hi uh complete lesson uh on oscillation uh focused on Ed XL IAL unit 5 syllabus uh okay what is oscillation uh

oscillation means repeated movement or repeated motion is called oscillation uh there are examples many

examples we can uh tell for oscillation for example if you uh attach a small sphere to a

uh like a string and let it uh just push it to a small distance and leave it it will oscillate it will show oscillation

like that U even if you have a string attached to a mass the string will be at rest which is carrying a mass then pull

it through small distance and leave it then the spring will oscillate repeated movement it will show same way when a

spring attached in horizontal position to a trolley or something like that on a table and pull this uh trolley through a

small distance and leave it then it will oscillate not only that even a small object something like a test tube a

weighted test tube uh floating in water so it will be at rest push the test tube through small distance and leave it the

test tube which carries a small Mass inside that will perform oscillation so like that there are many different

examples we can show even bung jumber is a he also can perform uh oscillation so like that there are many different

examples we can tell or show for oscillation that means oscillation means repeated movement is called oscillation

okay so there we use different technical words first thing you must know uh equilibrium position so the equilibrium

position means the position at which the resultant for you know any system that when it performs oscillation there will

be uh force acting on it not only one force there with different forces or more than one forces acting on it so the

position at which the resultant force acting on it becomes equal to zero is called the equilibrium

position or we can say the position to which it comes to rest when it is not oscillating is also the equilibrium

positions either way uh we can show but the better way of telling U what is equilibrium position means position at

which uh the resultant Force acting on the uh oscillating system uh becomes equal to zero is called uh equilibrium

position so when the system is not oscillating it will come to rest at the equilibrium position then you should

know what is oscillator oscillated means the system that performs or shows oscillation is called oscillat earlier I

drew different diagrams for example a spring which is carrying a mass or a spring in horizontal position which is

attached to a trolley or test tube which is floating in water each of them are different oscillators so oscillator

Means A system that performs oscillation is called oscillator okay so you know the meaning

of oscillator you know the meaning of uh equilibrium position the next thing is time period of the oscillation what do

you mean time period of the oscillation means the time taken to complete one full oscillation okay what is one full

oscillation for example I'll take an example like uh a simple pendulum we call a small Bob attached to a string

okay it is the equilibrium position because it will be at rest at this position when it is not oscillating okay

so then pull it and leave it to the left or right you can pull it to the right or left okay for example you pull it to the

right okay it has come to this position now so from this when you leave it it will move from here to here then it'll

come back to this one this is called the time taken for this oscillation is called one period or if I consider from

the equilibrium position from the equilib you consider the motion so here I consider the motion from the extreme

position extreme position means where it has gone to the maximum position that I we call Extreme position so from there

when I consider the time taken to move from this extreme position to The Other Extreme position then come back to the

same initial extreme position uh the time taken for that oscillation is called uh time period we call this as

one full oscillation same way uh if I consider from the equum position so the time taken to move from here to the one

side of the extreme position then from here to the other side of the extreme position then come back to the same

initial point that is the equum this is called one one full oscillation and the time taken for this oscillation uh is

the time period okay so you must know the meaning of time period also and you should be able to identify one full

oscillation uh even if it is like a spring attached to uh Charly on a horizontal table when you keep it like

this uh it will be at rest at this position so you uh you pull it to the right so from here you you pull it to

this side and you are leaving it it will move from here to this extreme position where the spring will be compressed then

it will move back to this position so the time taken for this oscillation is the time period and this oscillation is

called uh one period the time taken for that oscillation is called uh time period like that you must be able to uh

know the meaning of one full oscillation and the time taken to perform one full oscillation is called time period it can

be any unit uh the notation is capital T okay the next thing you should know is frequency of the oscillation frequency

of the oscillation means uh number of OS ation completed during 1 second normally we consider 1 second or you can say

during unit time but normally we consider 1 second so the number of oscillations completed per second is

called or is defined as frequency of the oscillation that need not to be a whole number it could be a it could be 2.3 or

something like that also it also correct we can say the frequency so frequency means number of oscillations completed

per second is defined as frequency so its unit is uh per second sorry base unit SI unit is Hertz and

also you know you learned that In Waves also the period and frequency the relationship

betweene period and frequency uh T = 1 F or f = 1 / T the same relationship we here also use to relate uh time period

uh and the frequency okay uh then displacement okay what is displacement during oscillation displacement it it's

always measured from the equilibrium position displacement is a vector quantity you learn that in all even in

year 10 even in unit one displacement is a vector quantity but here the displacement is measured from the

equilibrium person careful uh it's a very common mistake St they make uh when we consider the uh displacement

displacement is measured from the equilibrium position so if I consider oscillation something like this

okay consider this diagram a sphere is kept on a smooth horizontal surface and it's attached to a spring uh o is the

equilibrium position so equilibrium position means the position at which the resultant force is equal to zero so from

this position uh the spring is compressible right from this position the C is pulled to the right up to this

position uh name it P1 uh the uh the sphere is pulled up to here then release so we'll take this is the time tal 0 at

which the sphere is released okay the sphere will move towards the equilibrium position because the spring will pull it

back due to the tension the sphere will move towards the equilibrium position so initially when t equal Z what is the

displacement the displacement is measured from the equilibrium position it's measured from the equilibrium

position okay the displacement is measured always from the equilibrium position so later when the sphere comes

to this position it has come to this position so at time tal T1 the sphere has come to this position tal T1 so

what's the displacement at T1 we use the word at at T1 that is not earlier one say if we say

earlier it was S1 now at time t or if I say initially X not initially when t equal Z the displacement is X when it is

at P1 now it has come to the position P2 at time T = T1 at time tal T1 has come to the position Point P1 so what is the

displacement means it must be measured from the equilibrium position so it should be written this way that is X1

not this one we don't measure uh the displacement uh from P1 to P2 we measure always from the to because here we see

