Physics 68 Lagrangian Mechanics (6 of 25) Simple Harmonic Motion: Method 1
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- Опубліковано 3 кві 2016
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In this video I will derive the position with-respect-to time and frequency equation of a simple pendulum problem using the partial derivative of Lagrangian equation.
Next video in this series can be seen at:
• Physics 68 Lagrangian ...
I learnt more in the last 9 minutes than 3 hours of studying my lecture slides. The fact that you explain each step and the reason for it creates a fantastically intuitive method. Thank you!
I agree 👍
Absolutly 100% true
that why I only take topic names from slides and then delete it. then I search those topic on reference books and youtube channels like this
Thank goodness you made these lectures.
I don't know why every physics course just skims over Lagrangians like they are so obvious.
I mean, I'm not saying they are hard, but you need them to progress through to so many other topics in physics that it is insane how little literature there is on the topic.
Couldn't agree more!
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Hours of trying to figure out the simple basic reason why the KE for a pendulum is 1/2L^2theta^2 and no textbook or reference gives me the derivation for it. You sir just explained to me in two minutes. Honestly the best professor.
U r right
Very straightforward. Perfect for brushing up on my mechanics knowledge. Thanks!
A big respect to you sir, Thanks for those illustrated lessons and helpful videos.
You are a gifted teacher! I never tire writing the same over and over.
Totally enjoying these lessons. This is beautiful!
this Sir is helping me prepare for my entrance exam free of cost :) I don't have words to pa him respect
I studied this last year in mechanics. Today, before starting quantum , I thought I would revise Lagrangian and Euler Lagrangian.... Everything about Lagrangian came back to me in 9 minutes. Nice and precise video.. thanks
Watch the following video for the graph plotting of the same problem using python language:-
ua-cam.com/video/JEC6X-OlMmM/v-deo.html
You are my favorite Physics teacher since high-school. Big thanks!!!!
Happy to hear that! Glad you found our videos! 🙂
Like the desert blooming after a spring shower. This is really good stuff.
I do appreciate your lecture, whereby the many times i had tried to read and failed to understand. Now its clear about Lagrangian and Hamiltonian equations of motion of a pendulum. Thanks
Glad you found our videos. 🙂
you r the best teacher keep up the good work i love you my brother
Thank you Sir! I learned more in 9 minutes of watching your video than a whole day of studying my professor's powerpoint slides. I'm now able to confidently answer my test in Classical Mechanics.
Brilliant Professor indeed❗️❗️🎗
This helped me so much, thank you!
Im a mechanical engineer student, as such i have taken dynamics which from i understand is analogous to Classical Mechanics physics majors take with probable less complexity than the physics major. That being said, i now understand so much more because in my class all these formulas were handed out as tools and i literally didnt understand how each "tool" worked or what it did. This helps sooooooo much understanding the mechanics.
I tend to agree with you that in many engineering courses, they teach your the "tricks of the trade", without fully explaining where they came from, why the work, and how to develop them. When I took those classes, I had the same problem, because I learned better if I was shown the context of what we were doing.
i learnt very simply by you sir,.......thank you.. four this
Such a nice teacher!!
Awesome. I became curious with the derivation of the equation of motion for a simple pendulum using Lagrangian Mechanics. This didn't disappoint.
Glad it was helpful!
Same but with the double pendulum
These videos got me through the bulk of my college degree
excellent, can you explain the inverted pendulum which is connected to a load gear.
Sir you are great your videos are very helpful .thanks for these videos sir☺️☺️
SOOOOOOOO great the video!!!!Thank you sir~!
OMG! Thank you sm for this explanantion. It really helps me understand
thank you...amazing and simple
what will happen to potential energy if bob is suspended by a rubber band?plz tell
great video nice explanation, i love it
Thnak u sir u r method is very easy to understand
3:47 if you multiply mg (l-l*cos(theta)), u get mgl-mgl*cos(theta), and mgl is constant, and constant doesnt matter in potential energy so u can forget this term, and u are left with -mgl*cos(theta) . It simplifies things, but doesnt matter i guess, bcz u're gonna do derivative later, so it will drop out anyways
Method of teaching is superbbb😉
Wonderful, you are more deserving of my tuition fee to you rather than some of my professors. If I ever get rich, you will get a fat cheque in the mail as well earned payment for all I've learnt from you!
