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7.34+ Consider the well-known problem of cart of mass m mnoving along the axis attached t0 spring (force constant k) whose other end is held fixed (Figure 5.2) . If...

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7.34+ Consider the well-known problem of cart of mass m mnoving along the axis attached t0 spring (force constant k) whose other end is held fixed (Figure 5.2) . If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency @ = Vkfm Using the Lagrangian approach, you can find the effect of the spring mass as follows: (a) Assuming that the spring is uniform and stretches uniformly: show that its kinetic energy is M;?

7.34+ Consider the well-known problem of cart of mass m mnoving along the axis attached t0 spring (force constant k) whose other end is held fixed (Figure 5.2) . If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency @ = Vkfm Using the Lagrangian approach, you can find the effect of the spring mass as follows: (a) Assuming that the spring is uniform and stretches uniformly: show that its kinetic energy is M;? , (As usual is the extension of the spring from its equilibrium length. Write down the Lagrangian for the system of cart plus spring: (Note: The potential energy is still ykx? ,) (6) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency Vk/(m 1 M/3); that is, the effeet of the spring mass M is just t0 add M/3 t0 the mass of the carl;



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Consider the well-known problem of a cart of mass $m$ moving along the $x$ axis attached to a spring (force constant $k$ ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency $\omega=\sqrt{k / m} .$ Using the Lagrangian approach, you can find the effect of the spring's mass $M,$ as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is $\frac{1}{6} M \dot{x}^{2} .$ (As usual $x$ is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still $\frac{1}{2} k x^{2}$.) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency $\omega=\sqrt{k /(m+M / 3)} ;$ that is, the effect of the spring's mass $M$ is just to add $M / 3$ to the mass of the cart.

Okay, so we have a spring with mass m spring constant K equilibrium like l not The end of the spring is moving with the velocity V said we'll call Well, define another velocity you This is the velocity that changes along the length of the spring. So you have l equals V when little l Which is to find from the from the end here this is l equals zero l equals capital l wherever the spring extended Thio. So you've l equals V when l equals big l knew of l equal zero when little well, equal zero So at the end of the spring, it's moving with speed V At the beginning. In the spring, it's moving with speed zero. The next thing we needed to find is the linear mass density. So this is how mass changes would length row l. And this is just em over l the total mass divided by the total extended length. And we're gonna turn this into a differential so d m equals m over l d l. So as we move incrementally with little l r mass changes by big M Big M over big L, which is the total extended length of the spring. Our potential energy kay is going Eagle 1/2 M U of l squared. Let's take a derivative here, So d k legal 1/2 d m times u of l squared. Um, let's plug in what we know for u of l c U of l equals l over l a times V. Um, since we have our boundary conditions that u of l when little l equals big l has TV and has to be zero when little while equal zero. And a linear equation of Celeste fits all of our conditions. So we go and plug this in here, all right? And then we want to replace D. M with, um, this guy. Okay, so we'll combine all this and get 1/2 m the so sorry. This is little b squared over l cute times. Little l squared d l Now we can integrate the kinetic energy. So, uh, the integral of decay, which is gonna equal kay equals 1/2 Mm. The squared over l cubed times the integral of l squared rum. 02 big l. So we're integrating. Were instigating over our length of our spring from zero to the total length of this spring. So this is 1/2 m the squared over l cubed l cubed over three from zero to l. This equals 1/2 m the squared over l cubed times l cubed over three. It's just equals 1/6 Um, the square. So, you see, we get about 1/6 of the kinetic energy. I'm gonna go to a new page for the second part. All right, so we're gonna get the equation for the total energy e equals 1/2. Mm. You squared plus 1/2. Okay. X squared where X is defined as l minus l not. So it's a distance from equilibrium. We could drop the one halfs for simplicity because we're just going to solve because we're saying, is that the secret of a constant and differentiate it So we have m u squared plus K X squared eagles a constant. Now, if we differentiate both sides, we get to m d to emu Dutt once to k x dx DT equals the differential of a constant, which is zero. If you differentiate a constant, you get zero so we can replace Dutt with a and e x city with the Are you sorry in this case? So we have to m u a plus two k x, you equals zero. Take a you out. We have you to m a waas two. Okay, X equals zero. We, of course, could take the two out and divide by the two. So you can just get rid of it here on. And we see that furthest equal zero m a plus k x must equal zero or a equals negative k x over at which we know is true for simple harmonic motion as well. And finally we can do part c No. So omega equals k o gram and we're at the kinetic energy again is 1/2 m the squared. Um, but we know our kinetic energy is actually 1/2 of me. Sorry. 1/6 Bigham B squared. So? So if you compare the two equations, we see that little M would equal big M over three. So if we replace that into omega to see how this changes things, we get Omega equals route kay over Big M over three, which equals route three. Okay, Over M. So the, uh, the angular frequency increases by a factor of Route three. If we consider the mass of the spring