displacement at you can see that later when I start simple harmonic motion the equations are displacement at we don't

consider displacement during the time we don't consider displacement at an instant what is the displacement okay so

here the simple harmonic motion I haven't defined it I'm going to Define define means the conditions required to

have a simple harmonic motion but any whatever it is in your syllabus always not only your syus during this simple

harmonic motion we always consider the this displacement at an instant not the displacement during the in time period

so for example if I say initially this uh X not for example 10 cm when t equal Z later when this X1 for example 4

cm so what is the where the time taken T1 maybe I can name the time uh maybe 2 seconds okay so I can the displacement

initially when Tal 0 is 10 cm displacement at 2 seconds is 4 cm but if you ask a student who is doing unit one

what is the displacement 2 seconds don't say at or during he will say certainly initially the sphere was at P1 now it

has come to P2 then he will see the displacement is 6 cm towards left but actually here we will say the

displacement at 2 second at 2 2 second uh is 4 cm towards right it's moving towards left actually you can see the

motion is towards left when you release it from P1 is moving towards 4 but we say at 2 seconds the displacement is 4

cm towards right because we measure always the displacement from o we take it as a reference point O and uh we

consider the displacement at not the displacement during the time we don't consider okay this is the displacement

so this is little bit confusion but anyway uh that's the way the equations we are going to get later uh you would

have done it at school that equation also always gives the displacement at a particular instant okay that you must

know that okay so the next thing is uh amplitude amplitude means during the oscillation uh it must be when we C

simple harmonic motion the oscillation must be symmetrical on either side of the equilibrium position position that's

the way your you won't get any question where the the altitude varies or anything you won't get so we consider

the oscillation symmetrical on either side about the equilibrium position and the maximum displacement is called

amplitude so amplitude could be denoted by either capital A or X not either way we can use uh for amplitude uh sometimes

books they use capital A for amplitude uh some books they use x not for the amplitude Okay in your syllabus we use

capital a uh to indicate amplitude amplitude is the uh maximum displacement okay so the next thing is uh simple

harmonic motion what is the meaning of of what's the condition for simple harmonic motion all the oscillations

won't be simple harmonic uh to be a simple harmonic motion certain conditions must be satisfied what are

the conditions uh that must be satisfied if an oscillation bonds to be uh simple Harmony you must know that so the first

condition is uh the displac I already told that displacement is always measured from the equilibrium position

so it's a vector quantity if I take towards right positive towards left should be negative or if I take vertical

oscillation upward Direction positive then downward direction from the equilibrium position should be negative

uh always we measure from the equ position and displacement is a vector quantity it has direction right okay so

you know acceleration also uh a vector quantity so it also has Direction so the first condition if an oscillation wants

to be a simple harmonic motion the first condition the acceleration must be directly proportional to displacement so

it can't be a constant acceleration the acceleration is depending on displacement and it must be directly

proportional to the displacement the second condition the direction of the acceleration should be always point or

Point directed towards the equilibrium position so for example um imagine this is the equilibrium position uh particle

is oscillating the oscillation is due to the motion oscillation of a particle a particle is oscillating on either side

of p and Q so p and Q are the extreme position o is the equilibrium position a particle is oscillating on either side

so if the particle is here and if the particle whether it's moving towards o or moving towards Q we don't care

whatever the direction of motion the acceleration must be directed towards the equilibrium position ACC must be

directed towards the equilibrium position and we know that this placement is always measured from the equ position

so the displacement is measured from here x you know that X is going to be opposite to the direction of X is

opposite to the direction of a same way when the particle is on the left side uh left side of the equum position there

also whatever the direction the particle moves we don't care but the acceleration must be always always towards the

equilibrium position so if the acceleration is always towards the equum position the displacement is

always measured from the equilibrium position means here also you can see the direction of the displacement and the

direction of the acceleration are in opposite so whatever the direction of motion the particle oscillates or the

system oscillates we don't care the direction of the acceleration must be always directed towards the equilibrium

position and I already told that the displacement is always measured from the equilibrium position therefore always

the direction of the displacement and the direction of the acceleration will be opposite among them but the direction

of the acceleration and direction of the displacement opposite among them at the same time acceleration must be directly

proportional to the displacement so when two vectors are always opposite how do we denote we denote it with minus sign

so that's the reason we write this like a proportional to - x so the minus sign indicates the direction of acceleration

is always opposite to the direction of displacement or direction of displacement is always opposite to the

direction of acceleration okay so you know that if an oscillation wants to be simply harmonic

the oscillation should satisfy two conditions what are they the acceleration must be directly

proportional to displacement first condition the second condition the direction of the acceleration must be

always towards the equilibrium position so one by one we'll analyze uh certain examples and find out whether can it be

a simple harmonic uh Mo or can it perform a simple harmonic motion or can that oscillation be a simple harmonic

motion we'll analyze one by one the first one we'll imagine two Rams are kept like

this imagine two small ramps are kept as shown on the diagram they make the same angle so perfectly smooth no drag Force

no uh friction if you keep a small sphere something like a ball bearing and release it what will happen the

according to the law of conservation of energy there is no drag Force no loss of energy so the uh ball bearing will

accelerate downwards when it comes here it will have the maximum speed gaining kinetic energy loss in GP gaining

kinetic energy so it will move to to this side so when it move to this side it will come back to the same level

because both side it has the same angle the both ramps are making the same uh angle so the sphere will more the ball

bearing will downwards accelerate then decelerate then accelerate decelerate like that since there's no drag Force no

loss of energy it will oscillate forever okay so can this oscillation so it's an oscillation this is repeated movement

the ball bearing is going to move move repeatedly between this point and the same level we can imagine the point this

is P1 to P2 so if I C the same level somewhere here P2 is going to oscillate between P1 and P2 continuously okay can

this oscillation so this is the equilibrium position because uh when it is not oscillating it will come to rest

at this position so that is going to be the equilibrium position so I can name the lowest point of the two Rams where

they are touching each other that is going to be the equilibrium position of this oscillation okay can

this oscillation this we can say an oscillator the whole system can be taken as an oscillator can this oscillation be

a simple harmonic motion we'll analyze it okay so no friction if I think about the uh ball bearing it has mass it's

kept on a ramp at this position so the mass is M it makes an angle Theta with horizontal so the component of the

weight along the ramp down is mg sin Theta mg sin Theta so when I consider the equilibrium position this is equum

position I already explained why because when it is not oscillated will come to rest at this position so mg sin Theta is

pointing towards the equilibrium position mg sin Theta so if I use the Newton Second Law FAL M that is mg sin

