Sir, It is with great personal pride that I write this mail to you today. I have been a long time fan and subscriber of your RUclips Channel. It has helped me realise that my passion and soul mate is physics and Maths. I have been watching your lectures (8.01, 8.02 and 8.03) since I was in 10th standard. Now, at the end of 12th standard, I have cracked both JEE Mains and Advanced and now am headed for IIT Kharagpur within a month. I wish to thank you for your contribution in my IIT-JEE journey. Rightfully, the sanskrit Shloka declares: "गुरू ब्रह्मा गुरू विष्णु, गुरु देवो महेश्वरा गुरु साक्षात परब्रह्म, तस्मै श्री गुरुवे नमः" translated, it means the teacher is Brahma, the teacher is Vishnu the teacher is Shive who sustains knowledge and destroys ignorance. I forever bow to you and your eminence my teacher (guru). This mail may be too small a method to express my deepest and most sincere reverence to the greatest of physics teachers of our age. -Yours Sincerely, Arya Dev Mukherjee (A subscriber, student and now IITian.)
Holy shit. Bro congrats! Any tips for me so i can get into an IIT too? I am a 10th grader studying in an IGCSE board. I love Maths and Chemistry and despise Physics. Its just because there is no "how" and "why" I seriously cant focus on studies since I am either addicted to my mobile or i cant concentrate Hope I get a reply Thanks.
@Holy! I am a neet aspirant currently in class 11th! I just discovered this channel a few days ago! So does watching his videos are this much helpful??🤔 But I can't find like whole topic wise or chapter wise list in his channel it's just scattered! So can u help me??
I did remember a problem about the double pendulum, where I actually built one to verify the solution. As I came home this evening, I looked up my hand written notes back then and there you are: problem 112 it was! The solution I got is from a very old book by Tullio Levi Civita and Ugo Amaldi, Lezioni di meccanica razionale, Vol. 2, Part 2, page 72, where the general equation of motion are derived. In this special case, given w=sqrt(g/L) and theta1, theta2 the angles of the primary and secondary pendulum with the vertical and d2 the second derivative with respect to time: 2d2(theta1) + d2(theta2) + 2w^2 • theta1 = 0 d2(theta1) + d2(theta2) + w^2 • theta2 = 0 The natural frequencies are wp = w•sqrt(2+sqrt(2)) wm = w•sqrt(2-sqrt(2)) The ratio of the angles is sqrt(2) and -sqrt(2), hence (in the small angle approximation sin x ~ x) in phase: sqrt(2)+1 out of phase: sqrt(2)-1
My solution is: Omega- = sqrt(g/l)*sqrt(2-sqrt(2)) , Omega+ = sqrt(g/l)*sqrt(2+sqrt(2)) , A2/A1 = (2 - sqrt(2))/(sqrt(2)-1) , A2/A1 = (2+sqrt(2))/(-1-sqrt(2)) or approximately A2/A1=0.6/0.4 and A2/A1= 3.4/-2.4. Greetings from Belgium.
Hey Sir Walter Lewin, Have you ever formally met Richard Feynmann ? ? and can you give some tips on how to study physics in your approach ?? because there is alot to learn from you
you have 2 options option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Only convex lens form actual image,concave mirror can also form real image,it's obvious.The answer for this question is yes,because it's the nature of geometry.Virtual image is an mathematical(imaginary image) that has nothing to do with real life(physics and photons). Concave lens diverges the photon beams so there's no image can be formed on a focal plane.Only its virtual image in the opposite direction which is a mapping of the real diverged photons.
Sirr remeber me btw i understood emf - emf is the potential diifference of the cell across two terminals when the current is not withdrawn from the cell i also learnt terminal voltage it is also the pd of a cell across two terminals but when current is withdrawn from it i have learned much more but cannot write so thankssssss u sirrr❤
Worthy Walter lewin "sir" I am science enthusiast who loves physics, maths, biology and chemistry at the same time. Still space study is always awe strucking for me. I want to pursue Astrobiology in my further studies starting with an undergrad degree in BS-Ms course at reputed research institution in my country . I wanted to know if starting with your 8.01,. 02,. 03 is right along with course textbooks for college? For pursuing Astrobiology, what should I take up a major in BS MS degree - Biology or physics? I feel I am equally at ease with both of them. Could you please help me with it? BTW I just started with your classical mechanics and felt gross at other people's way of rote learning - cramming with superficial knowledge. Thank you and stay blessed Hope we ever meet in a physics conference in our lives!!
