good morning everybody today we will continue with periodic motion what we have this in this chapter we will describe the oscillations in terms of amplitude period frequency and angular frequency we will talk about the periodic motion in addition to that we will study the ideal cases and we will learn how to apply the ideas of simple harmonic motion for different physical situations and then in the next lecture we will analyze the motions of a pendulum and we will learn what is the damping in oscillations so some oscillations will die out after certain time which is called as damping and we will also learn the damp oscillations in the next lecture in addition to that we will learn how a driving force applied to an oscillator at a particular frequency can cause a very large response or resonance we will also learn forced oscillations in order to keep the oscillation constant you have to apply some certain forces we will discuss all these things actually it is very simple lecture we will go step by step and try to stay on the line okay you will use your previous information okay why we are dealing with periodic motion why we are dealing with oscillation because many kinds of motion such as pendulum musical vibrations pistons in car engines also pistons in the industry okay it has many applications and this kind of motion repeat themselves this work is very important look at the piston here what do you see the piston is moving up and it goes back up and down up and down okay this is the movement direction of the piston and here you see a crankshaft which is rotating around certain rotation axis and it is also repeating its motion okay so we are talking about repetition of the motions balls piston and crankshaft repeat their motion okay so this is periodic motion or you can say oscillation so this kind of motion has many applications in industry in real life so very important topic but the basics of this motion is very simple you can easily understand in addition to that for the first coming chapters for example when we are dealing with waves mechanical waves or sound waves or when we are dealing with alternating electric currents or light we will use periodic motion information which we will learn today so now let me describe what is periodic motion what is the reason for this type of motion here you see a spring okay left part of this spring is attached to this object here and this part cannot move and right side of the spring is connected to a glider okay and glider can move on this frictionless surface and at the beginning this glider is staying in equilibrium position spring is relaxed okay so let's consider that this is the origin and equilibrium position and the mass and spring are at rest okay so if you pull this glider to the right side by applying some force it's considered you apply force to the right side on this glider then you stretch the spring okay so you remember from your old information if you scratch the spring you will store the elastic potential energy in the spring and there will be elastic force in the opposite direction to the applied force then this glider will come back okay if you release this glider here at some certain point so it will go to the equilibrium position if you release the glider then it will pass through this equilibrium position it will come back to this position and then this spring will apply another force in the positive x direction then the glider will move now in positive direction positive x direction so if the surface is frictionless and if you don't lose any energy in the system then this will have a periodic motion the glider will move between this expositions okay this is the periodic motion or you can say oscillation of the system the system will oscillate between these two points okay it will go back and forth okay so it will repeat this motion so this is periodic motion now let me discuss this situation in detail let's consider the glider here is located at some certain x point not in the equilibrium position it is far from the equilibrium position on the right side then here we have x is bigger than zero okay we are talking about this x here so then what about the force f x applied by the spring spring will apply this kind of force along the negative x direction f x here okay and due to this force applied by spring on the glider we will have an acceleration okay we have net force and we have acceleration and acceleration is also along the force along the negative x direction so then the stretched spring pulls glider toward equilibrium position this x is zero is the equilibrium position and this force applied by this stretch spring on the glider pulls the glider to the equilibrium position now let's try to write or draw the free body diagram of this glider what is the free body diagram of glider we have mg weight of the glider we have normal force they are in opposite directions so along the y-axis we have zero net force and along the x-axis we have only force applied by the spring on the glider okay what about the friction force this is a frictionless surface for this reason we don't have friction force okay a resistance is also ignored in this discussion so now let me continue just consider that due to this force applied by the spring on the glider this glider came to the equilibrium position equilibrium condition this is the equilibrium condition at the origin okay so now x is zero what about the force applied by this spring on the glider the force applied by the spring on the glider is zero here because the glider is in equilibrium condition okay and you also remember that this force this force applied by the spring is not constant it depends on the x it depends on the position of the glider don't forget this one remember your previous information okay so since spring exerts no force on the glider then the glider has zero acceleration so we don't have any acceleration here since we have zero net force again if you draw the free body diagram along the y-axis we have normal force upward direction we have mg weight in downward and then in along the x-direction x-axis we don't have any force so now let's consider that the glider displays to the left from the equilibrium position this is the equilibrium position where x is zero and now the collider is located on the left side of the equilibrium position this is the x okay so you store elastic potential energy in the spring and spring will apply a force on the glider and the direction of this force is along the positive x okay if the x is negative then the direction of the f x is positive along the positive x direction so then what we have we have positive force along the positive x direction and we have positive ax acceleration so this compressed spring pushes glider toward equilibrium position okay so then what about the free body diagram here for this condition this is the weight of the glider mg downward and we have normal force upward they are equal to each other and net force along the y axis is zero and we only have fx force applied by the spring along the positive x direction and moles acceleration and force are in the same direction if the x is negative so look at the first condition if the x is positive then balls f x and a x are negative okay so due to this force applied by the spring we will have a periodic motion okay then here i will try to explain some important terms in periodic motion so be careful here very easy to understand just stay on the line so here we have the first term amplitude amplitude in turkish gandik okay here we show it with capital a symbol which is the maximum magnitude of displacement from equilibrium so just consider that on the right side here we have glider it is staying in the equilibrium position okay where x is zero then i have compressed this glider in this negative direction okay so this is the x so amplitude defines this maximum magnitude of displacement from equilibrium you can also consider that just you stretch the spring you change the position of the collider for example to the positive side of the equilibrium then here you have x okay then this is the new position of the glider on the right side in the positive side okay so the maximum x gives us the amplitude okay the maximum magnitude of displacement from the equilibrium in the maximum displacement x is equal to the amplitude okay amplitude is this one it is related to the displacement and what about the period period shown me with the capital t it is the time for one cycle what i mean with the one cycle again let me explain on this graph just consider that this is the maximum compressible position for the glider where we have x is equal to minus a let's say and here we have maximum stretchable position for the x here x is positive a okay and let's consider that this glider is moving between this two x positions okay so it has periodic motion between these two positions of x the period here on the left side defines the time for one cycle what i mean with the one cycle just consider that first of all you have compressed the spring here at this position okay let's consider this is the starting point and then you release the glider then glider will move in positive x direction so it will reach to the maximum x displacement here positive side in the positive side then it will go back to the initial condition so this is the one cycle okay it started from this position and then it comes back to the starting position so this is one cycle and this is called as period how long does it take to reach to the starting point you can consider like this and what about the frequency frequency is the number of cycles per unit time so sometimes this periodic motion repeats itself very fast in very short time okay i'm talking about less than one second in microsecond in nanosecond in femtosecond okay so this