Yeaaaaayyy... I got the logic correct ... A concept related to the second twist part was in one of your lecture... From a bigg physics loverrr and your best friend...
@@paarthjagga7287 Well it depends my friend. If you are in the American Education System and started 10th grade : you can start taking AP physics ( I took AP phy 1 from India) And professor lewin's lectures are enough. On RU-vid take Classical mechanics and electricity and Magnetism from his channel. Books : ( Any one) 1. University Physics with Modern Physics by Young, Freedman & Lewis Ford 2. Fundamentals of Physics by David Halliday, Robert Resnick and Jearl Walker (IF YOU REALLY WANNA LEARN PHYSICS) ... Else if you are in India and wanna prep for Jee, then it's really complicated ... First you wanna have to think if. You really wanna give jee or not Example - Me - I prep for jee adv in 10th and 11th , gave sat in 12th and now I will be going to US with full tution scholarship ... So you have to think and sort out what you really want ... In India your love for physics or knowledge in this subject won't be awarded.. only your marks will be ... So think twice before you start any thing ... Ask if you have any other questions ... Better also consult a teacher. Blantly jumping into a physics topic or waiting for a teacher to start it in your class won't be a great idea ...
Love the problem and solution! There is a much easier way to avoid quadratic equations and the work-energy theorem. Just find the equilibrium position "Xo" if "m" is hanging still on the spring, using balance of forces [kXo = mg]. Solve for Xo = 20cm. Now, H = 2(d+Xo) = 48cm due to symmetrical pendulum motion. WE ARE LOVERS OF PHYSICS! :D
Don't know why I love to see you again and again, though I m not studying anymore. *Just one pray to God, May your all wishes come true in the same life.* Love u Sir, ❤️❤️ from 🇮🇳🇮🇳🇮🇳
I balanced the total energy at both points where v=0. So kd²/2 + mgH = k(H-d)²/2. The resulting quadratic equation is simpler since it doesn't have an independent constant: H² - (2d + 2mg/k)H = 0. Solving for H gives you the answer directly: 48cm. Thanks for the problem, professor! I liked solving it very much. Now, I'll tackle this new twist :) !
It makes the task very clear if you consider deformation of the spring Xo=mg/k = 20cm under the gravity without oscillation (release the weight slowly). Obviously, after you deform the spring on d = 4cm then, the weight will oscillate around the position of its static equilibrium with the amplitude of H = 2* (Xo + d) = 48cm. This gives immediate answer to the problem #78. In regards to the "twist" problem, the "h" is nothing else as: h = H - d (the whole peak to peak amplitude "H" minus initial deformation "d" ) h = 2*Xo + d h = 40cm + d, so exactly as you described in the end of this video. We love physics :) Thanks Mr Lewin!
That is one way to solve it. I thought about using conservation of energy but I know it will overcomplicate things. So, I solved it analytically; I start by finding equilibrium position which you get from solving for mg=kx• and I got x•=20cm and from that I know that you're compressing the spring by 20cm by holding it. And when you further compressed it by a distance d , it is not very hard to work out where the spring will stop which is obviously 24cm from equilibrium in opposite direction of compression if the Hook Law hold. So, adding two 24cms we get the value of H = 48 cm.
Yes, it's clear now. I seem to get easily tangled with springs. Yes, 40 cm is the distance gravity pulls it down alone and however much you push up, within the limits of the spring, is how much more you get in lowest point too, because that is what the spring constant does in accelerating and decelerating the mass, and these are symmetrical.
Really , thank you sir for your special interest to us, you are so keen to make us understand and love physics,,,, we love you sooo much , hoping you the best health and happiness,,,, one of your fans from Egypt, Mohamed Hassan
Sir I got the same answer and commented in the original video 3 days ago... I took the lowest level (v=0) as reference potential energy and used energy conservation... mgH + 1/2kd² = 1/2k(H-d)² (compressed) (at v=0 state) we get... H = 2mg/k + 2d which on substituting...we get H=0.48m (easily without using quadratic formula)
You provide very engaging lectures on physics. I have followed many of your earlier lectures as well as more recent ones and can only congratulate you on your ability to put across complex ideas in a way that are interesting and promote thought. Keep up the good work. I look forward to seeing more. 👍
Your solution is correct, but a lot harder than it has to be! For a linear system, the oscillation is symmetrical about the EQUILIBRIUM point. The equilibrium point is 20 cm below the relaxed spring position (calculated by taking 0.2Kg mass * 10m/s to get 2nt, & deflection of spring is 2nt/(10nt/m) = .2m = 20cm). The initial position of the mass at release is 4cm + 20cm = 24cm above the equilibrium point. So on release, it will go below the equilibrium point by the same amount. So the total excursion is 2 x 24 = 48cm. Who needs quadratic equations! I needed only 2 divisions by 10, 1 addition, & 1 multiplication by 2 to solve the problem! The secret is to work about the equilibrium point of the mass & spring together, not the equilibrium point of the spring alone. Then symmetry solves the problem easily! You will also make more friends!
