NEET Booster Week - SHM

Añadido: 2026-05-23

[00:00:39]Mama daddy.

[00:01:05]Oops.

[00:05:28]Yeah. Good morning my dear students. On this seventh day of physics of booster classes from MS we are going to have a topic by name SHM or oscillations.

[00:05:42]An important topic from which generally one or sometimes even two questions come in the neat entrance examination and as everyone knows the knowledge of this SHM concepts we use in sound and wave motion in physical optics we use even in electromagnetic waves we use so directly maybe one or two questions come from SHM indirectly its application will be there in another as many as four to five questions. So on a whole a very important topic. Okay. Now let us see the basic fundamentals as quickly as possible.

[00:06:26]Everyone of us know what do you mean by shm. Suppose there is a simple pendulum.

[00:06:30]Pull the bob of the simple pendum by small angle and leave it. It will oscillate there. This is known as the two and fro oscillation. Any tone for motion of this simple pendulum bob is called as shm. The basic points we require here is what do we mean by one osillation. So starting from let us say p there going from p to q and again coming back to p is known as one osillation and time for such an oscillation is known as a time period.

[00:06:59]Okay. Frequency is a reciprocal of the time period. N is 1x t and omega is 2 pi byt and 2 pi into n. Now what are the most essential formula under SHM? Okay, all the formula are compactly being written here there. Listen suppose this is a simple pendulum Bob or any body performing SHM.

[00:07:21]Let us start counting the time when the body is at the mean position. Everyone of you know that this is a mean position there and at when the bob is at the mean position the time gets started counting.

[00:07:34]So at t0 bob is at the mean position.

[00:07:37]Now in t seconds there in t seconds let the bob go from the mean position to any point like p having a position x. x is the position of the particle as measured from the mean portion. X is an important wording to be thoroughly understood. X is always the position of the body as measured from the mean portion. The time the body will have some velocity v some acceleration a when body is at mean position at t is zero. You know that x is equal to a sin omega t. What the velocity means? You have to go for differentiation. So a omega cos omega t as well as omega into under root of a square - x square. So this is a velocity time relation and this is the velocity and position relation. Both the formula we have to be good at. What are the formula for acceleration? For any body performing SHM, acceleration is always given by minus omega square x. So minus sign says that a and x will be acting in opposite directions.

[00:08:39]What do you mean by maximum velocity?

[00:08:42]The other name of maximum velocity is nothing but velocity amplitude. Instead of using a direct wording like maximum velocity, they can use little bit of technical wording like velocity amplitude. Velocity amplitude means maximum velocity of a body performing hm that is given by a omega. Similarly acceleration amplitude that corresponds to maximum acceleration that maximum acceleration is equal to a omega square.

[00:09:08]When the body is at any position x as measured from the mean position, its potential energy is given by half m omega square x and kinetic energy given by/ m omega square into a square - x².

[00:09:22]Potential energy kindinetic energy total energy is a sum of these two energy is there. If you sum them up, it comes out to be/ m omega square a square. Okay.

[00:09:32]The small point we have to make a note is that x is equal to a sin omega t.

[00:09:37]That formula is applicable provided the time is started counting when the body is at the mean portion. At t0 the body need to be at the mean portion. Now just say in some occasions there at t0 body may be at the extreme position. When the bob is at the extreme position that time you started the stopwatch and the time that moment is zero. At when the body is at the extreme position the then the time started counting. In such a case there x is equal to a cos omega t. In such a case x is equal to a cos omega t.

[00:10:14]The meaning of x is same all the time. x is a portion of the body as measured from the mean portion there. Uh then x is equal to a cos omega t. So by now you know that either x can be a sin omega t or a cos omega t or a general formula including this one. Okay. Now we are going to have the application of these patients here.

[00:10:37]Coming to the first bit, a combination of SHM plus chyntics.

[00:10:45]A particle performs SHM with a velocity amplitude. What do you mean by velocity amplitude means maximum velocity?

[00:10:54]Velocity amplitude means maximum velocity of 6.28 m/s.

[00:11:00]I want to know what is the average speed of the particle considering one complete oscillation.

[00:11:08]Yeah. So velocity amplitude means maximum velocity that is given as 6.28.

[00:11:12]What is the formula for maximum velocity? A omega a omega is 6.28. What is omega? Omega is 2 pi by t. So a into 2 pi by t is 6.28.