Great video but I am disapointed that the part of solving diffrencial equation was totally skiped
wow this was quit complicated but I understood it now I can solve the double pendulum problem
Thank goodness that I came across, Langrange has been my nightmare. Now I understand thank you Sir
Glad it was helpful!
Thank u it really helps me plz upload more videos
How theta double dot + omega square theta = 0 became theta in function t sir?
Perfect explanation
Can you please explain how the function will be the same when if we differentiate it with theta as you have said at 4:00
If the PE and KE are expressed in term of the angle, we can use the Lagrangian using the angle as the variable.
Please can you tell me how you solved the last equation to get theta = Axsin(wt)? I tried and I tried but couldn't have understanding it.
That is a standard differential equation of an oscillator and thus the solution is the standard solution. The best way to check is to use the result to get back the differential equation. Take a look at this video: Differential Equation - 2nd Order Linear (9 of 17) Homogeneous with Constant Coeff: Free Oscillator ua-cam.com/video/V9bl02Ffo_o/v-deo.html
Learn Laplace Transform, they are really easy and useful for ODE like this. The solution in the video is not the general solution, the general solution should be written as
theta(t) = theta(0)*cos(wt) + (thetadot(0)/w)*sin(wt)
(remember w here is not angular velocity, but sqrt(g/l) )
Another neat trick is to assign the roof where the pendulum is attached be the ground, or place where the potential is zero. The potential energy of the system is -mglcos θ, and the kinetic energy stays the same.
Interesting approach! 🙂
Really a useful short lecturer.
Glad you think so!
I love this video
Fantastic video
Your videos are great I love them
Glad you like them!
1:00 why dont we include additional kinetic energy? +mv^2/2?
this guy is amazing!!!!
Not really. Just a simple man enjoying learning how everything works, and then taking pleasure in sharing what I have learned.
Nice and easy to get it.
Easy to understand. Thank you.
Glad it was helpful!
I always find it impressive when people can just look at a differential equation and just know the general solution. That's where I trip up. I get a good bit through some math and then run into a differential equation and I am just not sure what the general solution looks like on top of my head. Thanks for the video!
That was the case for me as well as a student. "But how does he/she know the general solution". It turns out like almost everything else, after a while you learn to memorize the general solution to a number of equations.
Ahh okay that's good to know. I just tried to do this problem on my own several hours after watching it. I managed to get to the end but still confused on the solution to the differential equation. You said Sine or Cosine, and you went ahead with Sine? So I could have used A*Cos(wt) as my solution?
Thank you so much for your continued help over the past few years of my physics education. These higher level classes are humbling to the say the least.
You SIR are a legend! I just had one question, @ 7:15 I understand that sin(theta) is theta but I don't understand why we needed it to formulate the equation. Wouldn't we be happy with a final equation containg sin(theta)? Thanks a lot.
Using Lagrangian mechanics, or with differential equations it is easier to work the problem when we use theta as a variable instead of sin(theta).
Thank you sir
I didnt quite understand how we went from thetha(doubledot)+omega^2*thetha to thetha(t)=A*sin(omega*t)? is it just a thing you remember or is there a way to derive it?
Take the derivative and then the second derivative of A sin(wt) and you'll see that it is the solution of that differential equation. It turns out that is a very common differential equation and the common solution for simple harmonic motion.
this is so helpful
Just one word
Wow.........
Thank u so much sir thanks alot
Is it natural frequency omega_n = sqrt(g/l) and angular velocity omega = v/l ? They are both called omega in this so it is a little confusing.
Yes, some symbols are used many times over in physics. But in this case, there is a close connection between angular velocity and oscillation frequency.
A very good lecture, Thank you Sir!
Glad it was helpful!
@@MichelvanBiezen Really helpful. I understand how the equation to be derived. I have played the video, repeated , too and fro. Video will be for referance.
