As others have mentioned, the math for θ_ at 44:23 is incorrect. It should be Re[e^-iωt] = cos(-ωt) = cos(ωt), meaning his "general" form should actually just be e^Γt/2 * cos(ωt). The rest of the equation involving sin() does not follow from anything he did previously. The reason this happened was because the function he introduced as a potential solution, Z(t) = e^iαt, is *A* solution to the differential equation, but it is not THE general solution. In other words, it does not encompass all possible solutions. This is because the general solution to the simple harmonic motion equation (from the previous lecture) is A*cos(ωt) + B*sin(ωt) before initial condition constraints, while Ae^iωt = A*cos(ωt) + iA*sin(ωt) which only has one free term instead of two. Thus Ae^iωt does not encompass all possible solutions. (He later "corrects" it by claiming you get the sin() term from linear combinations of e^iαt. While it's true that this justifies the use of sin() in the general solution to the overall differential equation, it still does not justify that particular choice for θ_(t). He's just doubling-down on his mistake)
@Blue Raja I completely agree. I recommend people follow the explanation in the book used for this course, Georgi The Physics of Waves (the book is free online) section 2.1.2 but read the sections before that starting at top of section 2.1. It isn't long and the mathematics is clear and easier to follow.
If he said the latter, he’d still not be correct. He could only obtain x(t)=Acos(at)+Bcos(wt) as a linear combination of e^(iwt) and its complex conjugate e^(-iwt). Otherwise, he’ll always have some complex term in the solution. Once we have these two solutions, it’s not hard to show that for any c,d in R, there’s a complex A(c,d) and B(b,d) such that Ae^(-iwt)+Be^(iwt)=dcos(wt)+csin(wt).. So this bit of step skipping isn’t really helpful to what is more subtle than it may initially appear to be.
I was thinking the exact same thing as I watched his derivation and plugged the exact same question into Wolfram Alpha: Re{e(iwt)} = cos(wt) Re{e(-iwt)} = cos(wt) How? Because of the following identity: e^(+iwt) = cos(wt) + i*sin(wt) e^(-iwt) = cos(wt) - i*sin(wt) You essentially REMOVE the {i*sin(wt)} when performing Re{Z(t)} Edit: I think I see how they called that section: "Small Issues", so they can see the error, imo.
My university professor gave the same course in 7 classes also from Georgi's book. I couldn't understand almost anything, even after I read the book and took notes. I haven't had a real professor for around 2 years now. You're keeping me in love with physics. Thank you so much! You're amazing. I've managed to learn so much
Same Happened to me. My Professor at UPC (Catalan University) gave me this class in double the time and I did not understand a thing. Now, With this class I'm understanding everything right away
Thnaks a ton!!!!! you gave me another chance to study what i love but couldn't study before because of some financial issues. you made me, belive more that the science is free for all to know and study!
Thanks ! Professor LEE , it is really my luck to have the best teacher like you. Maybe human culture is passed along with good tutorials and insights you have been communicating.
It's how differential equations are solved, better look at cs 18.03 (ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/index.htm)
1:15:22 actually Energy (If I can call it like that) is conserved in a different manner, not in the form of T+U=E (T=Kinetic Emergy, U=Potential Emergy, E= Total Energy) but in the form: T+U - v•(δU/δv)= E, you have to add the velocity gradient of the potential Energy and multiply it by v). So you make E*=E+v(δU/δv) to rewrite the equation as T+U=E*, E* is not conserved because it depends on time. (Actually one defines an effective potential U* as U*=U-v•(δU/δv)) The potential U is defined to satisfy F=d/dt(δU/δv)-δU/δx. *You can find the proof in any classical mechanics book but if you want I prove it in a reply. In this case U(x,v)=(-bv+kx)x/2
Proof: The potential energy is defined in a wat such that it satisfies the relation (1)-- dU=-F•dr Being F•dr the scalar product of the force with the infinitesimal displacement r, which integrated gives the work. Now if V is some function of both the position x, and the velocity v (V=V(x,v)) then differentiating we have because of the chain rule for multivariable calculus: dV=δV/δx•dx +δV/δv•dv. Now we use the Leibniz Rule in: d(δV/δv•v)=d(δV/δv)•v+ δV/δv•dv. We rearrange this equation and get dV=δV/δx•dx +d(δV/δv•v)-d(δV/δv)•v Rearranging: (2)--. d(V-δV/δv•v)=(δV/δx-d/dt(δV/δv))•dx *Used the fact that dA=(dA/dt)dt and dx=vdt. Comparing (1) and (2) one sees that the potential U has to satisfy U=V-δV/δv•v (actually U is the effective potential and V may be seen as the original potential function). And the force F is of the form F=d/dt(δV/δv)-δV/δx. Now to estbalish that a quantity E is conserved (Energy), we part from Newton third Law F=ma, and multiply it by dx, so F•dx=ma•dx=ma•vdt=(mdv/dt•v)dt=m*d/dt(v^2/2)dt=d(mv^2/2). On the right side we have the Kinetic Energy & on the left side we have the potential energy (effective)!from its definition (1). Because we have -dU=dT, means that d(T+U)=0 which means there exists a constant E (the total energy) such that T+U=E. Finally, rewriting in terms of V: T+V-δV/δv•v=E.