Theta = to M so the acceleration is towards the equilibrium position which is um G sin Theta same way after it

passes the equilibrium position it's going to move to the left side until it reaches the point P2 there the velocity

is from o to P2 but the acceleration is due to the compound of the weight when the sphere is here the compound of the

weight here also will act towards the equilibrium position that is mg sin thet so there also the acceleration will be

towards the equilibrium position which is a isal to G sin Theta okay so the acceleration is whatever the direction

of motion the sphere or the ball bearing performs or in whatever the direction either left or right the acceleration of

the ball bearing is pointing towards the equilibrium position o but can the is be a simple harmonic

motion so here also the acceleration is always directed towards the equilibrium position but can it be a simple harmonic

motion because here the acceleration is directed towards the equilibrium position but when you consider the

acceleration the acceleration is constant here G is constant T is constant sin thet is Conant so here

acceleration is directed towards the equilibrium whatever the direction of motion the acceleration is always

pointed towards the equilibrium position but acceleration is not directly proportional to displacement it is a

constant the acceleration is a constant acceleration therefore it is not a simple harmonic motion it's an

oscillation yes where the acceleration is always directed towards the equ you know that to be a simple harmonic motion

two conditions should be satisfied first ACC must be directly proportional displacement the second condition the

acceleration must always Point towards the equili position this oscillation is satisfying the second Condition it's

pointing towards the equilibrium position but acceleration is not depending on displacement acceleration

is a constant here therefore this is not a simple harmonic motion okay so all the oscillations are not simple harmonic the

conditions must be satisfied I already told several times a proportional to X oxygen directly proportional to a

displacement that is not satisfied here but the second condition satisfied here therefore it won't be a simple harmonic

motion okay now I'll draw a few examples uh more examples and we'll check whether uh that type of oscillations can they

perform simple harmonic motion okay we'll consider this second example a spring which has spring

constant K and the spring is compressible you can compress the spring so that such spring is attached to a

trolley of mass m and kept on a frictionless horizontal surface so initially the trolley is at equili

position or you pull the trolley towards right through a small displacement and you release it so when the trolley is on

the right side I'll draw it here when the trolley is on the right side what are the forces acting on it so this is

the equilibrium position I'll draw a vertical line for the equum position so what are the forces acting on the

trolley the weight is acting downwards mg at the the normal reaction from the surface is acting up so Ral mg they

balance each other we don't need it okay the spring is extended so spring will exert a tension on the solid that is

towards left we know that we learned that tension is directly proportional to

extension according to Hook's law so T proportional to extension that we know that t is

equal to KX okay so the displacement is measured from the equilibrium position so the

displacement is measured from o towards right when the trolley is here when the trolle is moving to the left or right we

don't care but the that's nothing uh but it's not a matter but we measure the displacement from the equum position

that is towards right X the only force that performs the oscillation or motion is the unbalance Force unbalanced force

which is the tension because R and mg balance each other so the only one unbalanced force is there that is T so

the t is acting towards left you can see the T the tension which is unbalanced force acting towards left so according f

m when we consider the tension is towards left that is KX F = KX is = to M A so a = k m into X look at here K is a

constant m is a constant so K is a constant m is a constant so AAL a constant into X therefore a proportional

to X here the acceleration of the trolley depends on the displacement or the displacement is measured from the

equum position so actually X is both displacement as well as the extension both measured from the equum position no

so a proportional to displacement X here at the same time you can see that the direction of the displacement is

towards right the tension is towards left therefore always the direction of the tension or the direction of the

force that is the always force and acceleration are parall to each other resultant force and acceleration are

parall to each other according to Newton Second Law the direction of force the resultant force and the acceleration are

parall vectors so the resultant force is towards left means acceleration is always pointing towards the equum

position you can see that acceleration is always pointing towards the equilib position because tension is always

pointing towards the equ position when the trol is on the right side of O okay if the trol is on the left side of O the

spring will be compressed so when the spring is compressed I'll draw the trolley here when the spring is

compressed the compressive force will act on the trolley that will be towards right the compressive force that also

tension that is acting towards right so we know that t = to k x x is the displacement also that will be the uh

compression of the uh spring that is measured from the equilibrium position from o it's

measured towards left X that is the compression as well as that is the uh displacement also because only one

spring is attached to the solid so the compression or the displacement is towards left the tension which is

unbalance force is acting towards right so here also the direction of the unbalanced force is opposite to the

direction of uh displacement means direction of acceleration is opposite to the

direction of uh displacement also the tension is pointing towards the equum position therefore acceleration also

will Point towards the equilibrium position when the trol is on the left side of the equium position so here tal

k x same way I can derive here also f m so in of T that is KX = m a a = k MX when it's on the left side K and M are

constants so the whole thing is constant so a proportional to X also a is pointing towards the equilibrium

position because tension is pointing towards a% so this will perform simple harmonic motion so when you attach a

trolley or sphere or whatever any Mass to a spring and when the mass is kept on a smooth surface and when you let it

oscillate that oscillation will be single harmonic okay so we'll consider this

type of oscillation a spring is attached to a trolley and the trolley is kept on a smooth ramp that makes an angle Theta