[00:11:24]You know that pi is 3.14. So 2 pi and 6.28 to 8 get cancelled. From these things, a by t comes out to be one.

[00:11:31]First step, we have to just leave the step here and go to the next point there.

[00:11:36]Considering one complete oscillation, you want to know what is the average speed. You have to just make a note is he asking average speed or average velocity. Average speed is asked.

[00:11:46]Average speed is defined as total distance covered divide by total time.

[00:11:51]Average speed for one oscillation means average speed for one osillation is total distance covered. Yeah. What are the total distance covered for one oscillation? As we discussed at the beginning, one oscillation includes this two and fro motion to and fro. During monosulation, what are the distance covered? Distance covered is defined as actual path covered by the body. Actual path covered by the body. Without bothering about the directions, what the actual path covered? All of you know that this is a this is a like this it will cover 2 A. Like this it will cover 2 A. So total distance covered comes out to be 4 a and time for one osillation is time period t time for one oscillation is time time period t. So it comes to be 4 a by t out of which a by t already we have found 1 from the first step we have found a by t1. So 4 into 1 equal to 4 m/s. Okay.

[00:12:44]A simple but an important model.

[00:12:51]The next bit.

[00:12:54]A body of mass 2 kilogram performs shm starting from mean position very clearly said. Okay. The motion is started that means at t0 body is at the mean position. The mass is given as 2 kg. So mass of the body is given as 2 kg there and the motion is started from the mean portion. The position x and time t are related by x = 8 sin by 12t cos by 12t.

[00:13:26]When the motion starts from the mean position there, the general relation for x is equal to x = a sin omega t. What does it mean? We must have a single sign. We must have only one sign. But here intentionally both s and cos terms are given there. Okay. X being expressed in terms of sine as well as cos. What is our duty then? Our duty should be to convert this sine and cos into a single sign term. Single sign term. Okay. Then you have to recollect the good trigonometry. Which trigonometrical formula will tell will make you to convert sin theta into cos theta into a single s term. Okay. Then before we go to that one there, you have to recollect the basic trigonometrical formula that sin 2 theta is equal to 2 sin theta cos theta. Okay, what is the importance of that formula here? Using this formula, you can convert sin into cos into a single sign. Sin into cos you can happily convert into a single sign.

[00:14:29]Okay, listen carefully. X is equal to 8 sin by 20 cos by 12t. There this is the equation given. Now just for convenience sake this 5<unk> by 12t I am calling it as theta there. So this<unk> by 12 t is again theta. Now this is of the shape x is equal to 8 sin theta cos theta. So we have got sin into cos. How do you convert sin into cos into a single time means sin theta cos theta can be written as sin 2 theta* 2 is this can be written as this complete term is this term sin 2 theta by 2. This 8x2 get cancel as 4 x= 4 sin 2 theta. So x is 4 sin 2 theta.

[00:15:10]What is theta comes out to be? Pi tx 12.

[00:15:13]If you just do simplification there x= 4 sin this 2 and 12 get cancel by 60. So we have successfully converted this sine into cosm into a single sin term. This is what we got it. Now you compare this one with the standard S shm relation X= A sin omega T. You know that A comes out to be four. If you compare XX compared naturally is equal to is equal to compared A is equal to 4 sin sign compared TT compared omega is<unk> by 6.

[00:15:47]So A is 4 omega is<unk> by 6. Now what did he ask in the question? Find the momentum of the body when it is at the mean position. We want to know what is the momentum of the body when it is at the mean portion. What are the formula for momentum? Momentum is the product of mass into velocity. Momentum at mean portion means mass into velocity at mean portion. Velocity at mean portion is the maximum velocity is at mean portion. You know that speed is maximum, velocity is maximum. So velocity at mean portion is the maximum velocity a omega. You know the value of a already we calculated we know the value of omega. Just substitute there you get the momentum. See that units you write properly the momentum mass into velocity kilogram m/s.

[00:16:36]Okay.

[00:16:40]The next bit you please differentiate between the previous question and this bit.

[00:16:49]A particle performs shm starting from rest. This is the key wording. Okay, that should uh be clarity observed. A particle performs SHM starting from rest. Okay, particle performs SHM starting from rest means at t is equal to0 whenever the motion starts the time we call the time as zero. Whenever the motion begins that time t is zero. So starting means starting time zero. At starting time zero, its velocity zero. At starting time zero, it is starting from rest.