system this periodic motion repeats itself so it is moving very fast between these two x positions okay so then let's say within one second we have 10 cycles or within one second we have 100 cycles so the number of cycles per unit time gives us the frequency of the motion and a psi unit of frequency is given with hertz one heart is given with this symbol and it is equal to one cycle per second and it is usually shown with this one one cycle per second okay and we have also another term the angular frequency which is shown with the omega is given with 2 pi times the frequency here we have omega omega is given with 2 pi f f is frequency and you know of 2 pi we have learned in circular motion i will go into detail related to this angular frequency but just keep in your mind that we have relation between angular frequency and frequency of the periodic motion and they are completely different things okay be careful when you are dealing with periodic motion in the problems if angular frequency is given then don't confuse with the normal frequency okay they have this type of relation here i would like to give you information about angular frequency angular frequency omega is a useful quantity it represents the rate of change of an angular quantity not necessarily related to a rotational motion that is always measured in radians okay so the unit of angular frequency is given with radian per second okay this is the unit of the angular frequency and remember that the unit of the frequency is given this cycle per second and if you put here cycle per second if you put here radian per second for the omega then for the two pi you can get this one radian per cycle for the two pi okay so i will also discuss the omega in my forthcoming transparencies here i will not go into detail so here we have defined four terms amplitude period frequency and angular frequency now let me give you information about the relation between period and frequency so frequency is given with the reciprocal of period or period is given with the reciprocal of the frequency so they have this relation frequency can be written with one over period period can be written with one over frequency period is the time and frequency is the cycle per unit time you can consider like this and what about the angular frequency and period in the previous transparency we have given that the angular frequency is given this four pi f and here we have frequency instead of frequency you can use this expression one over t so then you can write the angular frequency in terms of period t let me give you one bio application wing frequencies of the birds the robbie trotted hummingbird hummingbird is synecushu normally flaps its wings at about 50 hertz what is the meaning of 50 hertz look at the definition of the hertz hertz is one cycle per second okay this is the frequency so here we have 50 hertz 50 hertz means 50 cycle per second okay this bird can flap this wings 50 times per second 50 cycle per second it is huge but if you compare this to insects it is much smaller just have a look on insects insects can flap their wings at even faster rates from 330 hertz 330 cycle per second for a house fly karasinic and 600 hertz for a mosquito to an amazing 1040 hertz for the tiny biting mitch this is another type of fly and it has this frequency and mosquito silversonic it has this frequency 600 cycle per second it is huge frequency let me continue with the example 14.1 from the book what we have here this is an ultrasonic transducer used for medical diagnosis it oscillates at 6.7 million cycle per second or 6.7 megahertz or 6.7 times 10 to 6 hertz how long does each oscillation take and what is the angular frequency so frequency is given in the question it is given with 6.7 megahertz and period is asked and also angular frequency is asked within the question so period is given with 1 over f just put the frequency here is hertz so remember the hertz hertz is this one one cycle per second so you can take it here a second then here we have microsecond this is the period is the time then we have unit for the time microseconds okay and what about the angular frequency angular frequency is given with this expression just put here two pi radian per cycle and just put here the frequency cycle per second this cycle will cancel this cycle and then we will have radian per second for the angular frequency so what do you see here very rapid vibration with large f and omega angular frequency and small period so each cycle each cycle is completed within 0.15 microseconds and within one second we have 6.7 million cycle so it is very rapid vibration in ultrasonic transducers they are used in hospitals maybe maybe you already know maybe you have seen in the hospital they are used for the medical diagnosis do you have any question here okay now let me continue with the simple harmonic motion generally when the students see this name or hear this motion simple harmonic motion they think that this is very complicated phenomenon no it is not true it is just a perfect ideal periodic motion okay we will talk about the periodic motion simple harmonic motion is simplest kind of oscillation simplest kind of the periodic motion okay which occurs when the restoring force fx is directly proportional to the displacement from the equilibrium so what i mean here we have glider again it is stretched to the x position here we have x is equal to zero this is the equilibrium condition and here we have x is equal to x the spring is stretched a little bit and then we have a restoring force this one f x applied by the spring on the glider so what about the f x f x is given with this expression minus k times x x is the position and this is the force applied by the spring this is also called as restoring force okay we have discussed this one in the previous lectures so what do you see here here we have k force constant here we have k force constant okay so this constant is constant for certain material we have discussed this one we have also given some tables okay so if you increase the x then this spring will apply bigger force on the glider if you decrease the x then the spring will apply a smaller force on the glider if the x is equal to zero then fx will also be equal to zero in the equilibrium condition okay so what we say here that if this restoring force this fx is directly proportional to the displacement from the equilibrium then we have simple harmonic motion okay this happens if the spring in the figure here if the spring here is an ideal one that obeys hooke's law okay we have discussed in chapter six so then we can write this expression this f x is the force applied by the spring on the glider or called as restoring force exerted by an ideal spring and this is the force constant of the spring and this is the displacement x on the right side then this restoring force is directly proportional to the displacement so then we have simple harmonic motion but here again i would like to remind you that we have considered that there is no friction on the surface between the glider and this air track okay we considered that this is frictionless surface and we have also ignored the air resistance so mechanical energy is conserved here okay we are talking about the ideal case so then this this motion will be kept okay in certain amplitude and it is called a simple harmonic motion so what about the relation between the restoring force and this placement here we have fx in this axis restoring force fx and here we have displacement x on the right side this axis is the displacement and this axis is the restoring force and we have this type of relation if x is positive restoring force the force applied by the spring is along the negative x direction then it has this type of relation what do you see here the force is directly proportional to the x i increase the x then force increases i increase the displacement then the restoring force increases in negative direction i'm talking about the increase in magnitude of restoring force and if i increase the displacement then i will increase the force okay so the force restoring force applied by the spring on the glider is directly proportional to the displacement from equilibrium if we have an ideal spring okay if you compress this spring if the x displacement is negative then we have force positive what is the meaning of that the supreme pushes the glider to the equilibrium condition it applies a force along the positive x direction so we have this displacement then we have this force this displacement and this force okay so you increase the displacement then the forces increase so then the restoring force is directly proportional to the displacement from equilibrium then for this condition we have a motion the glider is moving back and forth between positive x negative x positive amplitude negative amplitude then this motion is called as simple harmonic motion so you see the definition of simple harmonic motion is very simple okay so nothing needs here for you then let me define the acceleration in simple harmonic motion so what we have discussed here we have a net force in simple harmonic motion okay so if we have net force along the x-axis then we should have an acceleration right so what about this acceleration i will discuss this one so if you take the time derivative of x displacement you can calculate the velocity if you take the second time derivative of the x then you can calculate acceleration okay and it is given this f x over m so net force is given with f x which is given with m times ax and ax is given this this expression okay then you can write this expression but what is fx fx here is equal to minus k times x right then if you take acceleration from this expression then you can write the acceleration in this form minus k over m times x force