Hello Professor Lewin, I am a Middle School student and an admirer of yours. I have beeen struggling with research projects lately and wanted to seek your guidance regarding the same. I just want to ask you one question, what according to you makes a research "an in depth resaerch" ?
Hallo hallo hallo, Geachte meneer Lewin. Ik wil u ontzettend bedanken voor uw colleges. Ik studeer fiscaal-Recht in Leiden, maar ik vind natuurkunde veel leuker (logisch natuurlijk). Ik wil dit jaar indien mogelijk een minor-traject volgen aan de TU delft in kern fysica. Ik ben bijzonder geinteresseerd in het ''enhanced electron screening effect'', ik denk dat hierin veel vooruitgang te boeken is voor kernfusie onderzoek(misschien zelfs de doorbraak ligt). Uw college over ''charge density'' en hoe deze hoger is als de ''radius'' kleiner is, is echt bijzonder behulpzaam geweest (college met de kookpan) en ik wil u echt persoonlijk bedanken hiervoor. Mocht u ooit weer een keer naar Nederland afreizen dan krijgt u een biertje van mij ;) Of campina yoghurt als dat uw voorkeur heeft, mits het natuurlijk geen vrijdag is. Ik wens u een hele fijne dag verder.
I have been 4 times in India. My last 6 week visit was in 2014 when I gave 8 lectures there. Given my age and the pollution in India, my doctors do not want me to go back to India
Hi sir , i just solved the the unsolved problem, my solution for the one of the unsolved problem in history, navier stokes equation, i want check the correctness of the proof , can you help me sir ?
Why Walter Lewin the Lecturer is more popular than Walter Lewin the researcher? I have never seen any other physicist so popular for his teachings like you
Hi sir im from india.I have a small wish from my childhood that is I want to become a NASA Astronaut it is an unknown special interest for me. Any advice for that?
Adiabatic processes are often fast because they need to be thermally isolated, meaning the system can't exchange heat with its surroundings. If the process is fast enough, there isn't enough time for heat to change with the surroundings. For example, when compressing a gas in an engine cylinder, the process is often assumed to be adiabatic because it happens so quickly that little energy is transferred out as heat.
Sir ,love from india My question was that even after understanding the concept i am not able to do solve question in first attempt,it takes alot of time how to solve this problem in physics
I am trying a Lagrangian approach just for fun. Let the origin be in the pivot point of an x,y coordinate system. I will use q instead of θ as coordinates because I have not found how to write ”theta-dot” in unicode. For small angle approximation: sin q ≈ q and cos q ≈ 1- q²/2 Lagrangian: ℒ = T - V, where T is kinetic energy and V is potential energy. T = (1/2)mv² = (1/2)mẋ₁² + (1/2)m(ẋ₂² + ẏ₂²) ẏ₂² T = (1/2)mv² = (1/2)mẋ₁² + (1/2)mẋ₂² x₁ = l sin q₁ ≈ lq₁ ẋ₁ ≈ lq̇₁ x₂ ≈ lq₁ + lq₂ ẋ₂ ≈ lq̇₁ + lq̇₂ y₁ ≈ - l(1- q₁²/2) y₂ ≈ - l(1- q₁²/2) - l(1- q₂²/2) ẏ₁ ≈ lq₁q̇₁ ẏ₂ ≈ lq₁q̇₁ + lq₂q̇₂ This gives: T = (1/2)ml²[2q̇₁² + 2q̇₁q̇₂ + q̇₂²) V = mg(y₁ + y₂) V ≈ (1/2)mgl(2q₁² + q₂²) + C (The constant C is the zero point of potential energy It will dissapear when we take the dervatives) ℒ = (1/2)ml²[2q̇₁² + 2q̇₁q̇₂ + q̇₂²) - (1/2)mgl(2q₁² + q₂²) Lagrangian equations: (d/dt)(∂ℒ /∂q̇) - (∂ℒ/∂q) = 0 This gives the two equations: ml²(2q̈₁ + q̈₂) + 2mglq₁ = 0 => 2q̈₁ + q̈₂ + (2g/l)q₁ = 0 … (1a) ml²(q̈₁ + q̈₂) + mglq₂ = 0 => q̈₁ + q̈₂ + (g/l)q₂ = 0 … (2a) Let g/l = ω₀² => 2q̈₁ + q̈₂ + 2ω₀²q₁ = 0 … (1b) and q̈₁ + q̈₂ + ω₀²q₂ = 0 … (2b) Try the solution: q₁(t) = θ₁cos(ωt + 𝜑) q₂(t) = θ₂cos(ωt + 𝜑) Subtitute into (1b) and (2b) and divide the cos terms through. => Matrix M = [2ω₀² - 2ω² -ω² ] [ -ω² ω₀² - ω² ] M·[θ₁] = [0 0] … (3) [θ₂] [0 0] To find solutions, let the determinant of M = 0. => (2ω₀² - 2ω²)(ω₀² - ω²) - (-ω²)² = 0 => ω² = (2 ± √2)ω₀² … (4a) => ω- = √(2 - √2) ω₀ … (4b) and ω+ = √(2 + √2) ω₀ … (4c) Use one of the two equations in (3) to find the relations between θ₁ and θ₂. The second equation gives: -ω²θ₁ + (ω₀² - ω²)θ₂ = 0 => θ₂/θ₁ = ω²/(ω₀² - ω²) => θ₂/θ₁ = (ω²/ω₀²)/(1 - ω²/ω₀²) A₂/A₁ = (l sin θ₂ + l sin θ₁)/(l sin θ₁) A₂/A₁ ≈ (θ₂ + θ₁)/θ₁ = θ₂/θ₁ + 1 In phase: ω-²/ω₀² = 2 - √2 => θ₂/θ₁ = (2 - √2)/(1 - 2 + √2) = √2 => A₂/A₁ ≈ 1 + √2 Out of phase: = ω+²/ω₀² = 2 + √2 => θ₂/θ₁ = (2 + √2)/(1 - 2 - √2) = -√2 => A₂/A₁ ≈ 1 - √2
Sir,
It is with great personal pride that I write this mail to you today. I have been a long time fan and subscriber of your RUclips Channel. It has helped me realise that my passion and soul mate is physics and Maths.
I have been watching your lectures (8.01, 8.02 and 8.03) since I was in 10th standard. Now, at the end of 12th standard, I have cracked both JEE Mains and Advanced and now am headed for IIT Kharagpur within a month. I wish to thank you for your contribution in my IIT-JEE journey. Rightfully, the sanskrit Shloka declares: "गुरू ब्रह्मा गुरू विष्णु, गुरु देवो महेश्वरा गुरु साक्षात परब्रह्म, तस्मै श्री गुरुवे नमः" translated, it means the teacher is Brahma, the teacher is Vishnu the teacher is Shive who sustains knowledge and destroys ignorance. I forever bow to you and your eminence my teacher (guru).
This mail may be too small a method to express my deepest and most sincere reverence to the greatest of physics teachers of our age.
-Yours Sincerely,
Arya Dev Mukherjee
(A subscriber, student and now IITian.)
I originally wanted to write this an an email to you, but I could not find it.
Holy shit. Bro congrats!
Any tips for me so i can get into an IIT too?
I am a 10th grader studying in an IGCSE board. I love Maths and Chemistry and despise Physics. Its just because there is no "how" and "why"
I seriously cant focus on studies since I am either addicted to my mobile or i cant concentrate
Hope I get a reply
Thanks.
@Holy! I am a neet aspirant currently in class 11th! I just discovered this channel a few days ago! So does watching his videos are this much helpful??🤔 But I can't find like whole topic wise or chapter wise list in his channel it's just scattered! So can u help me??
1. System Description and Assumptions
Two identical pendulums, each of length L and mass m, are coupled.
