# Tag Info

48

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 ...

17

The tension of the rope is the shared magnitude of the two forces. Imagine cutting the rope at a point and inserting a spring scale in its place. The reading will show the tension. A rope with zero tension would be hanging loosely or laying on the ground, neglecting the rope's mass.

15

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 ...

12

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 ...

6

Strings are not described very accurately in popular science, because much of the physics of strings was only understood long after the mathematical theory was somewhat advanced, and an accurate classical analog for the string wasn't available until relatively recently. The classical analog people often use is a vibrating band of energy, but this is mostly ...

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

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 ...

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 ...

4

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 ...

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

Strictly speaking, tension is not the same as force, although it is sometimes described as the magnitude of the 'pulling force' experienced by an element (such as a rope). The important thing to remember when resolving forces in classical mechanics and to understand tension is to apply Newton's three laws of motion. They are: 1st Law: an object with no ...

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

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

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

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 ...

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

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 ...

3

When you pluck a string it does not start out like the fundamental above. The string is pulled into a bent shape of two straight lines and an angle and it may not be bent at the middle. Releasing the bent string causes a bunch of harmonics of various amplitudes depending on how far off-center it was bent. (It can not return to the bent angle shape and the ...

3

Imagine a finite segment of the rope, say on the left side. Suppose the tension at the top of the segment is $T_t$ and the tension at the bottom of the segment is $T_b$. Then the segment of rope feels a force $T_t$ up and $T_b$ down, or a net force $T_t - T_b$ up. Since the rope is massless there is no gravitational force so that $T_t - T_b$ is the net force ...

2

The assumption that the tension $T$ is constant is what's violated here. Because the cable has a weight, that isn't true. The shape of the cable is not a parabola but a catenary, i.e a shape descrbed in terms of the hyperbolic cosine $\cosh$. The way to solve for the shape of the cable is to find and solve a differential equation that describes the shape ...

2

You can only do the balance of forces between spatially separate points only if you know the slope of the rope at each point, and the weight of the rope between the points only. To get there you use a small section to derive the differential equations and then integrate over the range of $x$ values you want. Given a small section of rope spanning the ...

2

The response can be derived mathematically. Let $u(x,t)$ denote the displacement of a point along the string at $x$ at time $t$. The function obeys the wave equation in flat $d=2$ Minkowski space, $$\frac{\partial^2 u(x,t)}{\partial t^2} - v^2 \frac{\partial^2 u(x,t)}{\partial x^2}=0$$ If we pinch the string at the middle, this corresponds to a condition ...

2

For a string of linear density $\mu$ and tension $T$ and length $L$, the normal modes are given by $$y_n(x,t)=A\sin\left(\frac{\pi nx}{L}\right)\sin\left(\frac{n\pi}{L}\sqrt{\frac{T}{\mu}}t\right)$$ assuming initial condition $y_n(x,0)=0,$ and $A$ being the amplitude. Notice that at $t=0$ the string is flat, and so all energy is in the form of kinetic ...

2

When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey $$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n},$$ where $\lambda_0$ is twice the ...

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.

Only top voted, non community-wiki answers of a minimum length are eligible