constant of restoring force this k x is displacement m is mass of the object mass of the collider in this discussion okay then this is the relation between acceleration and displacement in simple harmonic motion what do you see here if the x is displacement is positive then acceleration is in negative direction because here in front of the x here we have negative sign and if x is negative then here we have a positive ax okay so these are the conditions and relations between the displacement and acceleration i have already discussed this one i have shown you here so look at this one so here the x is negative the spring is compressed and this is the direction of the force applied by the spring it is along the positive x then we have positive acceleration along the positive x direction or here if the x displacement is positive then the restoring force applied by the spring is in the negative direction then acceleration is in negative direction so this equation which we have calculated here will give us the same result okay this is mathematical representation of this physical result any question here related to the acceleration of the simple harmonic motion so here i would like to give you very important information related to the equation for simple harmonic motion look at the acceleration is this acceleration constant or non-constant in simple harmonic motion here we have non-constant acceleration because acceleration changes as a function of x when the x is zero force is zero acceleration is zero okay if you have bigger x then you have bigger acceleration in opposite direction if we have smaller x smaller displacement than we have smaller acceleration but in the opposite direction compared to the displacement so then this acceleration is not constant it changes as a function of the displacement then you cannot apply constant acceleration equations which we have learned in chapter two okay you cannot apply this equations these equations which we have discussed in chapter two because here we have non-constant acceleration don't forget this one here i also would like to give you another definition for the harmonic oscillator if a body undergoes simple harmonic motion it is called a harmonic oscillator okay so it oscillates what is the meaning of oscillation it repeats its motion with some certain amplitude and period and frequency okay so if a body has a simple harmonic motion then it is called as a harmonic oscillator what else so we will use this simple harmonic motion as an approximate model for many different periodic motions for example the electric current in an alternating current circuit for example at home we use ac current right we have 220 ac volt in the cables which electricity comes from the power plant station to the house okay we use ac current so it changes by time it is fluctuating in physics 2 you will learn that and this information which you have learned related to the simple harmonic motion will be useful for you in addition to that for very important scientific applications for example the oscillations of atoms in molecules or in solids this simple harmonic motion is also used so simple harmonic motion is an ideal case for the periodic motion okay do you have any question here then let me discuss the difference between the ideal case and real case what happens in real applications so in ideal case this restoring force applied by the spring is directly proportional to the displacement x okay then we have this type of curve this type of relation between force and displacement i have already shown you so you have displacement in positive x and then you have fours in negative direction if you increase the displacement then the magnitude of the force restoring force will be increased okay this is an ideal case if you have displacement in negative direction then we have positive force positive restoring force and the magnitude of the restoring force increases with displacement okay this blue curve here this dashed line is the ideal case which represents this equation but in real cases the restoring force deviates from hooke's law okay what do you see here it is shown with the red color here this one what do you see here that in real cases the restoring force deviates from hooke's law deviates from the ideal case okay it can be happened due to the spring properties okay it can be happen due to the friction force okay it can be happened due to the air resistance okay but in real cases the situation is a bit different any question here then let me continue with the relation between uniform circular motion and simple harmonic motion be careful it is very easy to understand if you understand this picture then it is very easy for you so look at this setup here we have a ball okay this is the ball and just consider that this ball is rotating along this circular path this is reference circle and just consider that this ball here is rotating around this circle okay then here on the left side you have light sources or single light source okay then you send light from this side from left side to the right side and here we have a screen okay let's say this is screen and what do you see here if you send light you will have shadows of the ball here on the screen right and if this ball is rotating in this circular path then the position of this ball will change by time on the screen okay so if the ball is here then you will see shadow of the ball here due to this light source on the left side okay now let's discuss this example here in this picture here this is the side weave of this experimental setup and here this is the top weave of this experimental setup here we have ball the name of ball is q this is the cue ball and it is rotating on turntable okay it has a circular pass with some certain angular speed it is rotating on this turn table okay this is the distance to the rotation axis of the wall and here we have a light source then you send light so if you send light you will see the shadow of the ball here just consider that ball continues its motion in this direction then it reaches to this point then now shadow of the ball comes to this point and just consider that ball continues here it's counterclockwise motion then just consider that the ball is here then the shadow on the screen is located here then the ball goes to its circular motion and it reaches here again let's say then the shadow again here about the origin and it continues to its circular motion then now the ball is located here on the right side then its shadow is located here okay so this cue ball has uniform circular motion on a rotating turntable it is rotating with some certain angular frequency okay then you are checking the shadow of this ball what do you see here on the screen look at the screen on the screen let's consider this is the initial position of the ball and it goes to the equilibrium it passes through the equilibrium and it reaches to the maximum amplitude in negative direction then from this point then it goes back it comes to the equilibrium and it goes to the maximum positive amplitude and it goes back and forth back and forth back and forth what do you see here this is more or less the motion of the spring a glider okay just consider here a glider attached to a spring and this is the origin this is the maximum amplitude maximum stretched foundation this is the maximum compressed condition okay this is motion of the glider with the spring okay this is periodic motion or simple harmonic motion so then we have relation between uniform circular motion and simple harmonic motion okay so now let me continue if everything is clear here please let me know if something is missing here the mechanism is very simple if you understand this figure then you can easily understand the force coming equations so remember the circular motion we have tangential velocity in circular motion okay and this tangential velocity is given with r times omega what was r r is the radius and here r is a okay a maximum displacement from the equilibrium okay then you can write velocity of the ball which is this expression speed of the ball linear speed of the ball is given with the amplitude times angular speed or radius times angular speed okay just keep this information in your mind because here we have uniform circular motion in addition to that this ball has certain acceleration this is given with a radial okay it is given with v square over r okay and what else here again instead of a radial i can write this expression v square over a because radius is given with a or instead of v square i can write this expression just take the square of this one r square omega square over a so what is r r is a here then i can write it with this one a square then a radial can be written with a times omega square okay so you know all this information from the rotational motion i will not go into detail so now let me come back to this transparency so here we have a cue ball and this cue ball is moving around this circular pass it has uniform circular motion with some certain radius with some certain radial acceleration okay here we have the diagram of this motion here we have a definition related to this circular motion the circle in which the ball moves so that its projection matches the motion of the oscillating body is called reference circle what i mean here here we have a circular pass and the ball has circular motion around this circle and here we have a periodic motion okay so if this circular motion can be expressed with this periodic motion then this circle is called as reference circle okay this circle in which the ball moves so that its projection matches the motion of the oscillating body is called the reference circle and this cue ball is around this reference circle so now what about the speed of this cue ball and what about the acceleration