Even this is the problem is if I spring having the mouth and here it is David. I spring off Sam and length and not It's a bull and it's fixed on the other end is free to move. We have to fight. It's kind of particularly if the velocity of a point where is linearly with distance from the extent. So here the velocity is zero. Then it gives increasing and becomes Maxim at the other end. Then we have to find that kind of technology and second part if mass image has to it. So by considering the energy concept, we have to find its angular velocity. If spring having the mask, then what will be the spring Anguilla relatively off oscillation that we have to calculate. Let us start doing it. Starting with part A. I'm drawing the diagram again for upon the friends. This is all this is a at all and zero. So velocity is zero. Well, the city is increasingly nearly velocity is deflection off. Can Britain has we? L upon and is not. It's total length of this unstitched length of this week, Matt. Unique, partly distributed over their spring so mass per unit length will be, um, upon and not or simply, l If it is given l so I should take and only let us start solving it. If we consider a very small element, the l distance from it, then it's a mass will be I am upon Elder on it is moving with velocity use culture the and upon. So he element off having that kind of technology You can find that kind of technology d m into you square How, um upon l deal and to you will square that is the square Ellie square upon elsewhere. So what kind of technology off this element? You may right, hop um el que into the square and square deal toe. Find that total. We have to integrate it. After integration, you will get M B square basics. So this is the answer off a part. Now be part energy can be defined. That's toe on being toe the V a quantity potential energy K X. Do you accept quantity? That must be true. So we can write AM a plus k x. It's called zero. So from here Omega, you can find a skull toe minus K upon them into X on age. Um, you guys square X so miserable Be route off the upon up by energy consideration because energy is constant. So it's the derivative is you Don't I have written Did this equation directly? If you would like, you can do it here like I am doing, total energy will be half and be sweat plus half X squared since total energy is always constant So it's derivative is Toby zero. If you differentiated, you will get and be duly appointed plus k x de except quantity. That must be zero. So that equation we have written there directly, so use it. But it from here. No sea parte if spring having the mosque, bank it, um, to be replaced by m three. So omega is card too. Three. Give I am Or you may right que upon MDs. Um, this will be, And by three, that's all for it. Thanks for watching