[00:17:28]What do you mean by rest? At t0, its velocity is zero. There's the meaning of rest. Rest means velocity zero. When starting time, starting time is zero.

[00:17:37]Velocity is zero means the body is at extreme position as everyone of us know that. So at the extreme point velocity is zero. Here the velocity is maximum.

[00:17:50]So at t0 velocity has to be zero.

[00:17:53]Velocity has to be zero means the body has to start from the extreme point. At t0 body is at extreme point. At t0 bodies at the extreme point means we have to go for the relation that x is equal to a cos omega t. x is equal to a cos omega t. So if you have a proper short notes there that point will be clear. At t0 the body is at the extreme position x is equal to a cos omega t as we discussed.

[00:18:22]Now pay attention. So uh now you have to depend on the relation x is a cos omega t. Now in this particular question there instead of giving a single cause there he is giving cos square and sin square terms. Okay, we want the relation to be just a s simple cos but intentionally it is given as cos square and sin square.

[00:18:46]So x is given as 12 into you can take 12 common 12 common 12 into cos square p<unk> t by 12 - sin square p<unk> t by 12 the 2 in cm there what is the acceleration amplitude that's what being asked okay x = 12 this p<unk> t by 12 we call it as theta this is also theta so this is of the shape x= 12 into cos square theta - sin square theta what is cos square theta - sin square theta then simple trigonometry will tell you that cos 2 theta. We need to be thorough with the two formula there. Sin 2 theta is 2 sin theta cos theta. cos 2 theta is equal to cos square theta minus sin square theta. This cos theta - sin square theta can write as cos 2 theta.

[00:19:29]x= 12 cos 2 theta already we know pi t by 12. So 2 and 12 get get cancel six times. x= 12 cos by 60. You compare with the standard relation X is equal to in this particular case as already we discussed X is equal to A cos omega t. So comparison tells you that A is equal to 12 and omega is<unk> by 6.

[00:19:52]You know both the values of A and omega.

[00:19:54]But what is being asked there? What is asked means? We want to know acceleration amplitude. Acceleration amplitude means maximum acceleration.

[00:20:03]How to calculate maximum acceleration? A omega square. Maximum acceleration is a omega square. So what is the amplitude?

[00:20:11]Already we know 12 amplitude is 12.

[00:20:13]Omega squared. Omega is<unk> by 6. Omega square means p<unk> square by 36.<unk> square by 36. So 12 36 will get cancelled as many as 3 * 8 12 36. You must be knowing the value of pi square.

[00:20:28]Pi is 3.14. Pi square mean 3.14 whole square that comes to be 9.9 clos. So almost we can call it as 10. Pi square is 10. So many times in different topics of physics we approximate p<unk> square as 10. So p<unk> square is 10 by 3. 10x3 means it comes to be 3.33. Remember that all these are given in CGS system of centm. So the answer comes to be 3.33 cm/s squared. Since it is given is x is given in cm. It is centm/s squared. So when to use x is a sin omega t. When to use x is a cos omega t and how to use this trigonometrical formula sin 2 theta cos 2 theta. The these two questions will tell you the next bit.

[00:21:20]One of the highly repeated question in the entrances for a particle performing SHM the potential energies are E1 and E2 when it's positions are X1 X2. So when the position is x1 potential energy E1 when the position is X2 potential energy is E2.

[00:21:43]If the position is something different 2 X1 + 9 X2 then what is the potential energy?

[00:21:51]Okay since decades this is a important model many times the question has come in the real entrances. Okay. Now the first thing is whenever you are talking about potential energy of the body the first step is recollect what is the formula for potential energy. Potential energy of a body perform SHM is half m omega square x². Okay, you must be knowing that for a body performing shm there mass of the body will always be constant. It's a time period and angle of velocity will be constant. Half is anyhow constant mass constant and omega is constant for a body performance system. Same time period will be there all the time for every any oscillation.

[00:22:31]So all these are constant terms but you know that position keeps on changing from the mean portion closer farther and so on x changes and potential energy changes. Now these constant terms can be replaced by proportionality sign. So potential energy is directly proportional to x².