of this curable so we know that as 0.2 moves around the reference circle with constant angular speed we are talking about omega vector or q here this one this vector or q vector rotates with the same angular speed so this q vector is rotating around this circle and such a rotating vector is called as phaser we have learned this one in the previous chapters and this phasor information will also be useful in the physics too okay so this is the phaser and now let's define the terms here this is the radius of the circle or maximum amplitude of the periodic motion simple harmonic motion okay this is the theta the angle between this vector or q vector and the x axis and here we have p point projection of this q during this circular motion okay and here we have x what is x displacement displacement in periodic motion so x is zero when the system is in equilibrium and it is moving along the positive x direction it reaches to its maximum amplitude and it goes bad and it comes to the maximum amplitude in negative direction okay so this is the x displacement it can be positive or negative in periodic motion here along this x-axis i am shoving the periodic motion of this true ball okay so since this cue ball is moving in this circular path this uniform circular motion then i can express this motion in terms of periodic motion simple harmonic motion okay then i am writing x x is given with a times cosine theta what is a a is the amplitude or radius of the circle here we have theta and x is given with a times cosine theta is it clear any question here then let me continue ball moves in uniform circular motion around this circle and shadow of the ball moves back and forth on x-axis in simple harmonic motion okay here so now let me continue so we have x which is given with a times cosine theta so what about the velocity what about the acceleration okay this is the cure ball it has uniform circular motion and this is the tangential velocity of the cuboid okay which is given with v q and this tangential velocity has this component along the x-axis and this component along the y-axis okay so then when the q is at q ball is at this position its projection p will have this velocity which is equal to the x component of v q okay the x is equal to the x component of the v q so what is the x component of dvq here we have theta if you apply the geometry then you can calculate vx is equal to minus vq times sinus theta y minus because the direction of the velocity is in negative direction in this case okay this is the velocity do you have any question velocity of the projection of the cue ball along the x-axis so since the point q is in uniform circular motion it has uniform circular motion the acceleration vector is given with this one okay this is a radial or a q okay and what about this a radial i have already calculated this one how to calculate this one let me show you again what was the a radial remember from uniform circular motion it is given this v square over r and r here is given this amplitude v square over a and what is v v tangential velocity v q v q tangential velocity is given with r times omega in uniform circular motion right and i can also write it like this a times omega amplitude r is equal to amplitude and just to put this v curve here instead of v then you can calculate a radial or a q radial acceleration of the cue ball which is equal to omega square a okay this one here so now we have x position of the projection of the cue ball we are dealing with the projection of the cue ball on the x-axis so we have displacement of the p point of the cuboid we have velocity of the p point of the cube ball we have acceleration of the p point projection of the curveball okay so we have calculated the x v x and ax along the x-axis so then we have found a relation between uniform circular motion and simple harmonic motion do we have any question up to this point related to the definition of x v x and a x any question here i have forgotten something let me also discuss this one we have calculated radial acceleration of the cuboid which has uniform circular motion but what about the x component of this acceleration x component of this acceleration is given with minus aq which we have calculated here times cosine theta this minus sign is showing the direction of the acceleration then let's consider cue ball is here then the projection will be here projection will be here then in this condition we will have this vx and we will have this ax along the positive x direction so if you understand the physics others are mathematics then let me continue we have calculated aq this this one omega square times a a is the amplitude or radius of the circle okay then ax x component of this acceleration is given with minus aq times cosine theta okay so what is aq acceleration radial acceleration of the cue ball just take this one put it there we have calculated aq then here we have omega square times a this is the radial acceleration of the curball then here i have a times cosine x a times cosine x what is this a times cosine x this is the x okay this is the x displacement so instead of a times cosine x you can write x then x component of the acceleration is given with minus omega squared times x remember that in the previous transparencies we have calculated this ax in this form we have minus k over m times x so here we have ax here we have ax so by using these two expressions you can get relation between omega square and k between omega square and m or you can get this expression omega angular frequency is given with square root of k over m this is the force constant this is the mass of the object do you have any question here okay now let's discuss the characteristics of simple harmonic motion actually i will summarize what i have found here we have found this relation by using these two equations for the accelerations then we have angular frequency for simple harmonic motion which is given with square root of k over m this is the mass of the object this is force constant of restoring force okay and frequency for a simple harmonic motion is given with this expression omega is given with 2 pi f if you take frequency it will be given like this omega over 2 pi so instead of omega just use this one then you can write the frequency in this form one over two pi square root of k over m and since the period is given this one over f just put here f then you can calculate the period in this form so what do you see here this angular frequency is independent from the amplitude we don't have any amplitude in equation of angular frequency look at the frequency the centifrequency we don't have amplitude it is independent from the amplitude look at the period it is also independent from the amplitude okay amplitude is not included within this equations so what about the real applications here we have times with different sizes with different masses this is a time with the biggest mass and biggest size okay then it has low frequency frequency is 128 hertz and here we have the smallest time of signs okay it has the smallest mass and smallest size so due to its small mass it has high frequency so look at the equation of the frequency force constant of restoring force if these tines are made from the same material they will have same k and only m affects the frequency so here we have small m then you have high frequency here we have bigger m than since m is bigger than the frequency is lower okay low frequency let me finish this part with this transparency and then i will continue with the characteristics of simple harmonic motion after the break so this is the example 14.2 from the book angular frequency and period in simple harmonic motion period is time so as spring is mounted horizontally this is the spring here in figure a with its left end fixed as spring balance attached to the free end and pulled toward the right indicates that the stretching force is proportional to the displacement and a force of six nifty causes a displacement of 0.03 meter so this is the equilibrium condition for the spring where x is equal to zero this is the stretched condition and this is spring balance spring balance in turkish yali contact okay so you apply force six newton to the right side and it causes a displacement of 0.03 meter we replace the spring balance with a 0.5 kilogram glider pull it all point or two meter to the right along a frictionless air track and release it from rest so you remove the spring balance from the system and instead of this spring balance you use a glider okay then you stretch this spring to this condition and you release this system from this condition so since it is stretched the restoring force applied by the spring will try to move the system to the equilibrium condition okay so then f spring will be in this direction so what is the question find the force constant k of the spring k is asked find the angular frequency omega frequency f and period of the resulting oscillation so when you release this glider from the rest it will move back and forth back and forth okay so it will have this type of motion so then this motion will have angular frequency will have frequency we'll have period okay so these are asked within the question so what about the force constant k we know that fx is given with minus kx so if you take k from this expression you can write like this what is f x the force applied by the spring here we apply a force in positive x direction by hand the magnitude of force is sixth newton and at the same time spring applies a big force okay it's the same magnitude but negative sign okay so then instead of f x just use this minus six newton and instead of x just use this 0.03 meter which is given here in the question and then you can calculate force constant of this spring 200 kilogram per square second and in the