with horizontal uh and this is the position we call equ position the trolley at rest

when the trolley is at rest already the spring is extended because you know the the compound of the weight of the

trolley along the slope down is mg sin Theta to balance that downward compound of the weight there will be a tension

upwards acting on the trolley so that tension will balance the weight of the trolley along the slope downward so when

the trolley is at equilibrium position o that when I draw the free body force diagram separately something like

this so the weight will act vertically down mg that we don't need the compound of the weight along the slope down is mg

s Theta that is balanced by a tension and that is T we can see the extension of the spring due to uh this uh force of

mg sin Theta is X1 so T is equal to K X1 where K is the spring constant of the spring so we can say

kx1 is equal to mg because it's at o where the resultant force acting on the trol is zero there a normal reaction we

don't need we don't need the normal reaction we don't consider that in this uh oscillation so R is balanced by mg

cos Theta what you learned in unit one so here kx1 is equal to mg then what we do from this position we pull the

trolley through a small distance you know o is St position so you are pulling it down through X I'm not drawing the TR

again so but you are pulling it toward right through X and you leave it so trolle is going to oscillate between

this point you can call P1 that point has P1 and P2 between P1 and P2 the trolley is going to oscillate the spring

is compressible you can compress the spring a lot we'll take that way uh the turns of the spring won't hit each other

and stop the oscillation the compressible spring so the spring is going to the trol is going to oscillate

on either side of O up to P1 and P2 so when the trol is somewhere on the right side at a displacement X or from o that

time the extension of the spring is uh X1 + x so when we consider that situation I draw the diagram again when

the trolley is at a position uh before P1 somewhere when it has a displacement towards right X from the equum position

the extension of the string is going to be uh x + X1 already it has extension X1 when it is at 4 now the extension is

going to be x + X1 so the tension will act on the trolley along the slope up that is K into x + X1 the compound of

the weight will be along the slope down mg sin Theta okay so we know that KX + X1 is

greater than mg sin earlier K X1 balanced mg sin Theta sorry mg sin Theta earlier kx1 balanced mg G sin Theta so

here k x + X1 so this is larger than X1 so the resultant force acting on the TR will be along the slope of the resultant

force is going to be k x + X1 - mg sin Theta this is the resultant force acting on it towards the

equilibrium position no so when I solve it's going to be KX + kx1 - mg sin th but I know that kx1 is mg sin Theta kx1

is mg sin Theta you can see that this kx1 is equal to I got it first equation kx1 mg sin Theta so it's going to be uh

this kx1 and mg sin Theta will get cancel each other finally we'll have KX but f is the resultant Force so that

must be equal to ma a so ma a is equal to KX so a is going to be k / M into X you can see X is the displacement from o

x is the displacement from o is it o is the equilibrium position the displacement downwards is X so a is

directly proportional to X because K and M both are constants so a proportional to X also you can see the resultant

force acting towards the equium position so according acceleration will be towards the equum position X is away

from the equ see the diagram X is away downwards it's X but FAL m f is acting towards the equ position the resultant

Force therefore the resultant Force towards the equ position means acceleration will be towards position

therefore a proportional to Min - x therefore it will be simple harmonic motion same way you can show when it is

towards P to somewhere here the extension will decrease by X1 uh decrease by X so I'll change that again

so same diagram I use it so when it go to the P2 this is I der when it is towards P1 from person the TR goes down

I derived it and show where a proportional to- X so if it is on the left side above o towards P2 when the

trolley is oscillating I told between P1 and P2 so will it be a simple that side or it must be but we show that it's

going to be X1 - x this condition is correct there also when it is at equilibrium position K X1 mg sin Theta

it's going to be X1 - x because xtens is going to decrease when the trolley moves upwards from o this

direction okay K X1 - x so the resultant force uh the resultant Force this is going to be smaller because kx1 balances

mg sin thet now the extens is decreasing so this tension is going to be smaller so the resultant force is going to be

towards the equilibrium position resultant Force f is going to be because this will be smaller than uh mg sin

Theta the tension so the resultant force will be towards o that is going to be mg sin Theta minus K X1 - x so Solve IT mg

sin Theta - kx1 + KX so here Min - kx1 kx1 is already mg sin Theta so here it's going to be mg sin Theta you can

substitute for kx1 you substitute that mg sin Theta and this mg will get cancel FAL KX what's the direction of this F

this direction of f is towards 4 because mty sin Theta greater than this so the resultant force is towards o that is

going to be m a a is going to be towards O So this shows a = k/ m x so here also K and M

are constant so a is directly proportional to X also a is pointing towards the equ position X is from o

upwards but a is towards o so it's going to be a simple harmonic motion so when a trolley attached to a string and in the

tring is kept on a slope where there's no friction when you allow to oscillate the oscillation will be simple harmonic

here also okay fourth example a trolley is attached to two identical Springs uh

spring Conant of each spring is K it's not necessary to be identical but for Simplicity I took it identical uh if

there are any questions related to oscillations with two Springs in your syllabus it will be identical because

otherwise K1 K2 two different spring constants it will slightly more work normally they won't ask that type of

questions I guess yeah so for Simplicity I took both Springs are identical when the trolley is at equilibrium position

each spring has the same amount of extension X1 so the extension of this spring is X1 this also has extension X1

okay so if I draw the free body Force this is the O equilibrium position so if I draw the free body force diagram for

the trolley when it is at equilibrium position the weight and normal reaction balance

each other this side there will be tension T1 that is equal to K X1 this side also there will be tension uh T1

same thing kx1 okay now uh the spring is pulled towards right no reason taking towards

right if you want you can take it towards left just for no reason I'm taking towards right through a distance

a small distance and allow to oscillate but we take this way the questions the spring the trolle is pulled towards

right such that the both Springs are always under tension that means the for example if X1 is 5 cm X1 is equal to 5

cm you pull the trolley towards right maximum five maybe four or three because then only if you pull it to the right

this extension will decrease by for example if you pull it towards right by 4 cm from o the extension of the right

side space will become 1 cm already it extended by five now you are pulling it to the right by four means it will

reduce by four the extension of the right side string will become one the extension of the left string will become