I have something to add, the rotational inertia I, shouldn't it be (1/4)*m*l^2 ? the formula isI = m*r^2 , and r is supposed to be (1/2)*l, right? so shouldn't it be (1/4)*m*l^2 instead of (1/3)*m*l^2 ?
17:16 I can tell bros are going through a lot at the same time. Even I'm watching some parts several times to understand what actually just happened at that moment. 😅😥
Respected Sir, At 16:32 , The direction of torque must be inside the plane , so R vector should be pointed along the length of rod(opposite to what you did) please clarify my doubts
Go to the full site on OCW ocw.mit.edu/8-03SCF16 for the course materials. The problem sets (without solutions) are available, but the exams have solutions. Good luck with your studies!
@19:03, alpha(t) is just like "a" as acceleration in formula a=F/m. It is a angular acceleration so equal to theta double-dot! That is rotation version of Newton's 2nd Law...
Is it correct to say "energy is not conserved"? I mean after all Energy is ALWAYS conserved. What could be happening is that for a driven oscillator and a dampened oscillator is that energy from the system gets transferred to the surroundings of the system (in this case the air I guess) Can someone correct me if I'm wrong please?
You are correct, of course. It's just if you're considering only the mechanical system itself, and not including the rest of the universe its energy is not conserved; as you said it's transferred to the surroundings.
Honestly speaking, from what I have watched so far in the first two lectures, this teacher is not as good as most teachers on MIT Opencourseware for 2 reasons. A: His explanation is not clear enough. He only copied the formulas from his paper on his hand and read out without referencing the previous contents on the blackboard which is a fatal behaviour in teaching. Many concepts are not well-explained and defined. B: I have to say his english proficiency is not fluent enough. During this lecture, he repeated using 'actually', 'OK' over 200 times along with other phrases. This is a typical lexical deficiency. However it is still generous of MIT that they devote so much effort and time to publish these qualified studying materials and videos for free to the world and nothing is perfect. I appreciate that! THANK YOU ^_^
As others have mentioned, the math for θ_ at 44:23 is incorrect. It should be Re[e^-iωt] = cos(-ωt) = cos(ωt), meaning his "general" form should actually just be e^Γt/2 * cos(ωt). The rest of the equation involving sin() does not follow from anything he did previously.
The reason this happened was because the function he introduced as a potential solution, Z(t) = e^iαt, is *A* solution to the differential equation, but it is not THE general solution. In other words, it does not encompass all possible solutions. This is because the general solution to the simple harmonic motion equation (from the previous lecture) is A*cos(ωt) + B*sin(ωt) before initial condition constraints, while Ae^iωt = A*cos(ωt) + iA*sin(ωt) which only has one free term instead of two. Thus Ae^iωt does not encompass all possible solutions.
(He later "corrects" it by claiming you get the sin() term from linear combinations of e^iαt. While it's true that this justifies the use of sin() in the general solution to the overall differential equation, it still does not justify that particular choice for θ_(t). He's just doubling-down on his mistake)
Would this be fixed if his solution were Ae^iat?
@Blue Raja I completely agree. I recommend people follow the explanation in the book used for this course, Georgi The Physics of Waves (the book is free online) section 2.1.2 but read the sections before that starting at top of section 2.1. It isn't long and the mathematics is clear and easier to follow.
If he said the latter, he’d still not be correct. He could only obtain x(t)=Acos(at)+Bcos(wt) as a linear combination of e^(iwt) and its complex conjugate e^(-iwt). Otherwise, he’ll always have some complex term in the solution.
