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61

The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^... 18 This source shows that for free beams like the bars of the glockenspiel, angular frequency \omega=2\pi f and wavenumber k=\frac{2\pi}{\lambda} are related by$$\omega^2=\frac{YI}{\rho A} k^4$$where I=\frac{1}{12}bh^3 is 2nd moment of cross-sectional area about a horizontal axis through the centre, and A=hb is the cross-sectional area. In the ... 1 The other answers saying that true plane waves don't exist and are mathematical idealizations are perfectly true, but you can certainly have waves that are near enough to plane in reality to give rise to the "problem" you allude to. This is where we meet a subtlety to the oft-cited, but somewhat mashed assertion that nothing can travel faster than light. ... 2 Plane waves are fully compatible with special relativity, since they are Lorentz-invariant objects:$$\psi_k(x) = e^{i px/\hbar} = e^{i(Et-\vec{p}\cdot\vec{x})/\hbar}$$You seem to be concerned with the fact that plane waves are spread through all space. But in fact, they're spread through all spacetime! They are perfectly periodic both in space and in time,... 1 Plane waves are not real, they are just a mathematical device. In quantum mechanics, particles are represented by wave packets, which do not have infinite amplitude and allow a collection of plane waves to group together by interfering constructively within a certain area and destructively outside that area. From Wikipedia Wave Packets In physics, ... 3 Loudness will not help. You do indeed have a (big) problem with reflection - the geometry is really designed to make the poor rat's task impossible. You need to manage the reflections. Since you state that modifying the surface is not an option, I would recommend tapering the walls - instead of being vertical, you might place them at a 15 degree angle (... -1 Quick note: Scientific inquiry doesn't "prove anything" but rather uses evidence to support hypotheses and theories. Interference shows that light behaves as a wave because of the fact that there are bands of High and low intensity. If you assume a plane polarized light wave traveling in some direction, lets say x, it can be described as a wave I_1 = ... 0 Yes, extremely low-energy "soft photons" exist and can have important effects even when their energy is too low to be directly detected. 2 You have the correct conclusion, and I think you have the correct analysis, but I don't fully understand your presentation. Think of it this way: If the displacement occurs only between "neutral" and "forward", then the average density (air molecules per volume, or spring coils per length) over the entire system (air chamber or spring) must increase. But ... 1 There is (in theory) no limit to the number of nodes N or anti-nodes A, except that these numbers cannot differ by more than 1. Also, standing waves in the pipe of length L can have other wavelengths \lambda besides those you have stated : for a pipe closed at both ends N=A+1 and \lambda=\frac{2L}{A} for a pipe closed at one end N=A and \... 2 The D'Alembert solution has a simple interpretation. Using your notation, it reads$$\phi(x, t) = \frac{g(x-t) + g(x+t)}{2} + \frac12 \int_{x-t}^{x+t} h(x') dx'. where $g(x)$ is the initial position, $h(x)$ is the initial velocity, and $v = 1$. The intuition is as follows: the first term above solves the wave equation with initial position $g(x)$ but zero ...

36

An animation is worth a million words:

11

This will be a purely mathematical treatment. It needs to be combined with some practical playing around to really "get" it. Traveling wave Let's start with the description of a harmonic traveling wave in one-dimension. Here "harmonic" just means the mathematical form of the wave is sinusiodal in both time and space. For concreteness we'll using talk ...

5

how is a standing wave related to the atomic orbit (It is my understanding that the atomic orbit is both a mathematical function that describes the probability of an electron being at a certain place, but it is also the image of this function in terms of real space, i.e. the actual 3 dimensional volume around the nucleus that a particular electron calls "...

2

A standing wave is basically two opposing waves of equal amplitude, as shown in the diagram below (where n is a positive integer): You can see this more clearly if you look at the top line where n=3, and follow it as it goes down, up, down. That's wave one. Then, if you look at the bottom line in the same case as it goes up, down, up, that is the opposing ...

1

The speed of sound is not constant. It is a function of the elastic modulus (or its closest analog in the material) and the density of the material. In gases, it noticeably depends on temperature. In solids, the speed of sound depends on the kind of wave; different kinds of wave have different elastic moduli. For purposes of estimating the speed of sound,...

1

Consider a box full of air in stady state (closed system). If the volume of that box is $22.4~\mathrm l$ there are $6.022\cdot10^{23}$ particles inside that have maxwellian distribution of speeds. Those speeds are randomly oriented and we can defocus from single-particle description to whole body desctiption supposing it is continuous and homogeneous. Then ...

