Online Physics Ninja Respect is always appreciated. You simply can not get the same solution without the top spring. If that were the case you'd lack the restorative force of this spring. And the forces of both springs, along with the fact that these are applied in different masses, each one being capable of moving independently has a lot to do with the non intuitive nature of the solution. It helps me to think that when one spring is pulling the other is pushing, but the friction force is always resisting sliding or relative motion between the two masses.
I'll freely admit I'm not even skilled at college level physics. But as far as I can see, this problem involves two independent objects that are only held together by gravity and friction. Without a spring on each object in equilibrium, it seems likely that the system would break down much faster. For instance if you suppose a 2k spring is connected to 2m, while m is free, than as the system approaches x max, m will ONLY have friction to hold it in place with no spring at all to provide a push to keep it in place with 2m. Basically a whole section of the equation that could cancel itself out, no longer does and m would go tumbling away.
I have been solving these problems from the beginning since the past few days and it is so fun. And I got this one right! It's such a good feeling to be able to learn so much from you. Thank you so much professor, for creating this brilliant playlist.
I do need to watch your lectures on simple harmonic oscillation again. This demonstrated perfectly, what I don't understand yet. It's easy to find the formulas, but knowing is not understanding. ω still remains an abstract concept to me, I don't know what it means. Thanks for the motivation again!
Hello Dr Lewin, thanks for the elaborate explanation and the “two cases” approach of this problem in the two video’s . The onze thing troubeling me is the 2*k in the explanation. In your 8.02 lectures you have an almost identical setup and use just k for the system with springs on both sides on the frictionless bank. But apart from that I am pleased I found the solution myself using your lectures...
I think that in the first part of your explanation, where you say you move the "whole system" to x1, it means you are doing so by moving either the 2m block or the 1m block, while depending on the frictional force to prevent the other from slipping. This is the only way to get the net force on m to be zero other than at the initial rest point. You could also get no slippage if you moved both blocks identically to some x less than the free oscillation maximum and then released them simultaneously. I was not surprised by the final result, as that is an analysis of the problem presented. I was surprised by the first result, as I didn't bother to calculate this case because moving one block with an external force was not the stated problem When the system is free to oscillate, you get the final result from equal accelerations at the limit position (of course, they are equal at all times if there is no slippage), independent of the fact that the oscillation is sinusoidal. The fact that the oscillations are sinusoidal IS critical in assuring that the acceleration is never greater than the limit value, and and is everywhere less than the maximum that occurs at the maximum displacement. The numerical value of the frequency of oscillation is interesting and good to know, but I think not essential to the answer.
>>>The numerical value of the frequency of oscillation is interesting and good to know, but I think not essential to the answer.>>> The diff eq for a spring (k) and mass (m) is ma=-kx, a is d2x/dt^2. therefore x = A*cos(wt). w=sqrt(k/m) please show me your solution without using the frequency w.
Knowing that the oscillation is sinusoidal, we know that the position, velocity, and acceleration are all sinusoidal (or cosinusoidal) and also that the peak acceleration occurs at the peak displacement (but has opposite sign). Then, examination of the equations for force and acceleration when the system is at maximum displacement will directly give the equations for the accelerations of the two masses m=1 and m=2, which must be equal if there is no slippage as I wrote in my solution. The smaller mass is accelerated by a force kx minus the frictional force ugm, and the larger mass is accelerated by a force kx plus mgu. The two accelerations are the net force on each object divided by its mass. We simply equate these two accelerations and solve for x when the frictional force equals mgu. So, you see, at this point, we have not needed to know the value of the angular frequency. The important information is that maximum acceleration is at maximum displacement, which depends on the sinusoidal shape of the oscillation and not on the value of its frequency.
Maybe using a sort of "seesaw"? That way we might exploit gravity and accelerate the system a little bit every time the blocks get over the middle point until we reach the maximum distance?
I'm a high school physics teacher. Sir, m is moved to a position 'x1' & it is at equilibrium. But there are UNBALANCED forces on '2m', which will cause acceleration! Plz explain this contradiction 🙏
@@lecturesbywalterlewin.they9259 Dear Sir, I request you to provide your email address so that I can attach my doubt. This will be a great help of you on physics community. *Thanks & Regards*
sir why the area of triangle comes out to be the same irrespective of the base chosen for the same triangle or in other words why is 0.5*base*height = 0.5 second base*second height=0.5*third base*third height of the same triangle or why these numbers should come out to be same. for me this is not intuitive as i was taught that area of rectangle is base*height by definition and then we use it to deduce the formula for area of triangle. Regards and sir i have already searched it on google
One of the simplest ways I can think is using the law of sines & calculating heights from the sides. give it a try. also there is this nice math channel of professor Wildberger ru-vid.com that will teach you how to find the answer by yourself, and more.