Effect of relativity on spring masss system lagrangian
Thanks for the excellent explanation. I have a question that confuses me, what if the pendulum moves horizontally. not vertical as exemplified in the video. What are the potential and kinetic equations? I can't find it on youtube, I hope you help me
There needs to be a storage of potential energy for a pendulum to work. If it is not gravitational potential energy, what would be the potential energy? One possibility is that is an object attached to a spring and the compressed spring contains the potential energy. We have examples of such a system in this playlist.
The best professor ❤❤ ,🇩🇿🇩🇿
Thank you. Glad you find our videos helpful. 🙂
sir may u please tell me where we take positive potential energy and where negative potential energy in finding Lagrangian .i am quite confused
Unless we define the down direction as positive, PE is positive in an upward direction.
thank you
Question: How do we calculate the time till that SIMPLE PENDULUM practically STOPS in air at STP? 'standard Temperature and Pressure' = normal day at a comfortable temeperature'
We have videos on the damping equation for simple harmonic motion. The decay equation will be in the form: e^-ct which then shows how long it takes before the pendulum comes to a stop.
@@MichelvanBiezen How do we calculate the time to full stop in an UNCRITICALLY DAMPED system?
Thank you sir.
Sir from where should I read the theory of lagrangian mechanics For even better understanding?
Check out the book "Introduction To Classical Mechanics" by David Morin.
That book has a chapter devoted to the Lagrangian method.
Thank you so much, sir !
Most welcome!
Interesting! Lagrange gives the same results as using vector notations and Newton's laws...
When computing the partial of L with respect to theta. Shouldn't you also get a theta dot term?
Since theta dot is essentially the angular velocity, the answer is no.
Oh. Yes, I see my error now (I was thinking about the partial with respect to time instead of theta). Thank you for the reply and the great video!
Thank you 🌷
Beautifully explained, sir . Thank you so much . Just a friendly suggestion , instead of Theta dot, writing theta dash avoids confusion .. in terms of reading
Cool lesson keep up
thanks you sir
Great video 👍🏻
Thank you
Plz sir solved forced harmonic oscillation lagrangian body
Sir I know that it may be not a question to this chapter (because I should now it) but how is it that at 5:30 while taking the partial derivative m and g haven't been differentiated but the cos has been ??? Thank You for Your great work :)
m and g are constants. but theta is a variable, therefore cos(theta) must be differentiated.
That makes sense ;) thank You very much
Thanks for your video, Sir. But I'm so confused. On your previous video (the SMH), we got the equation is x double dot + w2x = 0 and the solution is x = A cos (wt). But in this one, we got the similiar equation, which is theta double dot + w2 theta = 0. Those to are exactly the same but the variable, the one is x and the one is theta. But why the solution is different? The x one is cos and the theta one is sin?
Both the sin and the cos are solutions. The one you pick depends on the intial conditions (when t = 0)
Sorry to (probably) misplace this comment/request, but... I have been a model railway enthusiast since ever and have been bothered by the following problem.
Assume you have a train consisting of one locomotive pulling some freight cars: the first cars being flat deck cars with no load on them, followed by some much heavier box (closed-chassis) cars or even another loc. Now when the train takes a turn, in model trains the flat cars would most probably cut the turn and derail, while in real life the train would probably pass the turn successfully. Obviously, this depends on the radius of the turn, as well as on the mass distribution of the vehicles. Could also be that the type of wheels ("bogies") play a role. I am aware that the mass of the model rolling stock is often not to scale, but would you be able to explain all the dependencies/vectors involved in this problem? What are the critical points here for the train to pass the turn successfully? Thank you.
I used to be a model railroad enthusiast myself, and I always wondered about that same problem. It is mainly caused by the tightness of the turn. If you make the turn less sharp that will be less of a problem. It is for that reason that railroad tracks do not have sharp turns or the same thing would happen. Draw a vector of one car pulling the next and see what the magnitude of the component is perpendicular to the tracks. (the cars being relatively light also does not help).
Beautiful
very very nice
can't we take the potential energy as -mglcos(theta)??