0

Sorry all, but the impedance of free space is set by the choice of units. If you use natural or Planck units where $c$ and $\hbar$ and other constants are 1, the Planck impedance comes out to be about 30. More to the point, Planck units sets Coulomb's constant $\frac{1}{4 \pi \epsilon_0} = 1$. This represents a choice of units. To the point. If instead you ...

4

Do radio waves from the Sun reach Earth? Of course they do. It's just another form of electromagnetic radiation. If so, do they penetrate the atmosphere or are they reflected, absorbed, or scattered? That depends on frequency (or wavelength). The atmosphere reflects, absorbs, or scatters most incoming electromagnetic radiation. There's a window in ...

1

Offhand I'm not sure where to find information about how much is absorbed, reflected and scattered, but the waves certainly do reach Earth, and some, at least, penetrate the atmosphere and end up in solar radio observatory detectors, otherwise we wouldn't have so many of them.

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It does, but the effects are negligible in the regions we think about. If you think about a volume of air as a box of atoms bouncing around, you can apply an oscillating pressure gradient across that box and show that it behaves close enough to an ideal wave propagation medium that you can get away with using such an ideal model. The variations you are ...

0

I think the sound wave DOES spread out and dissipates over enough distance. Consider this: A small stick of dynamite explodes in a flat empty field. The compression wave of the air (which is what cuases sound) moves away from the source of the explosion at about 340m/s and propagates in a hemispherical manner outward. One second after the explosion, ...

1

I think this may be correct. The individual air molecules are indeed moving about randomly, and oscillating. When a large amount of energy is generated, these random motions are superposed by an alomost uniform behavior, making the propagation of sound possible. This is quite like the free electrons in a metal. Without a potrntial difference, the motion of ...

16

They do, its just usually negligible in practice. There's also scattering because the particles are not all the same (H2O, N2, O2, etc.)--but that, too, is usually negligible. Its mainly because there are so many particles in a single wave. Consider that the wave must be extremely short before it becomes noticeable (megahertz).

-3

The frequency can change ("still there is a frequency shift") while the wavelength remains unchanged ("the distances between subsequent pulses are not affected"): http://www.einstein-online.info/spotlights/doppler Albert Einstein Institute: "Here is an animation of the receiver moving towards the source: http://www.einstein-online.info/images/spotlights/...

1

Yes, two EM waves with the same wavelength would have the same frequency. Frequency and wavelength are inversely related, by the formula $\nu = \frac{c}{\lambda}$, where $\nu$ is frequency, $\lambda$ is wavelength, and $c$ is the speed of light. Therefore, if two waves have the same wavelength, they will have the same frequency. The intensity of the wave, ...

1

Let's first address the general question--do "waves" of any kind have (or can they have) inertia? I suppose here by "inertia" you mean "resistance to changes in velocity." This is certainly true of waves in, say, water--you've certainly felt resistance to your hand if you sweep it through water to make a wave; the destructive force of a tsunami is a more ...

-4

TDSE TISE Here you have the time dependent and independent Schrodinger wave equations, respectively. These relate to the energy of particles, but the trident symbol, Psi is representative of the actual wave equation I believe you are referring to. While De Broglie and schrodinger and others like them do describe particles as behaving like waves, they ...

2

My theory is that since water, and indeed all liquids, are incompressible, they both form at the same time. As the pebble hits the surface of the water and pushes it down, the surrounding liquid is pushed up (relative to the original surface level of the liquid), as you stated. Any displacement of a liquid must be accounted for.

0

No, they can't. In fact sound waves would lead you to Galilean relativity, and in all the cases where it differs from Einsteinian relativity they would lead you to the wrong result. This is a catastrophically bad model for Einsteinian relativity. As a simple example of where things go most horribly wrong consider a supersonic aircraft, and that the ...

1

No. In "STR" the speed of light is the same value no matter how the measurer is moving relative to anything else. For sound waves, the speed of the wave is measured relative to the medium through which the wave travels. Consider a sound source and receiver at rest relative to each other. The source and receiver have coordinated watches and have agreed on ...

4

This is a lot more subtle problem than is indicated in any of the comments. The problem is not just the issue of how the sum of non-causal signals can approximate a causal one, but how is it possible that while all real-life signals must start and stop at some time they must also be band-limited beyond some frequency, but as we know these two are ...

2

The Poynting vector is just an intensity vector, it is just $\vec S = \vec E \times \vec B$, one might need some factors of $c$ in there, I have been in $c = 1$ land too long â€¦. Due to the time dependence, the amplitudes of the electric fields are time dependent. If one would call that flickering, it would be the frequency of the light. I don't think that ...

6

I just wanted to add to a previous (very accurate) answer: you can think of it as an Fourier expansion of the actual (physical) wave profile. It is not a real life process, it is a mathematical approximation. The wave pulse can be thought of as a superposition of plane waves, which happens to interfere destructively in entire space, except for the localized ...

