several examples earlier so let us now solve a couple of problems one involving a periodic function and

00:43:25 the other one where we calculate the period of a particle performing periodic motion so in the first problem a function given as a function of time is equal to cosine of two pi T plus cosine of 3 PI T plot the function and find its period so let us see how this function looks if I were to look at cosine of 2 pi T right this has a period of T equals 1 how do I know that because cosine of 2 pi T at T equal to 0 is 1 and next becomes 1 is at T equals 1 because then it becomes cosine of 2 pi alright and cosine of 3

00:45:03 PI T at T equal to 0 is 1 and cosine of 3 PI T at T equals 2/3 is again equals cosine of 2 pi so it's time period is 4 2/3 what is the time period of the two functions together when they are added together I could plot the two functions so cosine cosine of 2 pi T is going to look like this where the period is 1 so this is T equals 1 and cosine of 3 PI T is going to look like its period is 2/3 so this is 0.5 this is here so it's going to look like this and the net result is the sum of two

00:46:29 it's not very obvious what the period is going to look like what the net function is going to look like but if you just play around what you're going to get it's going to look like something like this and you can see this is going to the period it is not very obvious what the period is going to be so for that I'm going to play a trick and write F T equals cosine of two pi T plus cosine of 3 PI T as cosine of 2 pi plus 3 PI by 2 T plus 3 PI minus 2 pi divided by 2 T that's cosine 3 PI plus

00:47:36 cosine of 2 pi plus 3 PI over 2 T minus 3 PI minus 2 PI over 2 T and that's cosine 2 pi T right so this is cosine of 2 pi T and this one is cosine of 3 PI T and therefore this becomes cosine of 5 PI by 2 T plus PI by 2 T plus cosine of 5 PI by 2 t minus PI by 2 T and you add them together you will get 2 cosine of 5 PI by 2 T cosine of PI by 2 T that's your function and now you can find the period easily so f at 0 is 2 times 1 times 1 which is 2 and I want to find at what time T is 2 again so 2 times cosine

00:49:00 of five PI T by 2 cosine of PI by 2 T and you notice ft is equal to cosine of 5 PI by 2 T times cosine of PI by 2 T so T equals 2 yes f2 equals 2 cosine 5 pi times cosine of pi and both are minus 1 which gives you 2 again so it's repeating after T equals 2 notice that is the smallest number T equals 2 that it repeats itself over so time period of this function is 2 and that's the answer for the second problem I'm going to take a particle moving along the x axis in a potential which is

00:50:12 let's say 1/2 m KX where m is the mass of the particle okay as a constant and on this side it is 1/2 m KX x equals 0 so potential it is moving in V X is equal to 1/2 m KX for X greater than 0 equal to 0 and is equal to minus 1/2 m KX for X less than equal to 0 in short the potential can also be written as 1/2 M K modulus of X you can see if I take a particle and leave it from one side it's going to come down go up again come down coop again and is going to perform a periodic motion so a particle with energy e is

00:51:28 going to perform a periodic motion notice that the motion is going to be periodic but not simple harmonic motion because for simple harmonic motion you need the potential to be of the kind one-half KX square because the force is linear so for simplicity I'm going to take e equals one-half M V 0 square so this immediately tells you this immediately tells you that V 0 is the speed when VX is equal to 0 Y my energy conservation e equals one-half MV square that is the kinetic energy plus one-half

00:52:28 M K mod X when mod X is zero potential is zero the one-half MV square is going to be total energy so V naught is the speed when the potential is zero and all the energy is kinetically so what we have now is that one-half and V 0 square is equal to one-half MV square plus one-half M K mod X I can cancel 1/2 m all throughout and I have V square equals V naught square minus K mod X now as this particle performs motion back and forth at any X a velocity V at X is given as square root of V naught square minus K mod X therefore if it

00:53:35 travels a distance DX the time taken is equal to only for traveling distance DX the time taken is VX which is going to be DX over square root of V naught square minus K mod X all right let us see now the farthest the farthest that the part of the goes to it travels between these two points the highest point the rightmost point in the leftmost point at these points of velocity zero so when V is zero right X gives you that point so V zero implies one-half M V 0 square equals 1/2 M K mod X 1/2 M 1/2 M cancels

00:54:30 and mod X equals V naught squared over K are the two points where it reflects so the higher point is V naught square over K and the leftmost point is V naught minus V naught square over K so now it is clear that the particle is performing motion between minus V naught square by K and V naught square over K and time taken from minus V naught square over K 2 V naught square over K is this and this is going to be 1/2 the time period because it is time taken from one side to the other and the total time taken

00:55:10 will be exactly the same when it goes back so the time period divided by 2 is the time it takes from leftmost point to the rightmost point and this is going to be minus V naught square over K to 0 DX over square root of V naught square minus K mod X which for X less than 0 is plus KX plus 0 to V naught square over K DX over we notice co-head minus KX square root you figured out that the time period t by two is equal to minus V naught square over K to 0 DX over square root of V naught square plus KX plus zero to V

00:56:07 naught square over K DX over square root of V naught square minus KX you can simplify this further in this integral the first integral take Y to be minus X or X to be minus y so DX is minus DUI then you get T by 2 equals integral with a minus sign dy over square root of v naught square minus KY and this is minus V naught square by K will become plus V naught square by K 2 0 plus the second term remains the same 0 to V naught square over K DX over square root of V naught square minus KX which we can

00:56:59 write as plus zero to V not square over K dy over square root of V naught square minus KY plus zero to V naught square over K DX or dy doesn't matter because this is variable over which we are integrating so this T by 2 therefore is two times zero to V naught square over K dy over square root of V naught square minus KY now the integral is very simple you take y equals V naught square over K sine square theta therefore dy is two V naught square over K sine theta cos theta D theta and the limits are from 0

00:58:01 to PI by 2 so T by 2 becomes 2 times 0 to PI by 2 dy which is 2 v-0 squared over K sine theta cosine theta D theta divided by V naught cosine of theta so you get 4 v naught over K this cosine theta cancels and D Rho 0 2 pi by 2 sine theta D theta which is nothing but 4 v naught over K and therefore the time period is 8 v not only that is the answer and frequency of motion is going to be K over 8 0 in the next lecture I'm going to be focusing more on simple harmonic motion look at this equation of

00:59:08 motion based on what we have learnt so far [Music]