question in the second part of the question when you put a glider here with mass of 0.5 kilogram so what about the omega what about the frequency what about the period of the oscillation you know of the k force constant you know the mass of the glider then omega can be found in 20 radian per second in case of frequency you have calculated omega you know of the 2 pi it is given this radian per cycle then the result is 3.2 cycle per second or 3.2 hertz and period is the reciprocal of the frequency just put the frequency here then you can find the period 0.31 seconds do you have any question here okay in the previous part of my lecture we have discussed the displacement as a function of time for simple harmonic motion and we got this expression x displacement is given this a times cosine theta this is the a radius of this circular motion or the maximum amplitude in the simple harmonic motion this is the theta and this is the displacement and x displacement is given with this expression we have also calculated v x we have also calculated a x for simple harmonic motion of this cue ball okay now let's try to write this theta here in cosine theta here we have a theta and just try to write this theta in terms of omega and time so here we have theta which is given with omega times t plus phi so how can i write this equation remember the displacement is given this velocity times time right and you can consider that this theta is angular displacement okay so this one and this is the angular velocity this is linear velocity this is time this is time okay so by using this analogy you can write this expression so theta is given this omega times t and here we have plus five so what is the phi just consider that this arrow phasor is here and theta is equal to 0 okay and in the second condition another system and theta at the beginning is equal to a certain phi angle theta is different from zero now for the second phasor vector okay so here we have phi angle okay if the phi is equal to zero tan theta is given with omega times t okay but usually we define theta with this expression omega times t plus phi so here we have x displacement for a simple harmonic motion a cosine theta instead of theta just use this expression then finally you can get displacement in simple harmonic motion a times cosine instead of theta we have omega theta omega times time plus phi phase angle okay phase angle means that at the beginning you have some certain angle okay the system starts from this condition or phi can be zero here and the system can start from the zero okay so then this is the displacement if you take the time derivative of this displacement you can calculate the velocity if you take the time derivative of velocity you can calculate the acceleration okay so then i will do that so before calculating velocity and acceleration just look at this equation and just try to show this equation on a graph here we have time axis from origin to certain time and here we have x displacement okay here we have amplitude this is the maximum amplitude this is the maximum negative amplitude this is the origin equilibrium just consider this system the supreme attached to a mass okay let's consider this is origin x is equal to zero let's consider this is maximum amplitude let's consider this is maximum negative amplitude okay so this system is oscillating between these two positions it has simple harmonic motion it has periodic motion okay now what about the position displacement is a function of time so if you change the time the position will change so here we have position versus time graph okay so what do you see here when the time is zero then the time is 0 cosine 0 is 1 right just consider that at the beginning phi angle is 0 and time is 0 if the time is zero if the phi is zero here we have zero then cosine zero is one then x will be equal to a okay so what do you see here look at the graph x is equal to a okay then after certain time time changes then the position changes okay now the system is in equilibrium and system goes to the negative maximum negative amplitude okay so just consider that you have started from this position mass is here at the beginning okay then spring force is in negative direction and pulls the system to the equilibrium condition now we are in equilibrium after certain time and after certain time we reach to the maximum negative amplitude okay and it will go back to the equilibrium it will go back to the maximum positive amplitude it will go back to the equilibrium it will go back to the maximum negative amplitude equilibrium maximum amplitude equilibrium maximum negative amplitude okay so it will go like this so this was the starting point of the mass okay then it has back and forth motion periodic motion simple harmonic motion so this was the starting point and one cycle is completed within time of period okay the period t is the time for one complete cycle of oscillation so in order to reach to the initial position now we are in initial position here we have one period okay and in order to complete a second cycle then we have another t period in order to complete another cycle we have another t period okay this is the displacement x as a function of time in simple harmonic motion okay we have investigated the position of x displacement as a function of time we have investigated this one now let me continue so this is the x x is given with a cosine omega t plus phi and here we have omega and in the previous transparencies we have calculated that this omega is given with square root of k over m okay so this is the x this is the time okay so what happens if i manipulate m just consider that here we have a mass m m1 let's say and here we have a spring with some force constant k and here we have equilibrium condition x is zero okay so then it has certain simple harmonic motion around this equilibrium just consider that you are using the same spring everything is same but now you have a bigger m okay now we have m2 bigger m and the amplitude is same so we have same force constant of k everything is same you only change the mass of the object so if you change the mass of the object then you can change the angular frequency of the system if you change the angular frequency then you will change the x as a function of time so what do you see here mass m increases if you increase the mass then angular frequency will decrease then you will have different curves okay here i have blue curve this blue one with some certain mass and if i increase the mass then i have this purple curve if i further increase the mass i have this let's say green curve okay so x versus time graph changes if i change the mass in this example okay because if i change the mass then i change the omega angular frequency and if omega changes then x changes x t graph changes okay so what about the period period is given with this expression which we have calculated in the previous transparencies which is given with one over frequency it is also given with 2 pi over omega so omega is given with this expression then i can write the period in this form 2 pi square root of m over k so if you increase the mass then you can increase the period if you decrease the mass then you can decrease the period so look at the graph you can look at the xd graph what happens to the xt graph if you increase the mass so what is period this is the starting point just for the blue curve and it goes to the initial point in usual x condition so this is called as period t okay so look at the second curve period is increased look at the third curve period is increased why because we have increased the mass if you increase the mass of the object in the simple harmonic motion then you increase the period okay so this is the relation between mass and period mass and displacement do you have any question here okay here let me discuss the k just increase k force constant is increased here again we have x which is given with a cosine omega t plus phi this is the x t graph okay and we are increasing decay so look at this angular frequency equation if you increase the k force constant then angular frequency will increase if you decrease dk then angular frequency will decrease just consider that m is fixed it is constant during this example a look at the period if you increase the force constant then period will decrease if you decrease the force constant then period will increase so what is the meaning of different force constant here we have the same mass m and here we have a spring with some certain force constant k1 and let's consider this is the equilibrium condition here we have same mass mass is constant and we have different spring with different force constant now we have k2 okay so if you have different force constant if you have different spring here in this example then you will have different period of the system different period of the simple harmonic motion so here look at the xt graph look at the xt graph if you increase decay what do you see here at the beginning we have blue curve and period is this one this is the period t and in case of purple curve if you increase decay then period will be decreased now we have lower we have lower period okay this is the initial period and this is the second period if you further increase decay then period will be further decreased okay this is the relation between period and force constant of restoring force and the relation between displacement and force constant k now let me discuss the amplitude effect of amplitude on x and effect of amplitude on period so look at the x within the x we have a here and we have omega look at the omega k is constant m is constant in this example we only increase the amplitude so angular frequency is independent from the a period is independent from the a so change in the amplitude cannot