[00:22:48]Next important step is that this small idea you should get there. Why root is being made there? Why root is being made? Okay. Then here we are talking about the position x1 position x2 some other position there. So we don't want we want position not position square here the data in was position not okay not position square here you have got position square but we want position that's why uh just take square root taking square root e is proportional to x square already you understood taking square root square root of e is proportional to x that means even a proportional to b is also equally proportional to a so root a proportional to x it means x is proportional Ry proportion is sign replaced by a constant term. Here proportion is sign replaced by constant term. So this particular formula will rule this small question and you have to use that particular formula exactly thrice hereafter. So x is equal to k root k into root e. Now come on tell me what is the data given. When the position is x1 the potential energy is even. So that means when the position is x1 potential energy e1 when the portion is x2 potential energy e2 okay now when the portion is some new portion like x3 what is x3 means 2 x1 + 9 x2 when the portion is x3 like 2x1 + 9 x2 I want to know what is the energy okay so uh when the portion is x1 the corresponding energy e1 portion x2 corresponding energy e2 when the portion is x3 corresponding energy e3 under root of e3 3. So x1 roo<unk> e1 x2 root e2 x3 naturally root e3 naturally. Okay. But what is x3 given the question? x3 is given as 2 x1 + 9 x2. So x3 to x1 + 9 x2 is k into root e3. Do you know x1? Yes, definitely we know thoroughly this x1 there x1 is k roo<unk> e1. x2 is k root x2 is equal to k roo<unk> e2. So you have to write here there x1 is k roo<unk> e1 x2 is k roo<unk> e2 then uh so in place of x1 k root e1 in place of x2 k root e2 and uh uh this is k roo<unk> e3 you can see that k is common on both sides the k term get cancelled what are you getting you are getting that root e3 is equal to I'm writing this term first root e3 is equal to 2<unk> e1 + 9<unk> t2. There is a chance like this in the entrance examination when this question has come twice there. Twice then in one exam they have asked the root relations there root of energy relations. Then which of the following relations is correct means root E3 is equal to 2<unk> E1 plus 9<unk> E2. Okay, this is the answer in terms of root of energy in one in one particular entrance exam they did not give the options in this way but they are talking in terms of E3 E3. So when you know root E3 how do you get E3 means just square it square it.

[00:26:08]So already you know that root E3 is 2 roo<unk> e1 + 9 root e2 if you square on either side square on either side e3 comes out to be this complete term whole square. This complete term whole square means a + b whole square. Simple a + b whole square. A square + b square + 2 a b. So here a square means you get 41.

[00:26:28]Here b square means 81 a2. 2 AB means 36 root E1 E2. So there's a for there's answer you find there. Okay. 4 E1 + 81 A2 + 36 root E1 A2. So depending upon the way they give the options either answer can be in terms of root E or just in terms of E. Both the things you have to be good at.

[00:27:00]The next bit a simple pendulum bob performs shm. So this is a bob performing shm between the points B and D. So and understood this is one extreme point there. This is one extreme point. This is the other extreme point. A simple pendum Bob is performing SHM between these two points. It is not crossing these two points there. So these are the two extreme points B and D. Okay.

[00:27:26]from some selected origin there it's it is at a distance X point B is at a distance X and this point is at a distance Y from some selected origin this point is X and this point is at a distance Y now if the velocity of the body at the midpoint of B and D midpoint of B and D means mean portion midpoint of B and D means exactly the mean portion so velocity of the body at the midpoint of B and D is given as B then you want to know what is the maximum force acting on the bob and mass of the bob is given as m. What is the maximum force acting on the bob?

[00:28:06]Okay.

[00:28:07]So for the last time I repeat that a simple pendom bob is performing oscillations between what extreme points is it oscillating means between the points B and D and B is at what distance from a selected origin means from a selected origin of coordinate system it this point is at X this point is at Y very good and at this midpoint of B and D here the speed is given as V then you want to know what is the maximum force acting on the body Now you have to understand first what are the formula for force mass into acceleration. Mass of the bob is already given as a very good. Now I want to know maximum force. You know that mass is constraint. Force to be maximum means acceleration is to be maximum. What is the formula for maximum acceleration?

[00:28:55]There maximum acceleration comes out to be a omega square. Maximum speed a omega maximum accelation a omega square. Here mass is given amplitude is nowhere given. So you have to calculate amplitude. Okay. Nowhere given omega. So even omega also you have to calculate.

