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
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.
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?
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))
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 RU-vid 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.... 😕😕
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.