See how the mass inside of a spring affects the oscillation of that spring when a obvious masses attached and set into oscillation. And we're actually going to start with conservation of energy and see what we mean by effective mass. So our total energy looks like our kinetic plus potential and we have one half M b zero squared for the oscillating velocity. Um and that's associated with the velocity of the mass a block. We will later generalize this to the velocity of individual pieces of mass that are inside of the spring. So we're going to imagine the spring divided up into tinier pieces. But then we have the potential energy of the spring which is one half K. X. Squared where X. Is the displacement of the spring from its equilibrium or it's it's difference in length from its unstrap etched, uncompressed length. And so in order to look at the oscillation, what we're going to do is use the fact that the derivative of the total energy in time needs to be zero as that energy is conserved and we can take the derivative of each term inside of there. Using the chain rule, the kinetic energy factor of two comes out for the v not squared. And we're left with feet not not dot And the same with the potential energy. The factor of two comes out and we're left with K X ex dot Yeah. Um and then we can set that equal to zero and we find out that mm v naught and we're going to substitute acceleration for the derivative of the velocity is equal to minus. And again we're going to uh set are derivative position equal to the velocity of that mass. And we're left with a very simple expression for the relationship between the acceleration of the mass and it's uh spring constant and inertial mass. So what we have is a is equal to minus K. X. Over M. And this K over em is what we're going to associate with the oscillation frequency squared. Okay. That may seem like a lot of work, but it's to show dynamically um if we just use took slaw, we would not necessarily have the sense of things how change in time. So now for how to treat the spring, but we are going to do is we are going to break up the spring into little tiny infinitesimal masses. So we'll look inside of there and just imagine a small chunk of the mass that we're going to called D. M. And it is at position will call it little L from the wall. And the kinetic energy of that little chunk is one half um B squared Dm. Okay, so we're saying that that little chunk has its own speed separate from the zero of the the big block. So that's important to realize is that that V. That we're writing down is a function of where we are inside of the spring. But anyway, like usual to find the total energy in the spring, we're going to integrate over that spring and we're going to assume that V varies with L. Um that it starts out at zero at the wall and it grows so that right at the attachment point, the V equals V. Zero. So this is a function of L. Um And we're going to assume a linear dependence. So V equals zero at the wall. That makes sense. That point is fixed. N. V equals V. Zero at block attachment point. That makes sense to we have sort of a boundary situation that we have to be aware of and then we can um put in for that function and integrate over the entire length of that spring from zero to L. So our total kinetic energy of the spring parts is going to look like one half V. Not over L. Times L. Squared. Um And then what we're going to write for D. M. Is it's the mass of the spring over the full length times little differential length L. So it's kind of like a linear mass density is equal to mess of the spring over L. And so we're assuming a uniform spring. This is in general true. If you want to take a small chunk of mass inside of a bigger piece of mass, then that small chunk should have the same density as the big mass provided its uniform. So we're going to integrate over the the linear mass density. Probably a better way to write that. So my derivatives aren't so messed up so mass over length time's D. L. Yeah. Okay. And that's fairly easy to do. We can bring out all our constants so called constants. Constance in the integral at least they may not be constants in time. And we're left with the integral 02 L. Over a little L squared D. L. And that is easy enough to integrate over this will give us one half, you're not squared, ill cubed mass of the spring times L cubed over three. And lots of things will kind of simplify down mainly the L cube will disappear. And we're left with one half mass of the spring over three times V not squared. And we can see then if we put this back into energy conservation. This will simply add a term onto our kinetic energy term. Um And it will add an amount of M. S. Over three to the inertia in the kinetic energy. And that means that our new, I mean it'll go through the math that we did above to find omega squared. But our new omega squared, including the mass, is going to look like okay over I am plus M. S. Over three, so that the mass of the spring will slow down the oscillation because more inertia is added in. And that does make a lot of sense that that's what would happen.

E no position of the peace of the spring position as and so any quarter zero, it is a fixed point, huh? Any legal capital is the moving end of the spring. How do we understand if something is there? One spring is attached to the site. So one scientists next on the other side is moving. So here the length a legal zero. And here the length Caldwell capital. This is how that is Define No, the velocity off the point corresponding to end The velocity is in ordered by you Now you off. L equal to be led by it That this when the spring is moving l b l b the function off time, then spring moves Then what happens? L a will be the function off time. What's it is dependent on time? Therefore, you is an implicit function off thing day. So in the A bob gm equal to capital m upon el into deal Therefore DK equals half d m you square because it was too half capital M b squared upon a que into elsewhere, do you? And then okay will do integration off decayed which is equals to em. These where a phone Tu que integration from the limit of zero to capital and square D l, which is equal to mm. These square once is I missed it. No, the beauty part. We have MB Devi upon DT last k x dx upon DT equal to zero or begin right and me last k X equal to zero. This is what we have. We'll be long to the sea Buck Emmis replaced by M by three Now therefore, neither will become Rudolph three k a one him there n dash will do and like


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