Love you from Kenya professor,you made me love physics
@Ceferov_825 ай бұрын
Yes me too
@duckyoutube63185 ай бұрын
You are not alone in that sentiment!
@World_Complex_The5 ай бұрын
I love your videos sir they are actually priceless❤
@gaming_with_bhadola5 ай бұрын
Love you sir , love from hills of uttarakhand ❤️❤️❤️
@garfielddexter62245 ай бұрын
Thank you for very interesting video🙂 very impressive 😃👍
@Ceferov_825 ай бұрын
Hello Mr. Genius. Why can't you eat yogurt on Friday? He said it happens every other day. Please answer the question when you see it. Thanks in advance
@DharaniDharan-rj6ps5 ай бұрын
Sir, i love your teaching and approach to physics❤❤... Currently I am preparing for NEET medical entrance exam 2025 so I have a complete year in my hand. but still, here in my language there is no proper physics lectures in KZbin and also I can't afford paid courses😢......can I watch your lectures for physics and solve your example problems and test for my preparation??.... I have been in this dilemma for watching your lectures for my physics preparation.... 😕😕
@lecturesbywalterlewin.they92595 ай бұрын
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, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@DharaniDharan-rj6ps5 ай бұрын
@@lecturesbywalterlewin.they9259 Thank you very much sir, for responding to my request💖💖..i will be grateful for your help✨😊
@AyanHaider_5mdcat5 ай бұрын
Thank u sir g I watched your videos interesting
@lecturesbywalterlewin.they92595 ай бұрын
Most welcome
@Nikhil-gh7qr5 ай бұрын
Sir please answer my query. The question goes as follows : A block of mass m is connected to another block of mass M by a massless spring of springs constant K initially the blocks are at rest and the spring is unstreatched when a constant force F starts acting on the block of mass M to pull, Find the max extension of the spring. If I try to solve it using centre of mass frame then the elongation comes out to be 2mF/(m+M) but when I solve it using inertial frame using equations, F-Kx = Ma ma = kx where x is max elongation then it is mF/(m+M) where the hell is 2 going?
@lecturesbywalterlewin.they92595 ай бұрын
I do not solve problems - I teach Physics. This is an easy high school problem
@carultch5 ай бұрын
The non-intuitive part about this question, is the blocks do NOT have the same acceleration when the spring is stretched at the maximum. The center of mass will accelerate at F/(M + m), but the blocks will oscillate indefinitely (for the ideal case). In the steady state solution in reality, they settle to a common acceleration. But at a point before this, block m will lag behind. For the steady state case, the spring extension settles to x_ss = F*m/(k*(M + m)). For the transient behavior, it oscillates between 0 and twice this value. Assign capital A as the acceleration of M, and little a for m's acceleration. In the lab frame: F - k*x = M*A k*x = m*a The setup in the center of mass frame, has to account for a force term of -M*a_cm and -m*a_cm in each equation on the left side, due to an accelerating system. The accelerations on the RHS would be replaced with (A - a_cm) and (a - a_cm) respectively. We need a kinematic constraint to relate A to a. To do this, relate A & a to the 2nd derivative of x. The acceleration of the spring length is the difference of the two accelerations: x" = A - a Multiply the 2nd equation by M/m, so we can generate (A - a), and replace it with x" when combining equations. This gives us the differential equation: M*x" + k*(1 + M/m)*x = F The solution with all initial conditions of zero, for x(t) is: x(t) = F*m/(k*(M + m) * [1 - cos(ω*t)] where ω = k*(M + m)/(M*m) Its maximum value occurs when cos(ω*t) = -1. Which is therefore: x_max = 2*F*m/(k*(M + m))
@pranav.mishra5 ай бұрын
Sir please make a video on electric dipole As its not covered in your playlist It will be very helpful
@lecturesbywalterlewin.they92595 ай бұрын
it is covered in my 8.02 lectures
@pranav.mishra5 ай бұрын
@@lecturesbywalterlewin.they9259 oh thank you sir I think I missed that part
@qinga85 ай бұрын
Great and beautiful, I will do it again
@lecturesbywalterlewin.they92595 ай бұрын
Have fun
@mrxzod205 ай бұрын
Walter sir can u suggest the best book for physics fir jee examination India
@lecturesbywalterlewin.they92595 ай бұрын
ask your teachers - they know you
@studytosuccess65015 ай бұрын
Sir , on full moon day, we can see full moon whole night but when it's crescent moon day we don't see it for longer period why?
@carultch5 ай бұрын
During a full moon, the moon is on the opposite side of the sun, and is visible to every part of the Earth that has nighttime. You can see a full moon at sunrise or sunset due to atmospheric refraction, but not during the middle of the day. During a crescent moon, the moon is on the same side of the Earth as the sun, at an acute angle from the line-of-sight to the sun. The vast majority of the land area that can see the moon at any given time, is experiencing daylight. During a waxing crescent, you can see it for a few hours after sunset. During a waning crescent, it rises a few hours before the sun.