[00:29:14]So in this question m given you can use amplitude term not given the question omega not given. So I have to find both a and omega and dump it here. Okay. Now how to calculate the amplitude? Just see all of you know that this is the extreme point. This is other extreme point and this is the mean point. Distance between the mean point and extreme point is known as amplitude. You know the meaning of amplitude. Distance between mean and extreme. Distance between this mean and extreme is another amplitude A. This is A. This is A. Okay. Now you can see what is the relation between these things. X + A + A. X + 2 A is equal to Y. You can see that from here to here X + 2 A is equal to Y. X + 2 A is Y. From these things 2 A is equal to Y - X. 2 A is Y - X. From that A is equal to Y - XX2.

[00:30:10]So whatever amplitude we want we have definitely found by simple from this diagram almost directly we get a. So we got A. Next step is how to calculate omega. Okay. Velocity at this mean position is given as V. All of you know velocity at the mean portion is nothing but maximum velocity. Velocity at the mean portion is given as V and that velocity is the maximum velocity. What is the formula for maximum velocity? A omega. What is A? We know already we know that A= Y - X2. We got it. So V= Y - X2 into omega. From that you have to calculate omega here. Omega comes out to be 2v by Y - 6. Okay. Now you want to know what is the maximum force. As already being explained maximum force is mass into maximum acceleration mass is given as m a omega square you have to write in place of a y - xx2 omega² omega square means 4 v² by y - x whole square then you have to do simplification 4x2 comes by 2 so you get it as 2 mv² by y - 6 this answer we get place okay this is the answer for the bit the next See here it is an important model many many entrances under under the name of this question.

[00:31:37]For a particle performing SHM, For a particle performing SHM, the velocity V and position X are related by 16 V² is 1 625 - 25 X² since it is velocity squared. The unit happens to be me squared by second square. You know velocity me by second v square means there is a unit. Once again I want to know what is the acceleration amplitude.

[00:32:07]Okay, acceleration amplitude means you want maximum acceleration.

[00:32:10]Maximum acceleration is nothing but a omega squared. So from this equation you have to calculate amplitude and you have to calculate omega and then uh you have to calculate maximum acceleration. First thing is whenever vx terms are being talked about what the best relation between them. So v is equal to omega into under root of a square - x square.

[00:32:33]So v square comes out to be omega square into a square - x square. If you expand if you expand v square is omega square a square minus omega square x square at the first step. Okay. Now this is the basic equation we have to totally rely on. Now what is given the question there? In the question instead of v square he has given 16 v². So 16 v square is given as 625 - uh 25 x square.

[00:33:03]I want v square. So get 16 to the denominator. Get 16 to the other side.

[00:33:08]So it will come to the denominator. So v square is 6 25 x 16 - 25 by 16 x². Now if you compare these two equations, you can see that v² v square compared is equal is equal to naturally compared.

[00:33:22]Then omega² a² is equal to 625 x 16 is here I have written you can see this green green comparison these green circled one omega square a square is 625 x 16 and if you compare these two omega square is 25 x 16 okay this red color circled omega square is 25 x 16 now we can see omega square is 25 x 16 from that omega is<unk> x 4. You know the unit of omega<unk> by 4 radian per second. The next step is take square root here there. If you take square root of omega square a square is omega a. So it comes out to be 25x4 out of which omega you know 5x4 from that a comes out to be 5 m. You want maximum acceleration. So the maximum acceleration comes out to be a omega squ. You know the value of a you know the value of omega. So a is five omega speed 25 by 60 in the next step you'll be getting the onset is it meas second sphere or centime per second spade. So you have to see this unit carefully.

[00:34:32]Okay all these are in SI system SI. So naturally here the answer comes out to be m by second square.

[00:34:52]The next bit a particle performs shm with the velocity amplitude p and acceleration amplitude q.

[00:35:01]Okay. I want to know what the amplitude of oscillation. There is a small part and frequently asked question for what position for what position means for what X he is asking potential energy is 3/4 of the kindinetic energy okay one routinely important question for a particle performing SHM the velocity amplitude is P intentionally the starting data uh slightly given in a different way so velocity amplitude means Maximum velocity that is a omega that is given as p. Acceleration amplitude means a omega square that is given as q. Okay. So a omega is given as p. From that omega is p by a. From this omega you can calculate p by a. a omega square is q. A omega means p by a. Now so omega square means p by a whole square is q. If you simplify it there if you if you simplify this particular one there this comes out to be the amplitude. If you just do simplification there, amplitude comes out to be p² by q. Yes, amplitude comes out to be p² / q. Now we want to know for what position x potential energy is 3/4 of the kindinetic energy. What is the formula for potential energy? Half m omega square x². You have to be thorough with the formula which we discussed and which you are already thorough with.