Once we have these two solutions, it’s not hard to show that for any c,d in R, there’s a complex A(c,d) and B(b,d) such that Ae^(-iwt)+Be^(iwt)=dcos(wt)+csin(wt).. So this bit of step skipping isn’t really helpful to what is more subtle than it may initially appear to be.
Thanks, this is what I was about to ask
I was thinking the exact same thing as I watched his derivation and plugged the exact same question into Wolfram Alpha:
Re{e(iwt)} = cos(wt)
Re{e(-iwt)} = cos(wt)
How?
Because of the following identity:
e^(+iwt) = cos(wt) + i*sin(wt)
e^(-iwt) = cos(wt) - i*sin(wt)
You essentially REMOVE the {i*sin(wt)} when performing Re{Z(t)}
Edit: I think I see how they called that section: "Small Issues", so they can see the error, imo.
My university professor gave the same course in 7 classes also from Georgi's book. I couldn't understand almost anything, even after I read the book and took notes.
I haven't had a real professor for around 2 years now. You're keeping me in love with physics. Thank you so much! You're amazing. I've managed to learn so much
Same Happened to me. My Professor at UPC (Catalan University) gave me this class in double the time and I did not understand a thing. Now, With this class I'm understanding everything right away
2:03 😂😂😂😂😂 It's a baseball bat professor... Some kinda rod
@Gerald Leonard no one does, try shoving this scam up your you-know-what u fool of a took
Thnaks a ton!!!!!
you gave me another chance to study what i love but couldn't study before because of some financial issues.
you made me, belive more that the science is free for all to know and study!
He is missing a l in the denominator. 19:43
52:00 Why the heck he takes the IMAGINARY PART for the negative theta, but the REAL PART for the positive?
Looks like he converted the sin to an euler identity and then pulled the negative out due to sin being an odd function.
this is a very nice warm-up for my physics 3 course next semester, thank you, Professor Lee.
Of which branch ??
@@rashmibajpai1607 your mom
Thanks ! Professor LEE , it is really my luck to have the best teacher like you. Maybe human culture is passed along with good tutorials and insights you have been communicating.
Thank .....plz don't stop
I might just be confused but shoudn't Re[exp(-iwt)]=cos(wt)? I know this wouldn't make much sense as a solution but mathematically isn't it correct?
It's how differential equations are solved, better look at cs 18.03 (ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/index.htm)
Vladimir, the link doesn't work. Could you please post a different link? Thank you!
@Paul Suroj cos(wt) is already real, so not correct to add Re(cos(wt)). Look at @Blue Raja explanation below, which is correct.
@@abigailfriedman2652 it is neither correct nor wrong. It’s just trivial. Re(4)=4 obviously.
Yes.
1:15:22 actually Energy (If I can call it like that) is conserved in a different manner, not in the form of T+U=E (T=Kinetic Emergy, U=Potential Emergy, E= Total Energy) but in the form: T+U - v•(δU/δv)= E, you have to add the velocity gradient of the potential Energy and multiply it by v). So you make E*=E+v(δU/δv) to rewrite the equation as T+U=E*, E* is not conserved because it depends on time. (Actually one defines an effective potential U* as U*=U-v•(δU/δv))
The potential U is defined to satisfy F=d/dt(δU/δv)-δU/δx. *You can find the proof in any classical mechanics book but if you want I prove it in a reply.
In this case U(x,v)=(-bv+kx)x/2
Proof: The potential energy is defined in a wat such that it satisfies the relation
(1)-- dU=-F•dr
Being F•dr the scalar product of the force with the infinitesimal displacement r, which integrated gives the work.
Now if V is some function of both the position x, and the velocity v (V=V(x,v)) then differentiating we have because of the chain rule for multivariable calculus:
dV=δV/δx•dx +δV/δv•dv.
Now we use the Leibniz Rule in: d(δV/δv•v)=d(δV/δv)•v+ δV/δv•dv. We rearrange this equation and get
dV=δV/δx•dx +d(δV/δv•v)-d(δV/δv)•v
Rearranging:
(2)--. d(V-δV/δv•v)=(δV/δx-d/dt(δV/δv))•dx *Used the fact that dA=(dA/dt)dt and dx=vdt.
Comparing (1) and (2) one sees that the potential U has to satisfy
U=V-δV/δv•v (actually U is the effective potential and V may be seen as the original potential function).
And the force F is of the form
F=d/dt(δV/δv)-δV/δx.