At first I was thinking about the problem for a while. Now I realised that it's actually not that hard. would love to be your friend!! your my biggest idol!! 😍😍❤️❤️ Please comment to confirm that you are also my friend ❤️❤️
This is a linear system, which will obey superposition. If you realize that this can be understood by superposition of a static solution and an oscillating solution, it becomes simple, and in fact you do not even need to solve the quadratic for the original problem. The static solution is the length of the spring with the weight hanging on it. It is 20 cm greater than the unloaded length of 60 cm (60 + 20 = 80 cm.) The oscillation will be symmetrical about the static point, so the end points will be +/- (d + 20) about the static point. Since h is measured from the 60 cm length, it will be (d + 2*20) = d + 40
The initial position of the mass is 24 cm away from the steady state position and when it is released, the mass will oscillate symmetrically around the steady state position with a peak-peak amplitude of 2 * 24 = 48 cm.
Regarding your last question, I think it's clear if we , again, think analytically. But lets compromise and use the mathematical equation which we got from thinking analytically which is H = 2(mg/k + d) and we get h = h - d = 2mg/k + d So, from that we learn that h is directly proportional to d.
When you suspend the spring-mass system, the gravitational force and spring force will be balanced at 20cm below L, i.e. the new equilibrium length of spring will be 80cm. Now when we lift the mass a distance d(4cm) above L, the spring will oscillate about its new equilibrium length, and the amplitude becomes 24cm. Hence, the H will be 48cm (24cm + 24cm). If we take d=7cm, the amplitude of oscillation will be 27cm. Thus, H will be 54cm (27cm + 27cm)
H = 2mg/k + 2d, so h = 2mg/k + d and indeed h depends on d. The spring + mass movement is linear with H, so the gravitational effect (2mg/k) and the harmonic effect (2d) are superposed. If the spring is placed horrizontally, the gravity effect is 0, and only the harmonic movement is present: H = 2d, or h = d.
OMG I now see that in my first quick calculation I got 48 cm, but then I complicated it too much and gave you the wrong answer. When the spring is hanging vertical, it will be stretched out 20 cm more with the mass attached compared with no mass attached. From the new equilibrium point you bring it up those 20 cm + 4 cm. So we are now up 24 cm from the equilibrium point and this should be the amplitude of the harmonic motion. This means that the mass will swing down also 24 cm below the eq. point. H = 2 x amplitude = 48 cm. In the case that d = 7 cm I therefore think that we get the amplitude to be 27 cm and thus H = 2 x 27 = 54 cm.
Dear dr.walter sir.im in sri lanka .this qustion .remember me our 2017 advance level physics mcq paper .qution number 50 .same theory to solve the 50th mcq. . I love physics.thaks for helping students.
Long Lived Newton! and Long Live Prof Walter Lewin! Image a 90 degree rotation of the figure so that a ball with a spring attached to it rotates about a fixed point on a plane. Some new variables introduced but the idea would be the same, right?
Dear Sir , I posted the Answer of This Question Just to check in comment section of two of your videos ie in Problem 78 Video and Hint Video. But Nobody did response to my comment. Inspite of this We Love You Sir :)
What I derived is h = 2mg/k + d So, its not a coincident that for d=4cm, h=44 cm. Also, I used conservation of energy and found right answer and commented on the question video on the day I watched.
@@lecturesbywalterlewin.they9259 Thanks for reply, when you commented that it is coincident, I really got confused because I wasn't able to find an error in mine solution. Thanks again for clarification. 🙏
That is a very complicated solution. If you simply calculate the length of the spring with the weight hanging loose and calculate what the weight is lifted up in total thats 0.5*H 0.5*H = L + m*g/k - L + d = d+m*g/k H = 2*(d+m*g/k) = 2*(0.04+0.2*10/10) = 0.48 meter For what kept you awake : h = H-d h = 2*(d+m*g/k)-d h = 2*d+2*m*g/k-d h = d+2*m*g/k h = d+2*0.2*10/10 = d + 0.4 Like you did it .. I'm pretty sure I could not. En die ABC formule is wel heel lang geleden :-). Ik moet hem opzoeken ... in het nederland en het engels ....
yes h = 40 cm +d (in cm) I will prove that in ONE line in my next video. However, I was not awae of that at night when I woke up after I had recorded the solution.
@@lecturesbywalterlewin.they9259 Yes : h = d+2*m*g/k. The reasoning behind it to write it down just like that is not so obvious to me. I guess it is to you. Curious for the next video :-).
Professor, I am a 10th grader aspiring for physics olympiad. Can you please make a video on how to proceed. I have solved RHK, Sears and Zemansky and David Morin Classical Mechanics.
Might not be the right place to ask. Professor Lewin is not an instructor of Physics Olympiad. However, I have the same goal as you and good luck to you my friend!
@@AbdullahKhan-jc7lr Brother, Professor Lewin is already 84 years old, and he cannot spend that much time on commments. But I believe he will treat you as a friend and part of his physics family :)
I am new to physics. Can someone please tell me why the sign of potential energy term mg(h+d) is positive in the first equation. I thought that if the object falls down it should lose its potential energy and hence that term should be negative
