Physics 68 Lagrangian Mechanics (5 of 25) Simple Harmonic Motion: Example
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- Опубліковано 29 сер 2024
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In this video I will derive the position with-respect-to time equation of a simple-harmonic-motion with spring problem using the partial derivative of Lagrangian equation.
Next video in this series can be seen at:
• Physics 68 Lagrangian ...
Note that the general solution to the (d^2/(dt)^2)x + w^2*x = 0 equation is missing a phase factor phi. It should be: x = A*cos(w*t+phi) . This is the reason why it indeed does not matter if it is sine or cosine, since both functions only differ by a phase factor of pi/2 .
the special solution is x(t)=x0*cos(wt)+(vo/w)*sin(wt), so if (v0=0) which is common then x(t)=x0*cos(wt) without using the phase.
@@blarnblarn8400 I don't think so.
@@collegemathematics6698 yes, I think it would be better if he gave the general solution. Easy and quick using Laplace Transform
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One Detail which caused a bit ofconfusion is that v(0)=0, which isn't mentioned in the task, but is implied in your solution (v(0)=Awsin(0)=0). If you have this information you can go ahead and use the euler Ansatz x(t)=Ae^(Yt)+Be^-(Yt) with Y=sqrt(-w^2) thus v(0) = Y(Ae^(Y*0)-Be^-(Y*0)) = Y(A-B) = 0 Thus A=B! With Y=i*w and C = 2A we can write:
x(t)=Ae^(Yt)+Ae^-(Yt) = C/2*(e^(iwt)+e^(-iwt)) = C*cos(wt)
Still great video!
sir - u r the best on internet on every topic
I appreciate your assistance, Now I understand Lagrangian .
You are welcome.
I think the special solution might be x(t)=x0*cos(wt)+(vo/w)*sin(wt).. great lectures thank you:)
🤩👌 your work is best
Thank you so much 😀
Please also make a video on Newtonian and Hamiltonian Mechanics. I loved your videos Sir.
We have some videos on Hamiltonian mechanics.
That was so beautiful!! Thank you for sharing it
You are so welcome!
You are so incredible honestly
We appreciate the positive feedback. Glad you found our videos. 🙂
For SHM, the differential equation is solved by any function, whose 2nd derivative is negative of original function - so sin and cis work. But f(x)= e^(ix) also satisfies the requirement. Why use sin and cos and not e^ix? Is it because this is introductory and exponential is reserved for a future exciting episode?
Very good input. At the moment we are preparing a new set of videos that show the derivation of the various equations for simple harmonic motion including damping. Stay tuned.......
@@MichelvanBiezen It has been over 50 years since I studied this. You are resurrecting/recovering old knowledge. Thank you
What I never understood is why don't we use the actual method of solving second order ODE's. Can't I just assume that e^rt is a solution and then use euler's identity to get a linear combination of both cos and sin? I felt like saying it was Acos(wt) was always cutting off info.
If you could clear this up that'd be nice. I'm taking my first physics class this fall so I want to get rid of any misconceptions.
Initially you want to learn the basic technique of solving the equations. After that you just want to learn how to apply them by using the standard solutions already worked out.
Technically you get two unique solutions as you would expect of a second order ODE, sine and cosine. However most of the time sine is taken into account by adding a phase shift to cosine so that all the boundary conditions are met.
Actually, the Acos(wt) solution *is* the e^rt solution. It's just that the "r" in e^rt is an imaginary number. Then you use euler's identity ... exp(iwt) = cos(wt) + i sin(wt) to get the cosines and sines from the exponential. The professor here is just taking the short cut in order to keep his lecture on topic.
@@Algebrodadio where did the isin(wt) go when we expressed exp(iwt). we know that exp(iwt) = cos(wt) + i sin(wt) but why did the professor just only wrote the cos part? thanks
The easiest way for these kind of systems is learning the Laplace Transform, spits out the general solution
x(t) = x(0)*cos(wt) + (v(0)/w)*sin(wt)
in less than 2 min
How do you handle situations where the energy in the Lagrangian is not conserved? For example, if you would add friction to the SHO in this example, how would you then approach the problem? How can you handle a damped harmonic oscillator using the Lagrangian framework?
good question.