Small angle approximation: sin(θ) ≈ θ
2. Derive the Equations of Motion
Top Pendulum:
mLθ̈₁ = -mgθ₁ + k(θ₂ - θ₁)
Bottom Pendulum:
mLθ̈₂ = -mgθ₂ + k(θ₁ - θ₂)
3. Simplify the Equations
Divide through by mL:
θ̈₁ = -g/Lθ₁ + (k/mL)(θ₂ - θ₁)
θ̈₂ = -g/Lθ₂ + (k/mL)(θ₁ - θ₂)
Let ω₀² = g/L and α = k/mL:
θ̈₁ = -ω₀²θ₁ + α(θ₂ - θ₁)
θ̈₂ = -ω₀²θ₂ + α(θ₁ - θ₂)
4. Matrix Form and Normal Modes
Rewrite in matrix form:
[ θ̈₁ ] = [-(ω₀² + α) α ] [ θ₁ ]
[ θ̈₂ ] [ α -(ω₀² + α) ] [ θ₂ ]
Assume solutions of the form θ₁ = A₁ e^(iωt) and θ₂ = A₂ e^(iωt):
[ -ω² 0 ] [ A₁ ] = [-(ω₀² + α) α ] [ A₁ ]
[ 0 -ω² ] [ A₂ ] [ α -(ω₀² + α) ] [ A₂ ]
This leads to the characteristic equation for the eigenvalues:
det [ -(ω₀² + α) - ω² α ] = 0
[ α -(ω₀² + α) - ω² ]
5. Solve the Characteristic Equation
[(ω₀² + α) + ω²]² - α² = 0
ω⁴ + 2ω₀²ω² + ω₀⁴ - α² = 0
ω² = ω₀²(2 ± √2)
6. Frequencies of Normal Modes
ω₊ = ω₀ √(2 + √2)
ω₋ = ω₀ √(2 - √2)
7. Amplitude Ratios
For ω₋:
(2ω₀² - ω₋²)A₁ + αA₂ = 0
Substituting ω₋² = ω₀²(2 - √2):
(2ω₀² - ω₀²(2 - √2))A₁ + αA₂ = 0
(ω₀²(2 + √2))A₁ + αA₂ = 0
αA₂ = -ω₀²(2 + √2)A₁
A₂/A₁ = -ω₀²(2 + √2) / α
A₂/A₁ = -√2
Adding 1 for total amplitude ratio:
A₂/A₁ = 1 - √2
For ω₊:
(2ω₀² - ω₊²)A₁ + αA₂ = 0
Substituting ω₊² = ω₀²(2 + √2):
(2ω₀² - ω₀²(2 + √2))A₁ + αA₂ = 0
(ω₀²(2 - √2))A₁ + αA₂ = 0
αA₂ = -ω₀²(2 - √2)A₁
A₂/A₁ = -ω₀²(2 - √2) / α
A₂/A₁ = √2
Adding 1 for total amplitude ratio:
A₂/A₁ = 1 + √2
∴
Frequencies:
ω₋ = ω₀ √(2 - √2)
ω₊ = ω₀ √(2 + √2)
Amplitude Ratios:
For ω₋, A₂/A₁ = 1 - √2
For ω₊, A₂/A₁ = 1 + √2
saw your JEE tip down below
and i am definitely gonna try that (especially the yogurt one >.
eat yogurt everyday but never on friday
omega minus= 0.76 omega zero
omega plus = 1.85 omega zero
C2/C1=1+sqrt2
C2/C1=1-sqrt2
It's all in Lect 6 (min 17) 8.03x, great lecture by the way.
Best teacher in universe❤
Love you sir,
I was not good at physics before,you mad me love physics
Thanks a lot ❤❤❤
Professor you are the man why I'm drowned in the love of physics ❤.
From milkyway galaxy.
keep it up
@@lecturesbywalterlewin.they9259 thanks professor for replying me it's give a positive sense to me .
🧡
Sir you are great
Much love from morocco
Good evening sir from india
No one asked you where you from .
I did remember a problem about the double pendulum, where I actually built one to verify the solution. As I came home this evening, I looked up my hand written notes back then and there you are: problem 112 it was!
The solution I got is from a very old book by Tullio Levi Civita and Ugo Amaldi, Lezioni di meccanica razionale, Vol. 2, Part 2, page 72, where the general equation of motion are derived.