affect the angular frequency cannot effective period cannot affect the frequency but this x will be affected with the change in the amplitude because here we have amplitude if you increase the amplitude you will have increased amplitude this blue curve is the first case we have amplitude a this is the period okay and this is the amplitude if you increase the amplitude period is same because period is independent from the amplitude but amplitude is increased we have different xt graph if you increase amplitude further then you will have this type of behavior of xt graph but the period will be same with the previous conditions you only increase the amplitude you see here we have a let's say we have bigger a here we have bigger a here do you have any question here now let me discuss the effect of phase angle phi on the xt graph and also on the period so look at the equation of period period does not depend on the phi angle angular frequency does not depend on the phi angle but x depends on the phi angle this is the phase angle i have explained okay in the previous transparencies if the phi angle is zero then let's consider time is zero phi is zero and let's consider time is zero this will also be zero then i will have a cosine zero then it is equal to a okay i will have this amplitude this condition for the displacement then i have this blue curve as a function of time this time x condition changes what we mean again let's consider we have a spring and we have mass here and this is the equilibrium condition and this is changing between these two amplitudes this is x is equal to a this is x is equal to minus a and this is x is equal to zero okay we have simple harmonic motion periodic motion less than this amplitudes so if the phi angle is zero then we start from this one when the time is zero okay then after certain time again we are in the same position this is the period okay the time to complete one cycle okay so just consider that phi angle is pi over four so instead of phi here just put 5 over 4 then this cosine function will change and then x will change then we will have this amplitude at the beginning okay and but of course what about the maximum amplitude maximum amplitude for the first case and second case are equal to each other but the system starts from this one if we have certain initial phi angle if the phi angle is pi half then it starts from this position okay here we have just look at this condition x is equal to a cosine omega t plus instead of phi just use pi half and when the time is 0 then this is 0 then i will have a cosine 90 degrees then it will have 0 then the position of the x is zero it is in equilibrium okay then by time it will have this type of motion the x will change so what do you see here amplitude is again same maximum amplitude it will oscillate within this range but in the first case when the phi is zero the mass is here it is initial condition for the mass when the phi is equal to pi over half the mass is here starts to oscillate between a and minus a from this position it will oscillate like this okay or you can start from this position so phi angle defines the starting position for the simple harmonic motion and it does not affect the period period is same it does not affect the amplitude it does not affect the angular frequency it does not affect the frequency it only affects the exposition by time do you have any question here then let me continue with the velocity and acceleration in simple harmonic motion if you know the displacement then you can easily calculate the velocity you can easily calculate the acceleration because velocity is given time derivative of this function and if you take the time derivative of this function here we have a here we have cosine omega t plus phi so time derivative of cosine function is given with minus sinus function and here in front of time here we have omega so time derivative of omega t will be omega minus omega a sine omega t plus phi and this is the velocity in simple harmonic motion and if you take the time derivative of velocity or second time derivative of the x function then you can get this expression the time derivative of sinus function is cosine function and the time derivative of omega t is omega here we have another omega then here we have omega square okay this is the acceleration in simple harmonic motion look at this expression acceleration in acceleration we have here a cosine omega t plus phi we'll get the x a cosine omega t plus phi then this is the x then this acceleration in simple harmonic motion can be written in this form minus omega square times x displacement look at this omega omega is given with square root of k over m if you take the square of this one then you will have your k over m x okay look at this k times x minus k times x this is f x okay then here we have f x over m or f x is equal to m times a ax okay nothing new here you always come to the same conclusion any question here related to the velocity and acceleration in simple harmonic motion okay now let's discuss this velocity and acceleration in vt graphs okay what do you see here this velocity depends on time this acceleration depends on time so just plot them on a graph here we have displacement and velocity for simple harmonic motion we have already discussed the displacement this is displacement as a function of time and displacement is given with this expression so we have this type of behavior okay we have already discussed the x changes between positive a and negative a this origin is the equilibrium condition and the x is given with this expression we have calculated in the previous transparency so it is given with sinus function and here we have negative sign when the time is zero we will have zero here okay sine of zero is zero here and if you have certain phi angle then you will have certain velocity and the maximum value of the velocity is given with this one for the positive side and this one for the negative side this is the maximum value of the speed okay in simple harmonic motion and what else look at the vxt graph so the graph shifted okay by one over four cycle from the xt graph so we have different graphs in addition to that this amplitude of this oscillation also changed okay now let's have a look displacement and acceleration in simple harmonic motion again here in the top figure we have xt graph and here we have acceleration versus time graph and here we have a cosine minus omega square a cosine omega t plus phi this is the acceleration in simple harmonic motion we have extracted this equation in the previous transparencies and this is the maximum amplitude of this oscillation this one in positive and negative directions and then this is the ax acceleration as a function of time what do you see here it is also different from the xd graph it is shifted by 1.4 cycle from the vxd graph okay this was the vxd graph and this axt graph shifted by one half cycle from the xt graph and this oscillation has different amplitude compared to the x okay so now let's discuss both velocity and acceleration in simple harmonic motion in this example here we have a spring and again here we have a glider okay and this is the equilibrium condition x is equal to zero this is the maximum amplitude maximum stretched position this is the maximum cost position positive a negative a and equilibrium condition so let's consider we are starting from the maximum stretch position this is a and this is equilibrium and this is the maximum compressed position negative a okay so we are starting from this position just consider the system is at rest here okay and the velocity is zero but about the acceleration acceleration is in negative x direction because this spring will apply a restoring force in this direction this is the f spring okay f spring so since force is in this direction then we have acceleration in this direction and velocity here because the system is at rest here then after certain time we have certain velocity and acceleration in the negative x direction velocity is negative x direction and here when the system is in equilibrium the force is zero right because force is given with minus kx when the x is zero force is zero okay system is in equilibrium okay so acceleration is also zero but the system is going along the negative x direction with the maximum velocity along the equilibrium condition we have maximum velocity so here after passing the equilibrium we have velocity in negative x direction but since we have compressed the spring here now the glider here we have compressed spring spring will apply force in positive x direction okay then the acceleration will be along the positive x and then the glider or box reaches to the maximum amplitude in negative direction we will have zero velocity the system is momentarily at rest and then acceleration is in negative direction okay because restoring force is negative direction so the force applied by the spring pushes the system to the equilibrium condition then here again velocity in positive direction acceleration is in positive direction when the system is passing through this equilibrium after certain time the acceleration is here in equilibrium is zero force is zero here when the system is in equilibrium then we have maximum velocity in positive x direction and whenever you pass through this equilibrium along the positive x direction we will have this positive v x but the acceleration will be negative because restoring force will be in negative direction again and here when the system glider reaches to the maximum amplitude again momentarily the velocity of the system is zero okay then we have maximum acceleration along the negative x direction so look at this one here here we have acceleration here we have acceleration is zero here we have acceleration is zero here we have