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157

This effect is known as inharmonicity, and it is important for precision piano tuning. Ideally, waves on a string satisfy the wave equation $$v^2 \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}.$$ The left-hand side is from the tension in the string acting as a restoring force. The solutions are of the form $\sin(kx - \omega t)$, ...


56

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 ...


23

When you pluck the string you excite many many overtones, not just the fundamental. You can observe this by suppressing the fundamental. Pluck the string while holding a finger lightly at the center of the string. That point is an antinode for the fundamental and all odd harmonics, but a node for the even harmonics. Putting your finger at that point ...


20

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.


20

In plain English, there is stiffness at the ends of the strings where they are fixed in place, which makes the string's frequency of vibration slightly higher (sharper)—effectively shortening the length of the string slightly, for all practical purposes. And the resistance to bending is dependent on the frequency. It behaves more “stiffly” with regard to ...


17

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 ...


16

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 "...


13

Because they are too long for the confined space. Reasoning: A very short string can't clot (say 1mm). So it needs a certain length before it can start clotting. If they could be placed neatly (say big pocket), they wouldn't clot because the ends are too far apart. They must be flexible. A steel bar doesn't clot. So you need certain elements to be just ...


13

This is the boring answer that your physics teacher would give you :-) The diagram above shows an idealised rope with all it's weight concentrated at the centre of the rope. If the rope mass is $m$ then the force downwards is just $mg$, where $g$ is the acceleration due to gravity. Suppose you're pulling on the rope with force $F$, and the angle of sag ...


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 ...


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 ...


8

There are three parts to this: In mechanical equilibrium, things go to their lowest energy state. A straight line is the shortest distance between two points. Whenever you've minimized something, it means that small deviations don't change its value (to first order). Let's start with the third point, which is mathematical, and then look at the physics of ...


8

Earbud cords form coils in your pocket. Even if you crumple the cord, it eventually finds its way into coils as you move and the cord is agitated. The coils lie in planes that are stacked. It's been found that agitated coils spontaneously braid and knot themselves. See this paper: http://www.pnas.org/content/104/42/16432.full, which found that formation ...


8

Carrying out the Fourier transform, I get a slightly different result for the frequency spectrum than 'knzouh'. I used $u$ instead of $y$ and $c$ instead of $v$, so the PDE becomes: $$u_{tt}=c^2u_{xx}-Au_{xxxx}$$ Fourier transforming the equation: $$F\{u_{tt}\}=F\{c^2u_{xx}\}-F\{Au_{xxxx}\}$$ Transforming $x$ to $k$: $$\hat{u}(k,t)=\int_{-\infty}^{+\infty}u(...


7

The most interesting version of this question (I think) is this: Suppose that the rope has been attached long enough to bring the two galaxies to a stop relative to each other. (d[physical distance]/d[cosmic time] = 0). For subsequent times, does the rope need to maintain any tension to keep the galaxies from starting to accelerate away from each other? If ...


7

What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end? As usual, this is in the context of a Schwarzschild black hole. First, outside the horizon, a object with constant radial coordinate 'feels' a constant proper acceleration, i.e., an accelerometer (think of a weight scale) ...


7

Tension is an internal force in a body, such as a rope, that resists any attempt to pull the rope apart. Simply, tension arises due to intermolecular interactions, and if it did not exist, ropes would fall apart the moment you pull on them. Now, it is necessary to distinguish between internal and external forces for a body. External forces are forces that ...


6

Amazingly, not only has someone recently done an analysis of the Tarzan swing, but they've posted it on the Arxiv. See http://arxiv.org/abs/1208.4355 for the article or the Arxiv Blog for a summary. To quote the Arxiv Blog article: In fact there is no simple rule for maximising the horizontal flight distance. It turns out this depends on a number of ...


6

To understand the physics here we should first consider the assumptions made, and perhaps try to justify them with some physical arguments. Assumptions of this simple pendulum: Newtonian physics apply. (ie any non-classical effects are negligible) The "string", or perhaps more accurately rod, has a fixed length $l$. (That is the string remains under ...


6

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 ...


5

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 ...


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.


5

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 ...


5

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 ...


5

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me. Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole ...


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given by ...


5

D'Alembert equation reads, for the considered case, $$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$ It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal ...


5

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 ...



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