In this special case, given w=sqrt(g/L) and theta1, theta2 the angles of the primary and secondary pendulum with the vertical and d2 the second derivative with respect to time:
2d2(theta1) + d2(theta2) + 2w^2 • theta1 = 0
d2(theta1) + d2(theta2) + w^2 • theta2 = 0
The natural frequencies are
wp = w•sqrt(2+sqrt(2))
wm = w•sqrt(2-sqrt(2))
The ratio of the angles is sqrt(2) and -sqrt(2), hence (in the small angle approximation sin x ~ x)
in phase: sqrt(2)+1
out of phase: sqrt(2)-1
My solution is: Omega- = sqrt(g/l)*sqrt(2-sqrt(2)) , Omega+ = sqrt(g/l)*sqrt(2+sqrt(2)) , A2/A1 = (2 - sqrt(2))/(sqrt(2)-1) , A2/A1 = (2+sqrt(2))/(-1-sqrt(2)) or approximately A2/A1=0.6/0.4 and A2/A1= 3.4/-2.4. Greetings from Belgium.
Way to hard. However, i remember seeing a coupled oscillator section in 801. Maybe I'll have a look.
Hey Sir Walter Lewin, Have you ever formally met Richard Feynmann ? ? and can you give some tips on how to study physics in your approach ?? because there is alot to learn from you
yes I have met Feynman several times. He was a student at MIT. I also met him at Caltech when I gave a colloquium there.
@@lecturesbywalterlewin.they9259 wow! thats amazing. really
@@lecturesbywalterlewin.they9259 sorry but, may I ask what was he like to you when you met him ??
@@lecturesbywalterlewin.they9259 sorry, but may I ask what was he like to you when you met him ?
I too have been baffled by this question. Thanks for your insight.
My pleasure!
ω± = 2 ± √2, in units of ω0=√(g/l)
(A2/A1)± = 1 ∓ √2.
Thank you sir from India 🕉️🕉️
Sir please give some tips to crack jee ans neet
you have 2 options
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@@lecturesbywalterlewin.they9259great reply in the history ❤⚡️
Thank you sir for helping...🙏@@lecturesbywalterlewin.they9259
Respected sir,
Can a biconcave lens form a real image in case of a virtual image?
Regards
ask Quora
Sir I had asked but I was getting conflicting answers.Some were saying it is possible while others were saying it was not.
Thank you Sir,
Regards
Only convex lens form actual image,concave mirror can also form real image,it's obvious.The answer for this question is yes,because it's the nature of geometry.Virtual image is an mathematical(imaginary image) that has nothing to do with real life(physics and photons).
Concave lens diverges the photon beams so there's no image can be formed on a focal plane.Only its virtual image in the opposite direction which is a mapping of the real diverged photons.
But why never on Fridays??
bcoz
@@lecturesbywalterlewin.they9259😢
Sirr remeber me btw i understood emf - emf is the potential diifference of the cell across two terminals when the current is not withdrawn from the cell i also learnt terminal voltage it is also the pd of a cell across two terminals but when current is withdrawn from it i have learned much more but cannot write so thankssssss u sirrr❤
Thank you sir 😊
I wish one day we meet sir
I thought you're in space
no I died 3 yr ago
@@lecturesbywalterlewin.they9259no we can't face that...
Sir who is talking in this video? Is that AI?@@lecturesbywalterlewin.they9259
😅😅😅@@lecturesbywalterlewin.they9259
Sir what is your daily routine?
sleeping, eating and drinking
Worthy Walter lewin "sir"
I am science enthusiast who loves physics, maths, biology and chemistry at the same time. Still space study is always awe strucking for me. I want to pursue Astrobiology in my further studies starting with an undergrad degree in BS-Ms course at reputed research institution in my country . I wanted to know if starting with your 8.01,. 02,. 03 is right along with course textbooks for college? For pursuing Astrobiology, what should I take up a major in BS MS degree - Biology or physics? I feel I am equally at ease with both of them. Could you please help me with it?
BTW I just started with your classical mechanics and felt gross at other people's way of rote learning - cramming with superficial knowledge.
Thank you and stay blessed
Hope we ever meet in a physics conference in our lives!!
yes that would be nice
Sir i want to bcome Mathematicians and physist like Eintein, tesla,and like u what i do
get educated
Love sir.. ❤
Hello Professor Lewin, I am a Middle School student and an admirer of yours. I have beeen struggling with research projects lately and wanted to seek your guidance regarding the same. I just want to ask you one question, what according to you makes a research "an in depth resaerch" ?