acceleration is zero so acceleration is not constant okay here we have non-constant acceleration why because this acceleration is given with this relation since this fx restoring force is non-constant it depends on the x okay then this is also non-constant acceleration it can be zero it can be positive it can be negative and also its magnitude can be changed by time due to the non-constant restoring force in this example so this is the velocity and acceleration in simple harmonic motion do you have any question here and let me continue with the amplitude and phase angle amplitude is a and phase angle is phi so how to calculate phase angle and amplitude in simple harmonic motion in the previous transparencies we have calculated velocity in simple harmonic motion okay so just consider that we have initial velocity when the time is zero initial velocity is v zero x when the time is zero and instead of v x just put v zero x when the time is zero so now we have this expression instead of t we have just put 0 then this is 0 and instead of vx we use v0x then the velocity when the time is zero is given minus angular frequency amplitude sinus phi okay if you divide this equation with x zero what is the x zero initial position so have to find the initial position x is given with a cosine omega t plus phi we have done it in the past and if time is zero then this term will be zero then x zero then the time is zero is given this a cosine phi right so you divide this term with x0 you divide this term with x0 just put here a cosine phi when the time is zero then finally we have here a here we have a you can cancel this one here we have sinus phi here we have cosine phi it is equal to tangent phi and here we have minus omega if you take phi angle here from this expression then you can calculate the phase angle in simple harmonic motion which is given with arc tangent minus initial velocity when the time is 0 over angular frequency times x 0 initial displacement initial position of this system okay and by using the same equations you can also calculate the amplitude in simple harmonic motion i will not go into detail if you wonder you can do it by yourself or you can look at the book so it is given in the book so you can get the amplitude in simple harmonic motion it is given with this expression square root of x zero square plus v zero square over omega squared this is the amplitude in simple harmonic motion now let me continue with this example example 14.3 describing simple harmonic motion in example 14.2 we give the glider and it has an initial displacement x zero okay open or 15 meter just consider and the initial velocity v 0 x 0.4 meters per second okay so this is the equilibrium condition and it has some initial condition which is given with all point of 15 meter and it has certain v0x it is in positive direction because given in the question 0.4 meter per second okay so find the period t amplitude a phase angle phi of the resulting motion period amplitude phase angle okay then write equations for the displacement x velocity and acceleration as functions of time okay this one so for the amplitude you can use this expression we have already learned okay if you know the initial position if you know the initial velocity if you know the angular frequency then you can calculate the amplitude so in the previous example in example 14.2 we have calculated angular frequency 20 radian per second just use this one here x zero is given in the question v zero x in the given in the question then i can calculate amplitude all point all 25 meter so this is the initial condition and the maximum amplitude is somewhere here okay this is the a which is equal to 0.025 meter and what about the phase angle it is given with arc tangent minus v0x over omega x0 just put all numbers here then you can calculate minus 53 degrees this is the phase angle or you can write it in radian minus 0.93 radian so what about the second part of the question write the equations for displacement velocity and acceleration so x is given this a cosine omega t plus phi we have calculated phi okay just put it there we have calculated omega just put it there we have calculated a just put it there so then x is given with this expression this is the amplitude this is the omega this is the phi okay and velocity as a function of time look at the formula for the velocity this is the formula for the velocity negative omega a sinus omega t plus phi we have calculated phi we have calculated omega we have calculated amplitude just put all calculated values then you can write the velocity as a function of time x as a function of time in the same way you can also calculate acceleration as a function of time do you have any question and let me finish this lecture with the last part energy in simple harmonic motion actually when we are discussing the collider and spreak system on a frictionless surface we have discussed the energy okay so if the surface is frictionless and if air resistance is ignored and if this spring is an ideal spring if all conditions are perfect if we have an ideal system then total mechanical energy of the system is conserved so locate the system along the y-axis we have weight mg we have normal force okay then net external force along the y-axis is zero and what about the net force along the x-axis so there is spring force okay but it is not external force i'm looking this system we don't have any friction we don't have a resistance okay so the vertical forces does not work because net force along the vertical axis is zero and we assume that the mass of the spring itself is negligible and we consider that this is ideal spring and since the only horizontal force on the body in simple harmonic motion is the conservative force exerted by an ideal spring okay this is given with the fx then mechanical energy is conserved then we have kinetic energy plus potential energy which is the total mechanical energy of the system what about the kinetic energy which is given with one half mv x square what about the elastic potential energy this one one half k kx square and this is constant right so here what do you see vx here you see x x is a cosine omega t plus phi we have calculated this is the x and here we have v x you can also write v x in terms of amplitude in terms of omega time and phi we have done it just put all the equations into this total mechanical energy okay then we will get an expression for this total mechanical energy before doing that i would like to discuss the conversion of energy from kinetic to potential in case of this simple harmonic motion so we have this relation total mechanical energy is constant it is conserved this is the kinetic energy of the system this is the elastic potential energy of the system and if this system is staying at the maximum amplitude just consider here we have a spring and this is the mass and this is the maximum amplitude and this is the equilibrium condition if the mass is here the velocity is zero but since x is maximum this is zero and instead of x you can write a then total mechanical energy can be written in this form if x is maximum if amplitude is maximum in this case total mechanical energy is only given with the elastic potential energy stored in the spring okay and it is constant so this value here one half k a square is the maximum value of the total mechanical energy in simple harmonic motion now let me discuss the change in the total mechanical energy as a function of time as a function of the position okay so let's consider that we started from this position x is equal to a maximum stretch position and this is the equilibrium condition x is equal to zero this is the negative a okay so here momentarily velocity is zero the x is zero and we have maximum acceleration along the negative x direction here since velocity is zero kinetic energy in the total mechanical energy is zero and we have the total mechanical energy is given only with the elastic potential energy look at this one this is the total mechanical energy kinetic energy is zero and elastic potential energy is maximum at this condition and whenever you release this system the box is moving with this velocity along the negative x direction we have this acceleration okay and the particle is moving in this direction so it has certain kinetic energy now kinetic energy is different from zero and some part of the elastic potential energy is converted to the kinetic energy mechanical energy is conserved it is same as the previous case but now we have kinetic energy and great elastic potential energy is decreased and when this system is in equilibrium when the box is passing through the equilibrium condition acceleration is zero because here force is 0 okay restoring force is 0 here and the object the box has the maximum velocity because the total mechanical energy is completely converted to the kinetic energy and the elastic potential energy is zero here since the kinetic energy is maximum in this condition we have maximum velocity and when you pass this equilibrium condition the system goes along the negative x direction and the box is located somewhere here but now whenever you pass the equilibrium the direction of the force and direction of the acceleration changes they are along the positive x because you compress the spring and kinetic energy some part of kinetic energy is converted to the elastic potential energy but total mechanical energy is conserved okay and when you reach to the maximum negative x position maximum negative amplitude the system is momentarily at rest velocity is zero we have maximum restoring force we have maximum acceleration and the direction of the acceleration direction of the restoring force are along the positive x direction and kinetic energy is zero here because velocity is zero and all the kinetic