[00:36:26]Potential energy/ m omega square x² is 3/4 of kinetic energy. What are the formula for kinetic energy? Half m omega square into a square minus x square.

[00:36:35]Yeah. Half m omega square term get cancelled. Some simple simplification there. You get this answer there. Okay.

[00:36:43]So instead of giving amplitude directly you have to calculate amplitude and go for it. In the world entrances there they given amplitude directly as a but here you have to have one extra step there. find out amplitude first and uh in terms of uh P and Q and get the answer.

[00:37:17]Okay.

[00:37:21]A particle is executing SHM along a straight line. Its velocities when its positions are x1 x2 are given by v_sub1 v_sub_2 that means when the position is x1 velocity v_sub_1 when the position is x2 velocity is v_sub_2 you want to know what is this time period okay the question says that when the position is x1 velocity v_sub1 when the position is x2 velocity v_sub_2 from these things our idea is to calculate time period this is a basic question among different formula in shm what is the correct formula to be properly recollected in time that if you think immediately you'll get it vx relation we want vx what is the vx relation v is equal to omega into under root of a square - x square now square it down a square it both sides v square is omega square a squareus omega square x square now here you can write v1 square is omega square a squareus omega square x1 square already we discussed that for a given shm the amplitude and omega will remain constant. So these circle terms are constant. Circle terms are constants. So v_sub1 square is omega square a square minus omega square x1 square. In the next case when the portion is extra velocity is v2. Here v_sub_2 square is omega square a square minus omega square x2 square. Okay. Now how do we get the idea that we have to do 1 minus 2? Sometimes we divide sometimes we subtract. What the character logic simple logic behind it?

[00:38:54]Means if you see the question there x1 given v_sub1 given x2 given v2 given v1 time period time period means omega omega is 2 pi by time period. So we have got interest and respect for x1 x2 v_sub1 v_sub_2 and omega omega but we don't want anything like t a amplitude a we don't want anything like amplitude a so we have to get rid of amplitude. How to eliminate amplitude? How to eliminate amplitude means you just subtract these two equations. If you subtract these two equations, amplitude term will get cancelled. So if you do v1 square minus v2 square there these two terms get cancelled and then you get omega square into x2 square - x1 square. If you subtract from that omega square you bring it here omega square comes out to be this much from that you can calculate omega omega is 2 pi byt from that you can be calculating the time period.

[00:40:02]Yeah, grabs questions are very important uh in any subject uh in in most of the chapters in physics and definitely in SHM.

[00:40:16]Once again, an entrance question here.

[00:40:18]There listen the position time graph the position time graph of a particle performing SHM is given like this. The position time graph is given like this.

[00:40:28]You want to know what is the acceleration of the particle exactly at this time 4x3 seconds.

[00:40:35]Okay. First of all from the in in any graph some data will be there is what do you mean by graph? Graph means data being represented by some diagram.

[00:40:48]That's the meaning of graph there. So some data will be there. Now how to use the data? All of you know that this is X. This this 1 cm is x maximum is if 1 cm is x maximum that is the highest possible value of x the maximum value of x itself is known as the amplitude. So from this amplitude of oscillation comes out to be 1 cm. Amplitude of oscillation is one. Okay. Now you have to use this data properly. All of you know that this is known as one cycle. This is known as one cycle two and fro motion. This is two. This is fro. And the time taken for one oscillation comes out to be 8 seconds there from here to here 8 seconds. This is two and fro motion together is one oscillation as we discussed as the first point please. And time period is 8 seconds. So from this data simply you have to calculate amplitude not 1 meter but 1 cm and time period is you have to see sometime they give the value in minutes there there's a time period in seconds s standing for seconds this is 8 seconds okay from time period you can calculate omega there from time period you can definitely calculate the omega there so the omega here comes out to be 2 pi by t so<unk> by 4 radian per second there So it comes out to be<unk> by4 radian per second.

[00:42:15]This the value we get. Then the next point is that I want to know what is the acceleration of the particle. What is the formula for acceleration?