Love from India ❤
i came back to check if you are still alive
Sir I'm reading in class 9th and I want to know about Compton effect
use google
Hallo hallo hallo, Geachte meneer Lewin. Ik wil u ontzettend bedanken voor uw colleges. Ik studeer fiscaal-Recht in Leiden, maar ik vind natuurkunde veel leuker (logisch natuurlijk). Ik wil dit jaar indien mogelijk een minor-traject volgen aan de TU delft in kern fysica. Ik ben bijzonder geinteresseerd in het ''enhanced electron screening effect'', ik denk dat hierin veel vooruitgang te boeken is voor kernfusie onderzoek(misschien zelfs de doorbraak ligt). Uw college over ''charge density'' en hoe deze hoger is als de ''radius'' kleiner is, is echt bijzonder behulpzaam geweest (college met de kookpan) en ik wil u echt persoonlijk bedanken hiervoor. Mocht u ooit weer een keer naar Nederland afreizen dan krijgt u een biertje van mij ;)
Of campina yoghurt als dat uw voorkeur heeft, mits het natuurlijk geen vrijdag is.
Ik wens u een hele fijne dag verder.
ik kom jaarlijks in Nederland maar ik drink geen alcohol
@@lecturesbywalterlewin.they9259 Een goede bak koffie?
Hello sir
Love from India
Sir Walter i am very good at understanding physics and i love physics but i just can't do well in my Physics exams do you have any tips for me
Sir, did you ever meet Albert Einstein sir?
no
What is a principle applied ? if we whirl a thick string with some angular velocity holding at one end and from side view ,it form a top shape.
Sir when you will come india ? I want to meet you sir
I have been 4 times in India. My last 6 week visit was in 2014 when I gave 8 lectures there. Given my age and the pollution in India, my doctors do not want me to go back to India
Hello sir...
Can you please launch a new series of physics for JEE ADVANCED???
☺️
Are really still alive?
I am from srilanka, i try too many times to understand physics but i can't understand it is so sad but i couldn't understand yet
Sir I'm so confused what profession I should choose after class 10th please give me some advice sir😢❤
I cannot give you any advice as I do not know you. Ask your teachers
SIr im the best fan of you but im only 10 years old. Is there any esson i can learn fro u sir . Thanks you sir !
age 10 - you are still too young to be able to understand my lectues and to do my physics problems
Sir why to eat yogurt everyday but never on fridays ? Give full answer sir pls
Amazing sir
Have u met American astrophysicist Neil Degrasse Tyson.?
Hi sir , i just solved the the unsolved problem, my solution for the one of the unsolved problem in history, navier stokes equation, i want check the correctness of the proof , can you help me sir ?
you should first publish it I will then read it. I never get involved in new ideas of my viewers, that is a matter of scietific honesty.
Why Walter Lewin the Lecturer is more popular than Walter Lewin the researcher?
I have never seen any other physicist so popular for his teachings like you
Sir can a image reflect light rays if not then how do we see image of image
Hi sir im from india.I have a small wish from my childhood that is I want to become a NASA Astronaut it is an unknown special interest for me. Any advice for that?
eat yoghurt every day *but never on Fridays*
Sir I have a question that why only in Friday?@@lecturesbywalterlewin.they9259
@@lecturesbywalterlewin.they9259why only not on Fridays
But why not only on Fridays?@@lecturesbywalterlewin.they9259
Eat yogurt everyday but NEVER ON FRIDAY
But sir why not on Friday...
Please answer this sir
Respect Walter lewin sir please tell me why Adiabatic process is a fast process than any other process
Adiabatic processes are often fast because they need to be thermally isolated, meaning the system can't exchange heat with its surroundings. If the process is fast enough, there isn't enough time for heat to change with the surroundings. For example, when compressing a gas in an engine cylinder, the process is often assumed to be adiabatic because it happens so quickly that little energy is transferred out as heat.
Good evening sir i am from india
a) ω-^2 = (2-sqrt(2)) g/L, b) ω+^2 = (2+sqrt(2)) g/L
c) for ω- : A2/A1= sqrt(2)-1, d) for ω+ : A2/A1= sqrt(2)+1
Sir ,love from india
My question was that even after understanding the concept i am not able to do solve question in first attempt,it takes alot of time how to solve this problem in physics
eay yoghurt very day but *never on Fridays*
@@lecturesbywalterlewin.they9259 does it mean to work hard and not to make excuse???
From which book you picked up this problem?
Winny the Pooh
@@lecturesbywalterlewin.they9259
Please 🥺
@lectur😂😂😂😂😂esbywalterlewin.they9259
're u'liv
First Viewer
I am trying a Lagrangian approach just for fun.