Now to estbalish that a quantity E is conserved (Energy), we part from Newton third Law F=ma, and multiply it by dx, so F•dx=ma•dx=ma•vdt=(mdv/dt•v)dt=m*d/dt(v^2/2)dt=d(mv^2/2). On the right side we have the Kinetic Energy & on the left side we have the potential energy (effective)!from its definition (1).
Because we have -dU=dT, means that d(T+U)=0 which means there exists a constant E (the total energy) such that
T+U=E.
Finally, rewriting in terms of V:
T+V-δV/δv•v=E.
I have something to add, the rotational inertia I, shouldn't it be (1/4)*m*l^2 ? the formula isI = m*r^2 , and r is supposed to be (1/2)*l, right? so shouldn't it be (1/4)*m*l^2 instead of (1/3)*m*l^2 ?
I think (l/2) should be multiplied by (b * theta dot) when he wrote the drag torque.
In the underdumped case, is theta minus the real part of zeta minus? Why?
It's so cool that physical laws are democratic in this class!
sir , please provide more lectures on all physics chapters thank you
I have seen that the name mit only enhances my focus of watching video 😅😅
Thank you so much sir
17:16 I can tell bros are going through a lot at the same time. Even I'm watching some parts several times to understand what actually just happened at that moment. 😅😥
I just stumbled apon you! Awesome dude!
Respected Sir,
At 16:32 , The direction of torque must be inside the plane , so R vector should be pointed along the length of rod(opposite to what you did)
please clarify my doubts
At 37:55, shouldn't is be Re(the left hand side)=0? So he is missing the Re?
Thank you so much!!!
Can somebody tell me how to get p-sets for this course
Go to the full site on OCW ocw.mit.edu/8-03SCF16 for the course materials. The problem sets (without solutions) are available, but the exams have solutions. Good luck with your studies!
Why is Lagrange method not brought up during these lectures? it sometimes makes it easier to solve then when using Newton's method
This is an undergraduate course
@@ameerhamza4816 I still learn it in my undergrad Classical Mechanics I class
if we use x= A exp (i (wt+¢)) then Total energy (TE) =0 can someone explain intuition behind this ??
wow really nice one and also entertaining and awesome one thank you sir
@19:03, alpha(t) is just like "a" as acceleration in formula a=F/m. It is a angular acceleration so equal to theta double-dot! That is rotation version of Newton's 2nd Law...
Where and how can i find Physics I and II, on MIT
Here are the all the physics course that we have on MIT OpenCourseWare: ocw.mit.edu/courses/physics/. Best wishes on your studies!
Here is a group for discussion on this course: facebook.com/groups/1088712164828659/
please join so that we can learn together!
Is it correct to say "energy is not conserved"? I mean after all Energy is ALWAYS conserved. What could be happening is that for a driven oscillator and a dampened oscillator is that energy from the system gets transferred to the surroundings of the system (in this case the air I guess) Can someone correct me if I'm wrong please?
You are correct, of course. It's just if you're considering only the mechanical system itself, and not including the rest of the universe its energy is not conserved; as you said it's transferred to the surroundings.
@@tenebreonlabs oh ok i get it now. It just felt weird to hear "energy is not conserved" thanks!
Nice Hat!
Right?! I've notived that too hahhahha what a random thing hahaha
40:26
Honestly speaking, from what I have watched so far in the first two lectures, this teacher is not as good as most teachers on MIT Opencourseware for 2 reasons. A: His explanation is not clear enough. He only copied the formulas from his paper on his hand and read out without referencing the previous contents on the blackboard which is a fatal behaviour in teaching. Many concepts are not well-explained and defined. B: I have to say his english proficiency is not fluent enough. During this lecture, he repeated using 'actually', 'OK' over 200 times along with other phrases. This is a typical lexical deficiency.
However it is still generous of MIT that they devote so much effort and time to publish these qualified studying materials and videos for free to the world and nothing is perfect. I appreciate that! THANK YOU ^_^
Fair criticism; in his defense I'd like to say I thought he was fantastic as an E/M recitation instructor :)
16:30
critical damping is not the best choice if we want the system to return to equilibrium in the shortest time
41:10 He's speaking chinese 这是 hhhhhh
سأعود الى هنا بعد ثلاثة اشهر عندما اصنع اسمي واكون الاول ..........
so
proffeser walter lewin has been given very GOOd lectures compare to this most worst lecture i had ever seen
learn some english