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1

I) There are already several good answers. OP is asking about the momentum of the non-relativistic string in the transversal model, which in textbooks usually is given as $$ {\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2}. \tag{1}$$ II) Let us fix notation: $\rho$ is the 1D mass density; $\tau$ is the string tension; $Y$ is the ...


4

There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and ...


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If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration. In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.


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I'm having a problem visualizing the transverse waves with 3 or 4 nodes you mention. All videos I've found show standing waves with only two nodes, the slackline moving up and down between the anchor points (or between an anchor point and the person walking the slackline). The rotation waves I saw (maybe torsion oscillation is a better name) also had only ...


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You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density $\mu$, is along with the $x\equiv ...


9

A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave ...


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The second derivation is correct, as explained by Diracology. However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems. A mass on a stretched spring. A mass sitting on a table. A charged mass next to another charge. These three systems have ...


9

The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small ...


0

Let's suppose first that the astronauts are tethered by a rope to their suits, and the rope is tight, in that initially there is no slack. Put another way, the length of the rope is the distance between them. They both grab the rope and pull with their respective forces and keep holding those points on the rope. Now astronaut A pulls with 20lbs of force ...


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The physics in the 1st approximation are worked out here: https://en.wikipedia.org/wiki/String_vibration with the result that frequency depends on the length, tension, and mass density as: $ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $ The 1st factor is why bass strings are long and treble are not (on a piano). The lower 3 strings on your guitar have a ...


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The existing answers are good, but i try to present an answer that encompasses all the concepts together. Step 1: Consider the astronaut B and the rope to be a system, and the astronaut A to be the other system. The astronaut A applies a force of $20 \text{ lb-f}$ on the other system, and hence, by Newton's Third Law, the other system applies a force of $20 ...


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Consider the frame of reference of the rope: Astronaut A pulls on the rope with $20 \text{ lb-f}$, so by Newton's Third Law, the rope pulls on Astronaut A with the same force. There are no other forces acting on Astronaut A. Therefore, Astronaut A will accelerate along the rope with an acceleration appropriate to this force and his mass. Similarly, ...


1

If you provide to the bottom an initial speed $\:v_{bottom}\:$ then the kinetic energy would be $\:E_{bottom}^{kin}=(1/2)\;mv_{bottom}^2\:$. If $\:E_{bottom}^{kin} >E_{top}^{dyn}=mg(2r) \:$ that is $\:v_{bottom} > 2\sqrt{gr} \:$, then the rock will execute this vertical nonuniform circular motion for ever. To the contrary, if $\:v_{bottom} ...


1

[...] which would end up giving tension a negative value. How is this possible? It isn't. If you set zero speed $v=0$ then you no longer has a circular motion, and the object is accelerating downwards. A non-zero speed $v$ is a requirement for a circular motion to happen, because a radial acceleration towards the centre is a requirement. Otherwise it ...


0

Tension in the string will never become zero, as long as rock is moving along the circular path. Also, speed of the rock also, will never become zero in this case.


3

Let me start out by saying this: subtracting the forces is certainly wrong. I have no idea why they would get the answer 20-8=12, other than wanting to subtract 20 and 8 and looking for an excuse (not exactly likely). A quick though experiment confirms what I say. If you extend their logic, then if both astronauts pulled at 20 lbs of force, then they ...


3

If Astronaut A were pulling with 20 lb of force then Astronaut B should be feeling 20 lb of force if he were just holding the line. If he is only "pulling" or "feeling" with 8 lb of force ,then he's actually letting the line slip through his gloves while using sliding friction to maintain 8 lb. of force. An insidiously tricky question if you ask me (and I'm ...


1

If the phase difference between the wave is zero i.e lies in the plane of wave motion the resultant displacement is equal to zero, Thus, $A = 0$, $(L,t)=0$, due to that fact, you can use $$A\sin(kx−\omega t)\to-A\sin(-(kx+\omega t))=A\sin(kx+\omega t)$$ for progressive wave, but nothing can happen when you use cosine rule because give us answer equal to zero ...


4

They are not equivalent. A way to see this is to find the normal modes associated to these solutions. Applying the condition $\xi(L,t)=0$ to $$\xi(x,t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t),$$ you get $kL=n\pi$. This gives $\lambda_n=2L/n$ and $f_n=nv/2L$. Which are the correct wavelength and frequency for the rope with both ends fixed. For the ...


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Yes, it is possible to have two different waves of different frequencies on a string (Imagine driving the string with a square wave; it consists of multiple frequencies with multiple amplitudes; See Fourier transform). However, the velocity of waves on a thin string are constant and depend only on the tension of the string and the density. Thus, you cannot ...


0

Just the longer version of the answer @xasthor Suppose the acceleration of $m_1$ is $a_0$ towards right. That will also be the downward acceleration of the pulley B because the string connecting $m_1$ and $B$ is constant in length. Also the string connecting $m_2$ and $m_3$ has a constant length. This implies that the decrease in the seperation between ...


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$a_0-a$ , here $a$ means the acceleration of two blocks with respect to the pulley. Hence, if $a$ > $a_0$ then also this relation perfectly works. They are assuming $m_2$ to be accelerating upwards because they have preasumed that $m_3$>$m_2$. A very simple way to assume this is that suppose the whole system was at rest. Then the mass $m_2$ would move ...



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