energy is converted to the elastic potential energy and total mechanical energy is only given with elastic potential energy and elastic potential energy is conserved so this is the energy in simple harmonic motion and energy conversion between kinetic energy and elastic potential energy do you have any question here okay let me continue with the energy diagrams for simple harmonic motion actually i have discussed all the things in the previous chapters so you will easily understand this is the x position and you know x depends on the amplitude and also omega and time are k so this is the maximum amplitude in positive x positive direction maximum amplitude in negative direction and this is the equilibrium condition and this is the total mechanical energy of the system total mechanical energy is constant this is the energy versus x okay when the system is here the total mechanical energy is only given with the elastic potential energy when the system is here on the left side in negative side total mechanical energy is only given with the elastic potential energy when the system is in equilibrium here the energy is only given with the kinetic energy elastic potential energy is zero here at this equilibrium condition and in between for example somewhere here the total mechanical energy is shared by the kinetic energy plus elastic potential energy and just plot the potential energy kinetic energy and total mechanical energy for a body in simple harmonic motion as a function of displacement again this is the displacement maximum amplitude in positive x maximum amplitude in negative x and this is the mechanical energy total mechanical energy which is constant where we have zero kinetic energy here we have zero kinetic energy and the mechanical energy is only given with the elastic potential energy and here we have mechanical energy is only given with the kinetic energy and here we have this green line we have the behavior of the kinetic energy as the function of the x position and with the blue one we have the behavior of elastic potential energy as a function of x okay so at this conditions here on the left side in negative x direction and here on the right side in positive x direction is kinetic energy and potential energy are equal to each other do you have any question let me discuss the vx as a function of x velocity of the glider for example here we have vx so the x changes as a function of x as a function of time so how to calculate vx as a function of position we can use this expression okay so just solve this equation for the velocity then you can calculate x velocity x velocity is given plus minus square root of k over m square root of a squared minus x square so just consider this is constant mass is constant then amplitude is constant then as a function of x position you can calculate the velocity and the maximum velocity happens when the x is equal to zero in equilibrium condition if you put 0 here then you will get the maximum velocity in the equilibrium condition which is given this omega times a look at the result remember the v omega r from the circular motion instead of r if you use amplitude we have omega a okay we have also done today so you have the same result this is the maximum velocity in simple harmonic motion and this is the velocity as a function of x displacement example 14.4 velocity acceleration and energy in simple harmonic motion find the maximum minimum velocities attained by the oscillating collider of example 14.2 find the maximum and minimum accelerations find the velocity and acceleration when the glider is halfway from its initial position to the equilibrium position find the total energy potential energy and kinetic energy so velocity is given with this expression we have given this one here maximum velocity is given with this expression if you know the force constant if you know the mass if you know the amplitude then you can calculate maximum velocity okay and you can also calculate the maximum acceleration remember the formula for the acceleration we have done it okay so if you know the k if you know the mass if you know the x then you can calculate the acceleration as a function of x and the maximum acceleration happens when the x is equal to a then if you put it there then you can calculate and what about the vx at some certain condition by applying this equation you can find this one and finally you can write the total mechanical energy and you can write the elastic potential energy kinetic energy by using this equations if you know the k if you know the amplitude then you can calculate in joule you can if you know the kx you can calculate elastic potential energy in joule if you know the mass if you know the vx for some certain exposition then you can calculate the kinetic energy in june let me finish with this last example and last transparency example 14.5 energy and momentum in simple harmonic motion here we have a block this the mass of capital m okay attach to a horizontal spring with force constant k it has force constant k is moving in simple harmonic motion with amplitude a1 okay it is moving between positive a1 and negative a1 the simple harmonic motion as the block passes through its equilibrium position so this is the equilibrium position when the x is equal to zero a lump of putty of mass m is dropped from a small height and sticks to it find the new amplitude and period of the motion so nev amplitude and period of the motion is s you can say this one and in part b repeat part a if the patty is dropped onto the block when it is at one or end of its pass so just consider that the mass box is staying here at the beginning and then you drop this lamp of party of mass with m okay at this condition when the x is equal to a1 in this one in a this party is dropped when the system is in equilibrium and in this condition when the system in a the party is dropped then the mass box is passing from the equilibrium but here in the second condition in part b the party is dropped when the system is at this condition okay so then you will repeat the same solution same equations in case of b so now what about the question neve amplitude and neith period of the system so velocity is given with this one this is the maximum velocity okay and here we have the energy energy of this system when it is passing through the equilibrium in the equilibrium condition energy is given with one half m v1 square okay in equilibrium total mechanical energy is only given with the kinetic energy because elastic potential energy is zero in equilibrium condition okay and v1 is given with this expression so then we have this v1 and during the collision the x component of momentum of the block party system is conserved momentum is conserved okay so just consider this is frictionless surface air resistance is ignored okay and the net force acting on the system is zero then the momentum is conserved okay remember the conservation of momentum and before the collusion the x component of the momentum is m times v one m is this one v1 is this one so we have calculated v1 and then this is the initial momentum of the system along the x-axis and what about the momentum of the party during the collusion this party has no x momentum okay because it is moving along the y-axis so then what about after the collusion after the collusion the momentum of the system is given with mass of the block plus mass of the party and the velocity of the system since they stick to each other then i can write it like this from the conservation of momentum initial momentum of the box along the x-axis initial momentum of the party and final momentum of the system then here we have this expression here we have v2 final velocity which is given in terms of v1 okay and what about the kinetic energy after the collusion so just after the collusion the total mechanical energy is still purely kinetic okay i'm talking about this time this condition so this party collides with this mass m and then they stick each other so before the collusion after the collusion they have same kinetic energy momentarily okay then the final mechanical energy after the collusion is given with the kinetic energy of the system so this is the total mass of the system this is the final velocity of the system after the collusion i am just dealing with the collusion happens in the equilibrium condition okay in a very short time then here we have v2 instead of v2 you can write this expression just put it there okay and what we have here we have one half m square over m plus mv1 square so if you take 1m from this one and if you take one half if you take v1 square then you can write this equation in this form and what is this one this is e1 okay because we have calculated v1 here e1 here so e1 is here this is the e1 and then what about e2 the maximum value of e2 is given with this expression won't have ka2 square okay so this is the maximum value of the energy total mechanical energy of the system and it is equal to the e2 here when the system is in equilibrium then you can get this expression this is the final amplitude of the system a2 this is the initial amplitude of the system okay then i have expression for the final amplitude of the system actually it is asked this in the question what is the final amplitude and what is the period of the motion period does not depend on the amplitude just remember period is given list this expression 2 pi m over k okay we have done it and after the collusion we have another m which is given this capital m plus small m this is the mass of the capital one is the mass of the box small one is the mass of the party then we have the final period of the system so with this one i have finished this lecture and next lecture we will continue with the applications of simple harmonic motion take care of yourself bye bye