[00:42:23]Acceleration comes out to be a= minus omega square into x. Acceleration is minus omega² into x. Okay. Now omega omega comes out to be<unk> by 4. You have already seen that this omega is equal to<unk> by 4. Omega square means p<unk>² by 16. What is x? x is equal to a sin omega t. Okay. So at t is 0 body is at the mean x is 0. At t0 x is zero.

[00:42:54]Then naturally x is a sin omega t. A sin already you know omega<unk> by 4 into t.

[00:43:00]Already amplitude is 1 1 sin by 4t.

[00:43:04]Okay. So acceleration is minus omega square x - omega square<unk> square by 16 x means sin by 4t. He is asking acceleration at what instant of time at t is equal to 4x3 seconds. So in place of t you have to write 4x3.

[00:43:19]This 4 will get cancelled. So this is sin<unk> by 3. Pi radians means 180° by 3 means 60. So it comes out to be sin 60 means <unk>3x2.

[00:43:31]So what the acceleration of the body here means? So minus<unk> square by 16 into <unk>3x2 or <unk>3<unk> square / 32 and uh all these are given in cm now. So it comes out to be cm/ second squared.

[00:43:46]So this acceleration of the body so much cm/ second square.

[00:43:50]Okay.

[00:43:54]The next thing generally about this particular question which has appeared nearly uh 9 10 years back there the students were wrong with uh one small point there. The point I want to emphasize a simple pendulum bob of mass 600 g. So mass of the bob is 600 g. You are doing everything in SI system. So uh convert g into kilogram.6 kilg. A simple pendum bob of mass 600 g is performing shm. I want to know what is a restoring force acting on the boba when its angular portion is 3°. When the angular portion is 3° okay now listen carefully. Suppose this is a simple pendulum performing session. At one stage the angular position is theta.

[00:44:44]Then I want to know what is the restoring force. Recollecting the basics of the simple pendulum there for the bob weight always act vertically down vertically down. This weight can be resolved into two components there.

[00:44:56]Okay. This weight can be resolved into components. If this is theta, this is theta there. This comes out to be mg cos theta. This comes out to be mg sin theta. And this particular force that is directed towards the mean position is set to be the restoring force. Okay. The fundamental point in simple pendulum is that the whenever the bob is having a angular position theta the restoring force acting on the body is given by mg sin theta. Restoring force is mg sin theta. So the restoring force acting on the body is mg sin theta. Okay. Now we want to know what is the restoring force when the angle uh angular position is 3°. The main point in the question is restoring forces mg sin theta. Okay. The two this theta is 3°. Any angle less than 6° we call it a small angle. Any any angle less than 6° we call it a small angle there. So 3° is a small angle. Whenever the angle is small you know that sin theta has to be written as theta. But the care to be taken is that sin theta should be taken as theta only when theta is being expressed in radians never in degrees. Never in degrees. So sin theta is theta provided theta is expressed in radians. So the first and foremost task here is the angle is given as 3°. Convert degrees into radians.

[00:46:23]Convert degrees into radians. 180° is pi radians. 180° is p<unk> radians. 3° is how many radians. So ratio and proportion 3x 180 into p<unk> by 60 radians. So 3° is nothing but p<unk> by 60 radian. So restoring force is mg sin theta. Mass of the bob is 6 kg. G is given as 10 in SI system 10². Theta in radians you have to write p<unk> by 60.

[00:46:50]Everything you have done in SI system now. So force comes out to be in Newton.

[00:46:54]Okay. You know that 6 into 10 is 6<unk>i / 60. This this 10 * the<unk> by 10. You know what is the value of pi? Pi is 22 by 7 3.14. So 3.14 divided by 10. So here the answer comes out to be 314 Newton. So this will be the answer for the question 314 Newton.

[00:47:19]So the main thing is when theta is small angle there sin theta has to be expressed sin theta can be approximated to be theta and theta has to be expressed in radians there's a there's a point to be highlighted uh when you are revising the things pay attention okay see very very standard question in a simple pendulum concepts Yeah, a simple pendulum bob of mass 100 g.

[00:47:56]This is a simple pendom bob of mass 100 g and is of length 40 cm. It is suspended to the ceiling of a lift. As you can see, there's a simple pendulum suspended to the ceiling of a lift and the lift itself is moving up with an acceleration 2 m/s squared.

[00:48:17]Now I want to know find the time taken for it to complete 10 oscillations.