5 + 4 9 if you pull it more than five then this SC will be compressed normally you won't normally there are no such

questions so normally what appeared in the old past papers also I consider and do so the charol is pulled towards right

and release so you pull it towards right such that the amplitude of oscillation is less than X1 okay so always the

Springs both will be under tension okay so it's pulled towards right and allowed to oscillate so it's oscillating between

the points P1 and P2 when it is on the right side the extension is uh sorry yeah the uh the displacement is X so

when the trol is on the right side I'll draw the diagram somewhere here x TR is here the X the displ M from the

equilibrium position this is the equilibrium position is it from the equilibrium position the displacement is

X so if I draw the free body force diagram for this uh situation if I draw the forces acting on it weight and

normal reaction will balance each other so the right side the tension now here the T1 early it was kx1 now the T1 will

become K X1 - x because the displacement from the equ portion is X x is less than x and always we take that with the

amplitude of illation is less than the initial uh extension of the Springs so T1 = kx1 - x here the t2 is going to be

now it won't be the same T1 it's going to be T2 now T2 is going to be K X1 + x okay we know that this is greater than

than this because X1 - x compared to that X1 + x is larger so the resultant force is going to be towards left so

when the trol is on the right side the resultant force will be towards left means it will Point towards o so the

resultant force is going to be K X1 + x - K X1 - x you solve it will be kx1 - kx1 kx1 + KX - kx1 + KX so kx1 kx1 get

cancel you'll become 2 KX careful 2 K comes in case if it is two Springs with different spring Conant K1 K2 then it

will become K1 + K2 X that's the only thing that's nothing K1 plus k2x will come since they are identical SPS it

become 2 KX so the force is pointing towards four resultant force is towards when the TR is on the right side of O

the resultant force is towards o that means f = ma a f = ma a so it's going to be Ma I can write ma a mass of the TR is

m m a = 2 KX so a is going to be 2 k/ m into X so the 2 k comes otherwise if there are two different strings K1 + K2

M into X we know that K is a constant m is a constant therefore a is directly proportional to X displacement X is the

displacement from the ATM position so the acceleration is going to be directly proportional to displacement and

direction of acceleration is along the direction of resultant Force which is pointing towards o therefore

displacement is measured towards right acceleration is pointing towards o therefore a proportional to - x

therefore it will be a simple harmonic motion the same way you can prove by taking the TR on the left side of O the

same results will come there it is will become K X1 + x it will become X1 - x same we get end up with the same result

so here also the oscillation will become simple harmonic motion okay the fifth example a spring

of uh which has spring constant K is hung uh in on vertical plane and when a mass an object or small mass is attached

to it there will be a static extension the spring will extend and come to rest that is called Static extension so the

static extension Let It Be X1 so with the system the uh spring Mass system when it is at rest this position

is going to be the equilibrium position so the position at which the spring Mass system comes to rest when it is not in

oscillation will be the equilibrium position so this is going to be the equilibrium

position okay so when it is at equilibrium position uh the extension of the spring is X1 so when we think about

the forces acting on the mass there will be two forces acting on it uh one is the weight acting vertically downward and

the tension which is k x 1 that is the tension so since it is at equilibrium kx1 will be equal to mg right now from

this equilibrium position we pull the mass further down and release it will oscillate on a vertical plane uh we

imagine when during the oscillation at a particular instant the displacement I already told that displacement is always

measured from the equilibrium position so if the displacement is vertically downward x uh

yeah this is the mass when the displacement is vertically downwards X the total extension of the string will

be X1 + x so that time when I think about the for when we consider the forces acting on the mass the forces

will be downwards mg upwards the tension that's going to be T is = to K X1 + x so when uh it is at the equilibrium

position mg is equal to kx1 but when it is below the equilibrium position the tension will become greater than mg so

the resultant force acting on it will be towards the equ position that is upward Direction so at that moment the

resultant Force when it is at this position with the displacement which is X from the equilibrium position at that

moment the resultant force will be upward Direction so that is going to be FAL uh

K X1 + x that is the tension minus mg okay so there K X1 + KX - mg so it's going to be of kx1 we can substitute mg

so that is going to be mg + KX minus mg that is it will be KX so the force will be equal to KX Newton Second if I use it

that is equal to m so the acceleration will be equal to k/ M into X this will be towards the

equilibrium position that's going to be towards the equilibrium position so acceleration and displacement X is the

displacement when we consider acceleration will be towards the equilibrium position because the

resultant force is towards equilibrium position equilibrium position so the acceleration will be towards the

equilibrium position which is directly proportional to displacement X but X is measured from the equum position which

is away downward Direction you can see the x is downward Direction but uh acceleration is towards the equilib

position so they are going to be in opposite to each other so there will be a minus sign in terms of vectors because

this gives the numerical relationship but acceleration is to towards the equilibrium position but X is downwards

from the equilibrium position which is downwards so the direction of acceleration and direction of

displacement they are in opposite to each other so a proportional to Min - x therefore K and M is constant therefore

I put a proportionality that's the reason it's going to be a simple harmonic motion it will oscillate in

simple harmonic motion okay same way if the mass is above the equilibrium position uh somewhere here if I consider

what will happen we don't need to do both sides but I'm just doing it in mathematics in Mechanics for example M3

we do it in a slightly different way but in physics we don't go to that detailed analysis but here we keep the mass

somewhere here like this so at that moment the displacement is upward Direction

X during that time if I draw the free body Force DM the forces acting on it will be uh weight is acting downward

Direction the tension is going to be when it is at the equili position the displacement is