Has anyone got an answer to this?
@@two697 Yes cause the Euler-Lagrangian equation is only for conservative forces, in the case of friction or other non-conservative forces one should add the Rayleigh dissipation function.
The way of teaching is so good,Thanks a lot
You are welcome
thank you sir great effort keep it up God bless you sir
Is there a specific reason sine or cosine doesn't matter? Intuitively that would seem to give different values for position for the same time
It depends on the initial position when t = 0.
@@MichelvanBiezenso that would mean that if you pulled the pendulum to its amplitude and let go, you'd use cosine since that would result in the amplitude×cos(w×0) yes? If thats the case, under what conditions would sine be used, I can only reason it be used at the EP but then thered be no forces moving it either direction, unless it assumes there's already some momentum to move it in a direction?
If you push the block the the left and then you let go and then you start the clock (at t = 0) when the object passed the equilibrium point, you would use the sin function.
@@MichelvanBiezen alright, thanks for clarification, I'm sure I'll need it throughout this series :D
awesome video professor
why was the second term ( partial differential of L with respect to x) taken as a positive quantity in the last video, but as a negative quantity in this video? when, in fact, it is negative in both cases. (-mg and -kx respectively)
L = KE - PE dL/dx = - kx if you make a comment on the "other" video, I can take a look there.
@@MichelvanBiezen hello! sir how are you.sir plz help me Lagrange motion of equation application sir google for searching just only sample pendulum example given ,sir plzz send the my email address sir teacher give the assignment,
let it go man
These videos are interesting but why not write the Lagrangian equations as dL/dx = d/dt(dL/dv) instead of their difference equals zero? Writing it that way in the first place goes more directly to the analogy of F = ma where F = dL/dx and ma = d/dt(dL/dv) (rather than, as in the video, having to constantly swap summands on either side of the equation to get it into the that aesthetic form.)
That appears to be the standard way of writing it. There is actually a reason why it is done that way and it is related with one of the methods of deriving the equation.
after 2:28 didn't get how u said that, as its solution?
It is a very common differential equation with the solution as given. The best way to check to see if that is correct. Go ahead and plug the answer back into the equation: - A w^2 cos(wt) + w^2 (A cos(wt) = 0 is indeed correct.
@@MichelvanBiezen tx sir i got it as i m not from maths background so i didn't knew generalized solution of DE means finding its primary function by integration
Excellent lecture 🙏🙏🙏🙏
Much appreciated
Hi sir. What if there is a friction force µmg ? How do we include it ?
Then we have a damping force and the frequency of the oscillations will change and the oscillations will eventually stop. We have videos on that in the simple harmonic motion playlists.
thanks
Helped a lot, thank you sir
Thanks sir.
Sir how do we handle the case when there is gravity? ( Vertical model's) In this way : T should be 1/2mxddot^2 and V= 1/2kx^2 + mgx. The problem is that in this way we will have mddotx + kx +mg that is not correct. Could you help me?
When simple harmonic motion is set up vertically we don't have to worry about the mgh term because you first let the object come to rest at the equilibrium point where the force of gravity is cancelled out by the force of the spring.
@@MichelvanBiezen yes but this is the equilibrium situation. Equation I want should model the total scenario (in order to include the transitional regime too) . And modelling without Lagrange I should have kx - mg = mddotx . With Lagrange I have the problem with sign of mg
I didn't get how do we get equation for x from the former equation
By solving the second order differential equation that we got from working on the Lagrange equation. Are you familiar with differential equations?
Sir, will it be different if the motion of the box is moving in the same direction and the spring is at an angle to the horizontal?
The motion should be along the same line as the axis of the spring.
@@MichelvanBiezen so that means you cannot use Lagrange when the motion of the box and the axes of extension of the spring are different?, btw thanks for the vids, I've used this channel since highschool.
@@MichelvanBiezen That depends on the attachment of the spring to the wall and how stiff the spring is. Gravity can bend the spring.
Thank you sir
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