using law of sines means that we are using trigonometry and trigonometry is based on similarity of triangles (basic proportionality theorem) and it is prooved by using the result(or axiom I DONT KNOW) that area of triangle must be the. Same when you say that sin(30deg)=0.5 you use the fact that all right angle triangle with one angle 30deg are similar and thank you for reply
Lectures by Walter Lewin. They will make you ♥ Physics. Sir I just joined edx ur lectures were not there(when I search Walter lewin it says no results found for Walter lewin)
All my lectures 400+ lectures and talks are on this channel 94 of them are my 3 MIT course lectures. 8.02 was on edX in 2012 and 8.01 in 2013. ru-vid.com/show-UCiEHVhv0SBMpP75JbzJShqwfeatured?disable_polymer=true
Hello sir. Could you explain how friction between blocks is on right side when released? Does it mean that it tends to slip forward? But that too how because lower body is going left and maybe friction too would be acting on left side. Sir pls help I am bit confused in this part. I have watched this video many times but couldnt understand that part
One thing I'm curious about: Does centre of mass make any difference in this equation? It feels to me like friction is only constant while m and m2 are horizontal relative to g, so if the springs were attached above the centre of mass of one or both objects it would seem likely to induce a rotational effect that in turn could substantially reduce friction.
the spring on the mass m should be pointing in the direction of the center of mass of that mass (we never mention that in problems but it is always understood). That is not needed for the spring on the left as the friction coeff with the table is zero.
Sir, in mechanics we use 0=F.v and in electricity we use P=V.I but in both cases P=dW/dt. So if we wanted to calculate power using P=F.v for an electrical appliance, how would we do that? Will F be the resultant force of all the forces exerted by the appliance and V be the resultant velocity of all particles the force is being acted upon???Thanks in advance.
there is an implication in your discussion of x1 that the initial displacement must be achieved by pushing on the bottom block. could one not simply grasp both blocks and displace them both to xmax initially?
no that will not work. You cannot push the small mass beyond x1 as it will then slip. When x=0 you can give the 3m mass a speed v such that x_max (where v = 0) will be 3mu*gm/k. 2*0.5k*x_max^2 = 0.5*3m*v^2. One eq with v as unknown.
I think I'm missing something then, as I'm not seeing how the initial displacement could not simply be xmax. Do not the two springs act simultaneously at the instant the release occurs?
yes you are missing something KEY. The moment that x=x1 and I hold the masses in my hand, a=0 on m. thus friction - kx = 0. But friction is then the maximum possible. Thus if you move you hand past x1 the mass m will slip. watch my soln again.
But I have two hands. Could I not press down on the top mass with my right hand, to create a temporary large friction force, while using my left hand to displace the masses? Then release both simultaneously? Or does that violate the problem statement somehow?
yes you could press on m with your other hand. You can then go to X-max and let them go. However, if you take your hand off m 0.01 sec before you take your hand off 2m, m will slip. And if you take your hand off 2m 0.01 sec before you take your hand off m, , m will probably also slip. BUT IN principle you are right. I could move 2m and m out to X_max and hold them both firmly at rest. Release them simultaneously and m will NOT slip.
Lectures by Walter Lewin. They will make you ♥ Physics. Actually sir I am a bit confused that what's the actual difference when we were holding the blocks and when we removed our hand... I understood the mathematical format how it comes 1/3 of max frictional force... But I wish to understand what actually causes this sudden decrease in friction... Because something must be causing this decrease ?!?!
>>>Because something must be causing this decrease ?!?!>>> YESSSSSSS watch my solutions! When the objects are in motion the acceleration has to be taken into account.The net force on the blocks is then NOT zero. When I hold the objects still at x1 the net force on the blocks is ZERO.
Lagrangian mechanics works as a mathematical shortcut to solving mechanics problems, that you could otherwise solve with Newton's equations alone. You could use Newton's equations and get the same answer, but it means a lot more steps.
Why is it not intuitive? Starting with the fact that for the X1 you used, you got 1/3rd the max friction force. Knowing the friction force is proportional to Xmax, it is obvious that Xmax must be 3 x X1.
We have assumed that x1 is the maximum distance upto which the smaller block will not slip but the amplitude comes out to be 3*x1 upto which the smaller block will not oscillate And trust me this thing has created a fight between maths and physics in my mind
Excuse me, sir, I accidentally found this video on RU-vid.I love to solve this problem. But I haven't got the question. I refused to watch this as I should solve the problem by myself. Could you please tell me the question?❤❤I would be much obliged if you consider this request. Please be kind enough.
Sir, I do have a question for you. In the video you indicate that you have a way to increase the mass displacement from X1 to Xmax while the system is in oscillation. How would you do that? Would you move the surface rather than the mass?...I'm interested...and curious...
you can do that with "wind" using a hair dryer. Use the wind to push on the system every time that the object (SHM) goe through equilibrium (maximum speed) - that will increase the amplitude in a very controlled way.
Sir but how do we get it started? 🤔 Holding both blocks together and pulling it together till 3x1? Is that how we get it started for maximum amplitude?🤔🤔🤔🤔
It's very easy. When x = 0 we give the 3m=(2m + m) a speed v such that it comes to a halt at x_max. 0.5*3m*v^2= 2*0.3*k*x_max^2. 1 eq 1 unknown solve for v!
Oww ok!!! Yes sir got it... Thank you for explaining like this.. many find me annoying and dont clear the doubts i get. Thank you sir for being kind enough 😊
@@lecturesbywalterlewin.they9259 sir isn't it true that the displacement from the mean position (where net force on the blocks is 0) is equal to the amplitude