18

In this case it's probably best to be pragmatic. A pulse can be described as a superposition of sine waves that extend infinitely into space and time. But it's just that: a mathematical description that is useful for your purposes. There is not necessarily a physical meaning connected to it. Nevertheless, in quantum mechanics the wave-description of ...

2

I'd like to add to ACuriousMind's Answer. His/Her answer emphasizes that Maxwell's equations and other equations that give rise to the Huygens spherical wave kernel in their Green's functions are inherently acausal and we must force causality by hand through the appropriate boundary conditions. In antenna problems, we simply discard the advanced wave part ...

0

Addressing your final question: "I am not understanding how his reasoning actually averts the possibility of formation of back-waves. Can anyone help me visualise what he is talking? How does his reasoning maintain the unidirectional propagation of waves? Can anyone explain his argument?" where 'his' refers to Huygens: Refer to "Treatise On Light", ...

4

The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by E=h*nu, where nu is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy. This is where ...

2

Concerning massless particles, don't forget that the spacetime of their lightlike worldline is empty (= zero). That means that the point of emission and of absorption are adjacent in spacetime, even if the space interval between them measures billions of lightyears. By consequence, there is no problem for the transmission of particle characteristics for ...

1

They can and do, just not commercial phones. Commercial phones were built on a cellular architecture because it's possible, and it provides the greatest capability to support a lot lot of simulataneous traffic with an efficient use of spectrum. Direct wireless device communications is both simpler and more complex. Bluetooth is a simple way, but can not ...

-1

A photon is a measurement on a quantum field. It's a "one time deal", if you like. Each photon has an energy and a helicity (sometimes confused with "spin"), but that's not enough to produce a "wavelength", which is a property of a classical electromagnetic wave. We only recover the wave by measuring many photons, which then approximate the classical wave ...

2

He is referring to the second paragraph of 31-3: However, we shall discuss the formula we have obtained, in various possible circumstances. First of all, for most ordinary gases (for instance, for air, most colorless gases, hydrogen, helium, and so on) the natural frequencies of the electron oscillators correspond to ultraviolet light. These ...

4

The battery on your smart phone only provides approximately 3 W of power to the phone's transmitter. This is enough power to reach a nearby cell tower, but totally insufficient to send a signal for several hundred miles. Due to this, the nearest cell tower is used to relay your signal to another cell tower that is located close to the party that you are ...

0

Intensity is energy per unit area; over short distances, the intensity can be considered constant. In a time period dt, the sound wave travels a distance c.dt, so the total energy passing through an area A will be equal to the sound energy present in a volume cAdt. (think of water flowing through a pipe, if it travels at 10 m/s through a pipe with a 0.1 m ...

3

In this type of problem one has to take great care in defining intensities. In this case there are 4 different intensities: 1. $I_{ss}$, the intensity received by the static observer as perceived by himself, 2. $I_{ms}$, the intensity received by the moving observer as perceived by a static observer. 3 $I_sm$, the intensity received by the static observer as ...

3

The maximum and minimum are "local" values. As you move closer to A (at 0.2 m you are MUCH closer to A than to B) the amplitude of A is much larger - so although there may be destructive interference between A and B at that point, this is by no means perfect interference, and the resulting amplitude is still quite large (lot of A minus a little of B). ...

1

The frequency must remain constant to avoid a discontinuity at the boundary. The easiest way to see this is to consider 2 ropes of different linear densities - e.g. a thin rope and a thick rope - joined in series. If you shake one end at a frequency f, then (transverse) waves will travel along the joined ropes. The waves travel slower along the thicker ...

1

With respect to the air the wavelength will be $\lambda = 343/f$ everywhere. With respect to the ground the wavelength will be $\lambda = v_{sound}/f$ but $v_{sound}$ will depend on position with respect to the source - picture your classic Doppler shift diagram with a set of not quite concentric circles. Downwind from the source the wavelength, measured ...

2

To add to the existing answer, I think there is a nomenclature issue. When you say "bass" people understand "low frequencies" but what you probably mean is "beat". Rapid changes in amplitude, like a beat, carry a lot of high frequencies. You do hear mostly the beat from other peoples' headphones, ans it's annoying. You can think about the extreme case: the ...

1

Formal note: you use "," in the picture but here the numbers look better with "." so I will use ".". I think your $-0.628$ is close enough to $-0.73$. The difference may be because of $x'$ error. Notice that the upper plot crosses the $x$ axis at $0.0061$ and $-0.0039$ rather than $0.006$ and $-0.004$ respectively. It may be even $0.00615$ and $-0.00385$, ...

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