# Tag Info

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This is a statics problem. Assume the cable is static, perfectly straight and horizontal. Pick any point on the cable and the sum of the forces on that point must equal zero. There is a force, due to gravity, "downward". So, there must be an equal, opposing force "upward". This upward force must come from the tension in the cable. But, if the cable is ...

13

Here's a slightly more mathematical answer to go with Vortico's excellent physical answer. Tension is not a force, and doesn't have a magnitude and direction. This is confusing, because we frequently talk about the "force of tension", which is a different thing from the tension itself. The tension is actually an example of a mathematical object called a ...

11

Imagine a heavy chord raised off the ground between two blocks. Rather than consider all of the mass pieces of the rope, and the forces on them, we can simplify the problem a little bit by considering a slightly different one. The chord can be represented by a heavy ball (in the middle of the chord) connected by two massless strings to the blocks. From ...

5

Waves on a string are transverse waves not longitudinal waves. They are not variations in pressure, but variations in the displacement of the string. The (average) displacement is greatest at the anti-nodes and zero at the nodes.

4

Nonzero fluxes are required because of some equations of motion linking them to a nonzero Euler character. Once they're there, they induce a superpotential that stabilizes some moduli, usually the complex structure moduli (the very "stabilizes" means that the allowed values of these moduli at which the total potential has a local minimum is discrete, ...

4

Concerning the second question, the Planck length is the Planck length and not Newton's length (yes, the OP has asked this question). Newton didn't know Planck's constant which was only discovered 2+ centuries later so he could discuss neither Planck's constant nor the Planck length and other natural units which are functions of Planck's constant. Max ...

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I did a bit of discussion on this subject in this thread on Music.SE. The fundamental doesn't necessarily have the strongest amplitude. As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each $\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is ...

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The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope. Imagine if this assumption were to fail in the following way: $$\frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1$$ Then near the origin, the rope would look ...

4

I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand. Note: it's a ...

3

It's pretty simple: the forces at the anchor points would be infinite because of the 90° angle ;-) An example: Imagine two pillars with the same height. If you attach a rope on both of them and try to tighten it, you will slowly increase the pulling force at the top of the pillars while increasing the angle between rope and pillar. To fully straighten the ...

3

There are many quantum field theory models which are exactly solvable in the Large $N$ limit, such that the $\mathbb{C}P^N$ model, the Thirring model, the $O(N)$ vector model etc. Please see the following review by Moshe Moshe and Jean Zinn-Justin covering many of these models. The main idea is that Feynman diagrams (for example the vacuum diagrams in the ...

3

The first resonant vibrational mode for a string clamped at both ends looks like: You should be able to deduce the wavelength from that diagram. The second mode looks like: Both of the images above are from http://www.clickandlearn.org/Physics/sph3u/Music/Music.htm and that site will spell it out in more detail for you. If your string length is ...

3

Depending on the mass region of $\Phi$, either A or B can be taken as source and the corresponding response (vev). If $B\neq 0$ when $A=0$, it means that the system can spontaneously have a nontrivial vev even without any source. That indicates a phase transition. In the case both $A\neq 0$ and $B\neq 0$, it doesn't mean any phase transition. If we treat ...

3

Dear James, there is no reason why $\psi$ should be periodic. First, if you have a problem, imagine that $\psi$ are just auxiliary variables but the true ones are the bilinears $\psi_i \psi_j$ which are still periodic. Only things such as the world sheet stress-energy tensor $T_{++}$ and $T_{--}$ have to be periodic and they are because they're bilinear in ...

3

Here is a lecture to CERN summer students on "what is string theory" so physicists should be able to get a gist of the idea. There are more lectures in the "summer student lecture program" if somebody is really interested. I used "strings" to search. String scattering is a "simple" extension from Feynman diagrams, for those who know Feynman diagrams, as ...

3

You ask: In String Theory it is predicted that as a result of the closed strings we have spin-2 gravitons. 1) How do we know there must be an excitation of spin-2 particles? String theories have been chosen and are being extensively studied because they can have a representation of spin two particles . One searches for such theories because when ...

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Since $\mathbf{h} + \mathbf{s} = \mathbf{r}$, there's no need to use $\mathbf{h}$ or any of its time derivatives. Physics doesn't care about something like $\dot{m}$. It cares about things like, "What's the energy?" and "What's the force?". But the kinetic energy of the bob is $\frac{1}{2}mv^2$ regardless of $\dot{m}$. The energy in the stretched spring is ...

3

The equation represents power = force times velocity. The force used is the restoring force of a small string segment ${\rm d}x$ which is equal to $-T \tan \theta = -T \frac{\partial y}{\partial x}$. The velocity of the segment is obviously $\frac{{\rm d}y}{{\rm d}t}$. You can do the math yourself to balance the forces for a small segment, and will arrive ...

2

You say: I understand how particles with certain masses can form to make atoms, which create rigidity in objects due to Pauli's Exclusion Principle and what have you. These particles actually have mass and to a certain extent clearly would produce rigidity in objects. Do you understand what rigidity is ? I would define it as the resistance of a solid ...

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Roughly explaining John's answer - but visually..! When you wiggle a sting (using a good resonator), you'd see something like this... The structure clearly reveals Transverse vibrations. If you still don't understand, Here's a good simulation for that...

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A cable will never be straight because it stretches due to the weight (its own or of the passenger). So in short, the reason is its elastic properties. If this doesn't convince you, think of a heavy rod. It's straight, or as straight as we want it. Why is that? But of course it's because it's much stiffer and less elastic. If your zip line was very stiff, ...

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I would expect that if you pluck a guitar string the mode with the highest amplitude is the fundamental, though this isn't necessarily the mode with the highest energy since the energy of a vibrating string is proportional to the square of the frequency (and the square of the amplitude). However you're not measuring the amplitude of the string vibration. ...

2

It all depends on the initial configuration. If the string's initial configuration (shape of string at $t=0$) includes a node at the center, the string will not vibrate at the fundamental frequency. This YouTube video is a demonstration of this phenomenon.

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The classical string equation that you are referring to, is formulated by making a number of assumptions, which include that the vibration of the string does not affect its tension. This makes Young's modulus irrelevant for results calculated from the idealized equation. In the real world, materials with low moduli of elasticity will follow the ideal ...

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I do not think it is that tricky. For each pulley you have an equation relating the speed of the rope in, the speed of the rope out and the speed of the pivot. The speed of the pivot is the average of the two speed ropes. Here are the kinematics of the system $$v_E = 0 \\ v_B = a t \\ v_C = v_B \\ v_B = \frac{v_D+v_E}{2} \\ v_A = \frac{v_C+v_D}{2}$$ If ...

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Let's look for solutions of the PDEs in a form $\eta(x, t)= \eta(z)$ and $\xi(x, t)= \xi(z)$, where $z= x-c_T t$. If we substitute these solution to the PDEs, they are reduced to ODEs. But to answer you question, we only need the second equation, which has the form:  c_T^2 \eta_{zz} = {\lambda \over \rho_0} [\eta_{zz} ({\tau_0 \over \lambda} + \xi_z + ...

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I'm not qualified to answer this question in detail...but I'd like to point out some things I've learned lately that may be helpful. Hilbert space is useful when you need an infinite dimensional space to characterize what you are studying and where each mode is orthogonal to the others. Great for Quantum Mechanics... Where I work, images, that is 2D ...

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