Height above ground = L - Lcos (theta)
@@marquez2390 yes that's correct but if we are taking the reference point as the fixed point about which the pendulum is oscillating then potential energy should come as -mglcos(theta),and i am asking that is it also a correct approach?
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Thank You!
Will force remain conserve or non-conserve?
Since it is working against gravity (a conservative force) the energy in the system will be conserved unless there is friction, then the force will not be a conservative force.
Sir when sin theta = theta,
Then can we use this pendulum as harmonic oscillator.
For angles smaller than 10 degrees, the two are almost equal (the angle must be in radians).
a great relief to south asia...
Thanks🙏🙇 sir
the general solution to the differential equation should include an arbitrary phase constant, Asin(ωt - φ)
No? why? the arbitray constant _is_ A
Sir, Why we are using Rotational kinetic energy in simple pendulum.. Qn..
You can use either rotational or linear. For small oscillations you should get the same answer
I thought ½Iω² only applied to a body that's rotating. Not a body on a circular motion. Could you please clarify?
Just like for linear motion KE = (1/2) mv^2 For circular motion KE = (1/2) Iw^2
@@MichelvanBiezen oh I thought it only applied to rotating body
It applies to any body that rotates about its axis or orbits a fixed point (or any other axis displaced from the center of mass)
Great .
I would rather take the potential energy in the first place, as the y component of the mgl which is mglcos(theta), which simplifies the thing.
That will work. 🙂
Thanks u so much Sir
You are welcome.
Thanks sir❤🙏
You are welcome.
Good job sir but you must discuss conditions and degree of freedom in details
Yes, that is a good point. We still need to cover a lot of examples with the Lagrangian.
Equation of motion of a compound pendulum..V=-mglcos© why v=negative
how to solve this differential equation ? theta ( double dot ) + (g / l )*sin theta = 0 for theta as a function of something.
It has to do with assuming the solution comes in the form of exp(r*t). You will eventually solve a quadratic equation and find two imaginary solutions for the constant r, and when you construct the meaning of exponential functions with imaginary exponents through Euler's formula, you will get a linear combination of A*sin(omega*t) and B*cos(omega*t), where A and B are the amplitudes of each component of the solution, and omega is the angular frequency, that will equal sqrt(g/L) in this example.
Through trigonometric identities, you can prove that this is equivalent to:
theta = theta_0 * cos(omega*t + phi)
This means that the solution comes in the form of a sinusoidal waveform with an amplitude (theta_0) and a phase shift term (phi) that depend on the initial conditions.
@@carultch so why does he not do this in video?
@@michaelempeigne3519 Because it isn't the focus of this particular class. In Differential equations, you learn to solve that equation.
In this particular class, I would assume that the students already took differential equations, or it is something that is considered a given, so that the students eventually learn to solve it when they do take differential equations. There might even be a proof for it in the textbook, that is simply not the main focus of the class.
For instance, when moment of inertia of a disk comes up in an introductory physics class, it is typically a given that it is 1/2*m*R^2. There is a proof in the textbook, or at least the background to do the proof in the textbook, if the students are curious. But knowing how to do the proof isn't necessarily the subject of the class.
Yes, it is assumed that the students is familiar with partial derivatives and differential equations. If not, we have videos on those topics as well.
but sir here length is constant then how can we consider it..
Not sure what you mean by "how can we "CONSIDER" it". The word "consider" can mean many things.
How can you just say sin(theta) = theta? What if you want the exact solution? My professor does not like approximations.
In small harmonic oscillations this approximation is perfectly acceptable and correct
It is not possible to solve this problem in closed form, and have an exact solution. You end up needing to use an infinite series expansion of sin(theta), and solve the differential equation accordingly, if you want to accurately capture the behavior of a large amplitude pendulum. Even then, your solution will not be an exact solution that you can easily evaluate. It will be in the form of an infinite series of sine waves that are harmonics of the fundamental, and there will be a pattern to the various amplitudes. The period will be an infinite series of polynomial terms. Hyperphysics has this solution available.
We are doing a first order solution to this problem by assuming the linearization of sin(theta) = theta. For amplitudes up to 22 degrees, you are 99% accurate with this solution.
@@carultch
Interesting. Thanks for the info.