X1 now it's reduced by X so the extension is going to be kx1 - x kx1 - x is the uh extension of the spring so the

tension is going to be kx1 - x okay when I consider the resultant force during this position when the mass is above the

equum position here also we assume for Simplicity the spring is not compressed at all always the spring is under

tension so always this x will be smaller than X1 or the amplitude of the oscillation is going to be smaller than

the static extension X1 if we take that way and when the mass is above the equ position the tension is going to be K X1

- x okay so when it is at this position when we consider the resultant force is

going to be uh towards the equum because we know that when it is at the equ position kx1 will balance mg now the

tension has decreased which becomes smaller than mg because the extension has become smaller than X1 so X1 - x is

smaller than X1 therefore the tension will become smaller than mg so the resultant force will be towards the

because mg is greater than tension when the mass is above the equilibrium position so the resultant Force f is

equal to mg - K X1 - x so if you solve it mg - K X1 plus KX so here also in of kx1 we can

substitute mg so kx1 we can substitute mg so will becomes KX so the resultant force is KX which is towards the

equilibrium position or downwards when it is when the mass is above the equilibrium position According to second

law it's going to be M A so a is going to be k/ m a constant into X which is going to be towards the equilibrium

position so K or M is constant therefore a proportional to X but X is measured upward direction from the equilibrium

position this is upward Direction acceleration is towards the equilibrium position because resultant force is

towards equum position so the direction of acceleration Direction EX are in opposite to each other therefore a

proportional to Min - x where k/ m is constant so a proportional to minus 6 means uh the system will oscillate or

perform uh simple harmonic motion okay so the sixth example about uh a child uh jumping on bouncing on a

trampoline so you know trampoline it consists of elastic m material on top of it like a elastic sheet on top of a

membrane on top of a uh top of it so when a child stands on it it will deform and the child will be at rest so this

position is going to be the equilibrium position so now the child is bouncing up and down such that it is always in

contact with the elastic sheet or elastic that material so when the child is at the lowest position uh it's moving

from the equum position lowest position the second diagram is third diagram is showing that so the child is at the

equium position so the Sheep will exert a tension on the child upward Direction so in that situation there are two

forces acting on it one is the weight of the child downwards mg and the elastic material the sheet will exert a tension

push upward direction that is T is going to be KX where X is the X1 we can say that is the uh

static extension from compare to the equilibrium position this is going to be the static extension this one when the

child is at equilibrium position okay when the child goes further down during the oscillation

below the equilibrium position the extent is going to be more it's measured from the equium position we can see the

displacement is downwards which is X displacement is downwards X so here the extension is going to be x + X1 EX

exactly same as uh vertically oscillating spring Mass system what is which is the fifth example what I

explained the same uh analysis we can do here also so at this position the child is below the equilibrium position so

there at this position the forces acting on the child there are two forces weight is downwards the tension is going to be

now T is equal to K X1 + x so at the equum position KX is equal to mg so it Bal bces each other now the tension has

become greater than mg so when the child is below the equilibrium position the resultant force is going to be upward

direction towards the equilibrium position so there are the resultant Force f is going to be uh tension minus

mg that is K X1 + x - mg that is K X1 + KX - mg but we know that K T this at the equilibrium

position therefore K X1 is equal to mg so inste of kx1 we can substitute mg so mg minus mg will get cancel so it's

going to be KX that's going to be Newton Second Law according to that FAL

ma FAL Ma so KX is = to m a a is = k/ M into X okay so when we consider this oscillation the system child and the

trolin the elastic material that forms the oscillator oscillating system so they acceleration is towards the

equilibrium position because resultant force is acting towards the equilibrium position the child is below the

equilibrium position so the acceleration is going to be towards the equilibrium position which is upward Direction but

the displacement which is measured from the equilibrium this is the equilibrium position from theum position their

displacement measured downward Direction so they are in opposite each other at the same time k/ m is constant so

acceleration is going to be directly proportional to displacement but direction of acceleration is always

towards the equilibrium position you can see that here displacement is downward Direction they are in opposite direction

to each other therefore a proportional to - 6 when the child is below I can show it is at equilibrium position uh it

performs simple harmonic motion sorry not equ position when the child is below the equili position it performs simple

harmonic motion same way the child is above the equum position I'll draw the next position the child is above the eum

position so I'll use the same diagram so I'll reduce the uh displacement like this and the tremol in the elastic mat

is above the equilibrium position so it's like this way so this is the equilibrium position

so the child is oscillating and when the trambolin sheet the trolin material becomes above the position where the

displacement is measured from the position is going to be upward direction is it during this time also there are

two forces acting on the child one is the weight other one the extension is now decreased earlier the static

extension is X1 when the child is at the equilibrium person from that it started to uh move up and down or bounce up and

down now the extension has become X1 - x at this position X1 - x okay we know that when the child is at K X1 balances

mg now X1 - x is smaller than X1 so the tension is going to be smaller than the mg so the resultant force will be

downward Direction so here the resultant force is towards the equilibrium position which

is downward Direction because tension is smaller than weight so there the resultant Force f is going to be mg

minus t so that is mg minus in of T we can say k X1 - x that's going to be mg - kx1 + KX

so kx1 is equal mg when it is at the equum position I can substitute inste of kx1 mg so mg and mg will get cancel it

will become KX f is equal to KX the direction of the resultant force is downward direction that is towards the

equilibrium position when the child when the the child or when the elastic material is above the equilibrium

position during the oscillation so f = ma a so f = ma a f is towards the equilibrium position so that is KX = ma

a a is going to be downward direction towards the equilibrium position k/ M into X okay you can see when the child

is above the equilibrium position acceleration is again pointing towards the equilibrium position acceleration is

going to be downwards means it's going to point towards the equilibrium position downwards the displacement is

measured from the equili position which is upward Direction so they are among them in opposite direction to each other

acceleration is pointing towards the equum position so a is going to be directly proportional because k m both