[00:48:23]Okay, he's not asking time period. He's asking what the time required for exactly 10 oscillations to be completed.

[00:48:30]And more important other bit is what is the maximum kindinetic energy with which it oscillates if its amplitude of oscillation is 2 cm.

[00:48:39]Okay.

[00:48:41]The formula for time period of a simple pendulum is 2 pi root l byg provided point of suspension is at rest. Okay. If this point of suspension is at rest then you can apply temporaries super root ly here when the lift is moving up this point of suspension will also move up.

[00:49:01]It accelerates up. Lift is accelerating up. Point of suspension is also accelerating up. In such a case, time period of a simple pendulum is given by Inuch case time period of a simple pendulum is 2 pi under root of L by G effect effective. Instead of simple G, it happens to be G effective. How to calculate effective acceleration due to gravity. Everyone of you know that the regular acceleration due to gravity always act vertically down. Regular gravity act down. Now this simple pendulum system is attached to accelerating system. Anything attached to accelerating system will experience a pseudo acceleration but in the opposite direction. So it will have another acceleration like a suda accelation a in the opposite direction. So what is the effective acceleration which it is oscillating g plus a this effective g comes out to be in this case g plus a after you conclude that particular small point there it becomes fairly direct there. So time period is 2 pi roo<unk> l by g effective. In place of g effective we have to write g plus a. Now here you have to take care that many students directly write G as 10 10 10 G should be taken as 10 only when the question set says so if you read the question once and twice no no nowhere it is mentioned the G is 10 so you have to be careful in applying G correctly as 9.8 okay so time period is 2 root L by G + A so G is equal to 9.8 8 acceleration is 02. So if you simplify these things time period comes out to be 4 pi by 10 seconds. Okay just simp after simplification you get this. This is not the answer. He is asking what the time per exactly 10 oscillations. Time for one osillation comes out to be 4 pi by 10 seconds. He is asking time for 10 oscillations. For one oscillation this much mean 10 osillations to multiply with 10. So 10 into 4 pi by 10 it comes to be 4 pi seconds. Then comes the point what is the maximum kindinetic energy with which it should be oscillating. So if the amplitude of oscillation is 2 cm what is the formula for maximum kinetic energy? Kinetic energy is half mv².

[00:51:11]Maximum kinetic energy means/ m into v max²/ m into what is the maximum velocity? A omega square A omega square half into mass of the body is 100 g convert g into kilogram. So half into.1 you know the amplitude of oscillation is 2 cm convert cm to meters there. 02 and already we have calculated omega here how to calculate omega means you know the time period now you know time period is 4 pi by 10. So omega is 2 pi by t 2 pi by time period. time period this 4 pi by 10 you have to write then you get the then you'll be getting this particular omega as 5 five rad/s²ared so you know amplitude you know omega already mass is given mass is given as 100 g converted into kilog amplitude he himself has given 2 cm 002 m omega value we have just now calculated five there five do simplification you get the answer in jewles take Okay, they may ask the answer maybe in terms of jo maybe suddenly in terms of okay so you know that one jol of seven there and you can properly get the answer there. So by being little bit alert all the time there okay during the exam time you can do well but more importantly you need to be much much more alert and definitely hardworking.

[00:52:42]We have seen uh very unfortunately many re needs and even uh I once conducted second time there. Okay. Then what the past history says that in any exam that is being reconducted exam papers will be definitely of higher standard. Exam papers will be of higher standard.

[00:53:10]Okay. Then the two after all we are human beings there so most of us may become lethargic we shouldn't be because of getting lethargic and being papers being higher standard there generally competition will become more and much more there okay so most of you have worked so sincerely for so many years there to have the better effect there you need to be sensible Right now see here this is a problem of reit not only for you for everyone for everyone there is a student of mine by name taki he got 700 marks in physics 700 marks total and 175 in physics okay definitely he got disturbed uh the moment he heard about this rene but by god's grace he could set himself ready for determined focused work hereafter.

[00:54:15]Okay. So there are plenty of students with the huge marks forced to uh write the exam there. So this is a common uh situation for you and everyone there. So how best you cope up with this particular challenge that's the main task there. Okay then I am sure you work very well in the coming days and get the better and best rank there. So all the best my dear students there. Good luck.

[00:54:46]I wish very good success in everything you do academically hereafter. Bye-bye.

[00:54:51]All the best.