Let the origin be in the pivot point of an x,y coordinate system.
I will use q instead of θ as coordinates because I have not
found how to write ”theta-dot” in unicode.
For small angle approximation:
sin q ≈ q and cos q ≈ 1- q²/2
Lagrangian:
ℒ = T - V, where T is kinetic energy and V is potential energy.
T = (1/2)mv² = (1/2)mẋ₁² + (1/2)m(ẋ₂² + ẏ₂²)
ẏ₂² T = (1/2)mv² = (1/2)mẋ₁² + (1/2)mẋ₂²
x₁ = l sin q₁ ≈ lq₁
ẋ₁ ≈ lq̇₁
x₂ ≈ lq₁ + lq₂
ẋ₂ ≈ lq̇₁ + lq̇₂
y₁ ≈ - l(1- q₁²/2)
y₂ ≈ - l(1- q₁²/2) - l(1- q₂²/2)
ẏ₁ ≈ lq₁q̇₁
ẏ₂ ≈ lq₁q̇₁ + lq₂q̇₂
This gives:
T = (1/2)ml²[2q̇₁² + 2q̇₁q̇₂ + q̇₂²)
V = mg(y₁ + y₂)
V ≈ (1/2)mgl(2q₁² + q₂²) + C
(The constant C is the zero point of potential energy
It will dissapear when we take the dervatives)
ℒ = (1/2)ml²[2q̇₁² + 2q̇₁q̇₂ + q̇₂²)
- (1/2)mgl(2q₁² + q₂²)
Lagrangian equations:
(d/dt)(∂ℒ /∂q̇) - (∂ℒ/∂q) = 0
This gives the two equations:
ml²(2q̈₁ + q̈₂) + 2mglq₁ = 0
=> 2q̈₁ + q̈₂ + (2g/l)q₁ = 0 … (1a)
ml²(q̈₁ + q̈₂) + mglq₂ = 0
=> q̈₁ + q̈₂ + (g/l)q₂ = 0 … (2a)
Let g/l = ω₀²
=> 2q̈₁ + q̈₂ + 2ω₀²q₁ = 0 … (1b)
and q̈₁ + q̈₂ + ω₀²q₂ = 0 … (2b)
Try the solution:
q₁(t) = θ₁cos(ωt + 𝜑)
q₂(t) = θ₂cos(ωt + 𝜑)
Subtitute into (1b) and (2b) and divide the
cos terms through.
=> Matrix M = [2ω₀² - 2ω² -ω² ]
[ -ω² ω₀² - ω² ]
M·[θ₁] = [0 0] … (3)
[θ₂] [0 0]
To find solutions, let the determinant of M = 0.
=> (2ω₀² - 2ω²)(ω₀² - ω²) - (-ω²)² = 0
=> ω² = (2 ± √2)ω₀² … (4a)
=> ω- = √(2 - √2) ω₀ … (4b)
and ω+ = √(2 + √2) ω₀ … (4c)
Use one of the two equations in (3)
to find the relations between θ₁ and θ₂.
The second equation gives:
-ω²θ₁ + (ω₀² - ω²)θ₂ = 0
=> θ₂/θ₁ = ω²/(ω₀² - ω²)
=> θ₂/θ₁ = (ω²/ω₀²)/(1 - ω²/ω₀²)
A₂/A₁ = (l sin θ₂ + l sin θ₁)/(l sin θ₁)
A₂/A₁ ≈ (θ₂ + θ₁)/θ₁ = θ₂/θ₁ + 1
In phase: ω-²/ω₀² = 2 - √2
=> θ₂/θ₁ = (2 - √2)/(1 - 2 + √2) = √2
=> A₂/A₁ ≈ 1 + √2
Out of phase: = ω+²/ω₀² = 2 + √2
=> θ₂/θ₁ = (2 + √2)/(1 - 2 - √2) = -√2
=> A₂/A₁ ≈ 1 - √2
you deserve an A++
@@lecturesbywalterlewin.they9259 Agreed
"Out of phase: ....... A₂/A₁ ≈ 1 - √2"
@@lecturesbywalterlewin.they9259 Wow 😀 Thanks!
@@KeithandBridget I made it difficult for myself this time 😄
Amazing sir
Amazing sir