are constant k m is constant so a proportional to minus so during the oscillation whether the child is above

the equum position or below the equum position the acceleration will be always points towards the equum position and

the direction of the acceleration will be opposite to the direction of displacement therefore that will perform

simple harmonic motion ACC is also proportional to displacement so it's going to perform simple harmonic motion

so during the oscillation If the child leaves the trampol so when the child there was a question in the past Pap If

the child leaves the trampoline during the oscillation that means maybe the trampoline would have come to this

position something like this maybe the temp have come to something like this horizontal position or something like

slightly upward Direction the child is thrown away from the uh trampoline thrown away from the uh the elastic

material then will that system perform simple harmonic motion no because when the child is in air above the equili

position the only force acting on the child is the weight mg so the only force acting on the child is mg means mg is

equal to M so a is going to be G that means the acceleration acting on the child is going to be constant which is

gravit ation acceleration it does not proportional to the displacement from the so this is the equum position the

displacement of the child from the equ may be X but acceleration is constant that is not proportional to uh

displacement there for it won't be a simple harmonic motion so when the child leaves the trampoline and when the child

moves freely away from the trampoline then the system which we include child and the trampoline will not perform

harmonic motion at that moment because the acceleration of the child becomes constant it won't be directly

proportional to displacement okay the seventh example is about oscillation of a weighted test

tube or small tube uh inside a liquid such as water so initially the tube is at equilibrium when it is at equilibrium

uh the length L of the tube is inside the liquid or water we'll say it so at that situation if you think consider the

forces acting on the system that is forces acting on the uh weighted test tube downward Direction uh weight upward

Direction up thrust I'll draw the whole system the weighted test tube has a small circle easy for that to draw so

the forces acting on the weighted test tube downwards weight mg at the equilibrium position and upwards the up

crust U up crust what you learned in unit one upus is equal to weight of the displaced

uh fluid so the length which is inside water is capital N so the up is going to be weight of the displaced fluid means

or weight of the displaced liquid water we take it as that uh we need to find the ESS volume we need to consider the

volume of the liquid displaced is going to be crosssection area if I say capital A the cross sectional area the length

into L that is the volume into density of the liquid row into G that is the up thrust so at the equilibrium position

the up thrust is equal to weight so a u is equal to mg at the equilibrium position so um from the equilibrium

position imagine you push the test tube little bit down and allow it so the test tube will oscillate in vertical on a

vertical plane so when the displacement this is the equilibrium position the blue color line what I drew so when the

test test tube is at rest so you push it down from the equum position and allow to oscillate so when the displacement is

downwards from the equum position the displacement x x during that situation the ESS is going to be U1 that is going

to be the total length inside so this is L already L was inside so this is anyway again L the equum position which was L

below the level of the upper surface of the liquid so the now the total uh up tress is going to be U1 is going to be

uh the U1 now or you can say even U whatever it is uh volume of the displaced uh liquid is uh cross-section

area into length inside the liquid is L + x into row G so the two forces acting on it one is the weight downwards and

the up now since the tube has moved inside the amount of the water displace or the volume of the liquid displace is

going to be more so up U1 is going to be greater than U so when I consider the resultant force acting on the system the

test tube it's going to be upward Direction because the resultant force is going to be upwards because the uppr is

more now compared to when the tube was at the equilibrium position so the resultant force is going to be upward

direction that is f is equal to U1 - mg so U1 means a l + x uh r g minus mg so that is if you multiply

inside a l r g a l r g plus a x r g minus mg a x r g minus mg so when we consider this Al r g and mg are same so

Al r g i can replace this Al r g by mg so mg plus a a row g x I rearranged it a row GX minus mg so mg mg will get cancel

it's going to be a row g x the resultant force that is acting upward direction towards theum position so new second L

this is going to be M so acceleration will be along the direction of resultant for you learn

that in unit one so acceleration also will be towards the equilibrium position that is upward Direction so if I make

the a subject a is going to be a row G over M into X which is upward Direction when the tube is below or the weighted

tube is below the equilibrium position so the acceleration is upwards means it's pointing towards the equilibrium

position but X is the displacement we measured from the equilibrium position so displacement is downward Direction

acceleration is upward Direction this all constant cross-section area constant density of the uh liquid constant G

constant total mass is constant so these all are going to be constant so a proportional to Min - x because they are

in opposite direction so when the tube is below the equum position what will happen uh a proportional to minus x a

acting towards the equilibrium position therefore it will perform simple harmonic motion when the test tube is

below the equilibrium position same way when the tube goes above the equilibrium position something like this if we

consider the tube goes above so there we can draw a diagram again like this it has gone up above the equum

position this waited to so this is the equilibrium position uh so the displacement is

upward Direction so the length of the test tube inside the liquid is now it's going to

be L minus this whole thing is L so it has gone up above the equilibrium position so that so you know that from

the surface of the liquid to equilibrium position is the L capital L so the same thing this is the capital L from the

surface of the liquid so from the surface of the liquid to this equilibrium position so equ is

drawn with blue color line This is L from here to here is L from this position to this position is L is it we

take that as L now the length actually inside the liquid is going to be this much that is going to be l - x so that

is going to be l - x is the length of the tube inside the liquid so l - x is smaller than l so what will happen the

uppr is going to be now l l - x now we can say the upas is l - x so when it is at the equilibrium position the upus is

equal to weight but now Theus becomes smaller than the upus when it was in the equum positions therefore the resultant

Force going to be upward Direction sorry the resultant Force going to be downward Direction because the upus has become

smaller now so when we consider the resultant Force it's going to be mg minus U1 when the tube is above the

equilibrium position so mg minus U1 when the tube is above the equilibrium position so that

is going to be mg minus U1 is a l - x r g so that's going to be mg - a l r g - a r g x a r g

x okay so inste of Al r g i can substitute mg so this mg and this mg will get cancel

mg - mg minus a uh sorry plusus Inus plus a r g x okay so that is going to be F the resultant force is going to be a

row g x which is the direction is downward Direction because mg greater than U1 so the resultant Force when the

test cube is above the equum position that will be downward Direction so that's going to be

M so the acceleration will be downward Direction so a is going to be uh a row g/ M into X acceleration is downward

Direction acceleration will be downward Direction but the X displacement is measured upward Direction so they are in

opposite direction to each other also you can see acceleration is directly proportional to displacement

acceleration is pointing towards the equ the weight test cube is above the equ position pointing towards the equ

position so a proportional to minus so it will perform simple harmonic motion so I separately did below the equ

person above the equ person make you to understand so whatever the direction it stays whatever the direction it is in

motion whatever it is the acceleration is towards the equum position and proportional to displacement such that

the displacement and ACC have opposite direction among them so it will perform simple harmonic motion

okay so I summarized all the examples I did the first example I didn't write because I told that uh when a small ball

bearing oscillates on us two incline planes like this which make Theta Theta that won't behave or that won't show

motion so I didn't draw that other than that all other six examples I drew it so in all these examples we came across uh

we gave a conclusion a proportional to - x a proportional to- X means k k m is constant k m is constant 2 k m because

here 2 K CES there are two Springs identical Springs attached on either side of the TR so here k m constant so

here trampoline also k m constant a g m constant so everywhere there is a constant there so generally I can

conclude in simple harmonic motion the acceleration is directly proportional to its displacement with minus sign

indicates they are in opposite direction to each other so a equal to Min - CX I can say there's a constant C the C could

be different for different uh examples for example here the constant C is 2 k/ m but in all these others k m k m k m

but here the constant is a r g/ m when weight at T cube oscillates in liquid anyway whatever the constant could be

different for different types of uh oscillators but there is a constant C Okay so what is the meaning of this

constant what does it mean what's the physical meaning of this constant we need to know that okay so to identify

actually what this constant C represents in simple harmonic motion we know that simple harmonic motion occurs or happens

in one dimension either up and down you can see all these things it's happening uh left and right one dimension this

also one dimension on a slope either along the slope or upward in a slope this is this is one dimension left or

right this one dimension up or down one dimension up or down this also one dimension up or down so the oscillation

simple Harmon motion happens or happens in one dimensional motion either up or down or left or right like that okay so

what is the meaning the physical meaning of c we want to find it okay so actually uh to relate or get a physical meaning

we need to relate the simple harmonic motion could be related actually simple harmonic motion could be mathematically

related to circular motion even when simple harmonic motion is not happening in circular path when we consider the

circular simple harmonic motion there is no circular motion actually it's happening in one dimension circular

motion happens in two Dimension but even there is no two dimensional Motion in simple harmonic motion that happens in

one dimension even that way it happens mathematically m matically we can relate the simple harmonic motion to circular

motion from that we can identify the physical meaning actual meaning of c what could be the effect of C for

example in this particular oscillation what will happen if the constant increases by increasing the spring

constant of the spring or by reducing the mass attached if I increase the constant what will be the outcome we can

find it same way here also by increasing the density of the liquid we can increase the uh constant C related to

this oscillation so what could be the outcome of that we can identify so to identify that we should know the meaning

of c okay that could be explained by using a circular motion mathematically okay so we'll consider a

particle is moving on a circular path of radius capital A Center is O moving on a circular path at constant angular speed

Omega for Simplicity no reason I can't take initially anywhere but for simple thing I took it initially the particle

is along the diameter in horizontal position at this position initially when t equal 0 later at time tal T the

particle has come to this position so when we draw a projector projector means a vertical line from the particle to the

diameter I consider the horizontal diameter we can draw any diameter but simple thing I we consider a horizontal

diameter so if you draw a projector from the particle to the diameter the base of the the the projector touches the

diameter which is in horizontal position at Point N so the distance of the n n is the projector drawn from the particle to

the diameter the distance of the N from o is X X which is directed towards right so in this right angle triangle we know

that X is going to be X is going to be a COS Theta but the particle is rotating moving at constant angular speed Omega

therefore we can say Omega = Theta / T from the Theta isal Omega T So X is going to be a COS Omega t Okay so X is

the position of the distance of the projector from o you can see that imagine that when the particle moves

from this position I'll name this position as initial position P this as Q when the particle moves from p in

anticlockwise Direction the projector when the particle is at P projector n will be at P when the particle moves

this way the projector will move towards o o when the particle is at exactly at this position exactly above o the

projector n will be at o when the particle moves further towards left the it will move from o to Q so initially

the projector was at this position here if I say this is O the pro will move from a p to

O then when the particle moves to O it will move from here to here to Q when the partic comes to this

this position but if I draw a projector again the projector will move from Q to O then again from o to P so that means

simply I can say when the particle completes one full circular path the projector will drawn to horizontal

diameter will move from P to q and again from Q to P so when the particle completes one full circular part the

projector will complete one full oscillation about o about o so the time taken for the circular motion and the

time taken for the projector to complete one full oscillation will be exactly the same okay so that's a comparison we are

going to do right so we know that x = a COS Omega T okay what could be the I told that the particle is moving at

constant angular speed Omega so when the particle moves at constant angular speed what could be the speed of the projector

in when we measured uh that's going to be differentiation anyway not in the physics but just for your information

that's how your knowledge I'm doing it when you differentiate x with respect to time that is speed is going to be DX

over DT when you differentiate this it will become minus a Omega sin Omega T when you differentiate cos it will

become minus s you add a different when you differentiate Omega to become Omega so minus a Omega sin Omega T that's

going to be the speed and don't worry you so that is the speed of the projector end when we measure when you

differentiate again we'll get acceleration of the N when the particle moves on circular path around it so the

acceleration is going to be d/ DT or d2x over DT to dv/ DT so when you differentiate the velocity again with