# Tag Info

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It can be done, but there's some trade offs. Larger speakers are better at moving longer wavelength (low frequency) waves. When you try to combine a bunch of small surfaces in different locations to recreate a single wave you end up with a some random interference where the wave is stronger or weaker (in 3d-space) (see phased-array antenna for some ...

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There are a few reasons I can think of: (1) The second order system is that it is time-reversible. If you let $t\to-t$, you get $$\frac{\partial^2f}{\partial(-t)^2}=\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2}$$ whereas the first order system has \frac{\partial f}{\partial(-t)}=-\frac{\partial f}{\partial t}=\pm v\frac{\partial ... 6 There has to be a few assumptions. Let's assume we are talking about a linear plane wave in relatively deep water. Because the the case where the bottom comes into play the upward hydrostatic force distorts the wave. Picking deep water or insuring the relative depth of d to L (d is average water depth and L is the wavelength of the wave) is d/L > ... 6 There is a fundamental theorem already conjectured by von Neumann but proved just at the end of 1900 by Solèr (in addition to a partial result already obtained by by Piron in the sixties) which establishes (relying on the theory of orthomodular lattices and projective geometry) that the general phenomenology of Quantum Mechanics can be described only by ... 5 This Washington State Department of Transportation page makes it clear that the choppiness is at the very least highly correlated with windstorms and high winds. This page is a good resource as well. The choppiness occurs on the upwind side of the bridge; thus, a south wind (which blows north) will make the south side choppy. The reason for this is that ... 5 There's nothing wrong with the first order wave equation mathematically, but it's just a little boring. If you want to use this equation to describe waves, it basically amounts to having a 1d solid with speed of sound v for left moving waves (say) and speed of sound 0 for right moving waves. It wouldn't surprise me if such a thing could be constructed ... 4 This is a neat question. Did you know that adding two Sine waves of the same frequency but different phase together always produces another Sine wave? Of course you can imagine two perfectly out-of-phase Sine waves that "cancel" by adding to a line but in that case you can just imagine the result as a Sine wave with 0 amplitude. Using gnuplot with the ... 4 Actually I have read (although I can't find a reference) that the subjectively perceived psychological notion of pitch itself, although very nearly wholly set by the sound wave's frequencies, is also weakly dependent on the intensity of the sound: that is, a higher intensity sound wave does seem ever so slightly sharper (higher in subjective pitch) than one ... 4 Are there experiments that could show that light waves resemble more say square waves than sine waves? Are there experiments that could show that light waves resemble more say square waves than sine waves? Temporarily looking at your example of a square wave, a square wave of spatial wavenumber k can be represented in a Fourier expansion as a ... 4 The definition of fundamental frequency should be: the lowest frequency of a periodic wave satisfying some boundary conditions. For example, in the case of the vibrating string: The lowest frequency is determined by the length of the string (top of the image), the tension in the string, and the mass per unit length. Of course, if there are no boundary ... 3 In general light that does not pass a barrier, a wall for example, is absorbed.The energy is turned mainly into heat and also chemical bond breaking etc. The part of the light beam that does not have the correct polarization for the polaroid will be absorbed in the same way. 3 It's a matter of convention. The complete wave function must describe a wave that propagates in the correct direction. Any function of the form f(kz - \omega t) = F(z - vt) describes a wave propagating to the right. For a plane wave, that could be \exp{(i(kz - \omega t))} or \exp{ (-i(kz - \omega t))}. An author is free to choose whichever he or ... 3 Like electromagnetic waves, mechanical waves, and (in fact) everything travels in a straight line until something acts on it causing it to stop going straight or until the medium it's travelling in changes. Light will bend due to gravity, refraction, reflection. Without an outside influence and in the same medium, everything travels in a straight line. ... 2 Yes, the waves do interfere when they are not parallel. In fact, there's no such thing as parallel waves: their direction is always fuzzy. See this image of one wave going through an aperture with size comparable to wavelength: You can see that it goes not only upward, but also expands to the left and right. Now let's add another source such that ... 2 Put simply, yes waves interfere even if they are not directly aligned. In fact all waves interfere of a given type. Interference is really just a re-statement of the superposition principle; that is, given 2 waves, the resulting pattern is simply the sum of the two waves at all positions in the space. The first figure you provide shows how, in 1D, 2 waves ... 2 The waves in the first picture are not necessarily parallel. Their amplitudes are simply projected onto a screen, which makes them appear to propagate only in the x direction, while really they could be propagating in any direction at all (still assuming they are confined to a plane, so no propagation in the 'z' direction. Without loss of generality we ... 2 Chris White's answer using an analogy to incoherent light pretty much answers the question; it's fundamentally a question of the statistics of how wave sources add. Here's a slightly different but equivalent rephrasing of Chris White's answer using matrices: Given N wave sources, incoherent waves add "diagonally" (I\propto N), ie, additively. ... 2 I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form ... 2 What does an aperture do? It "applies" Huygens principle to every point within the aperture, and ignores those outside the aperture because they are blocked. There are a couple of things going on when you consider a lens. Let's make sure we understand them. An aperture produces a diffraction pattern in the space of diffraction angles. Recall from the ... 2 Hydrodynamic perturbations = change in pressure due to a flow velocity (particles don't return to equilibrium positions). Acoustic perturbations = change in pressure due to the fact the particles undergo an elastic restoring force (for a compressible fluid) which causes perturbations to travel at the speed of sound. Any change in the pressure/velocity ... 2 The axes definitely matter. If you put light through a linearly polarized glass pane, the output light will be entirely polarized along the polarization axis of that pane. The intensity of the output light will beI_{\textrm{out}} = I_{\textrm{in}}\cos(\theta)^2$$where \theta is the angle between the polarizations of the ingoing light and the pane, ... 2 Electromagnetic waves don't always travel in straight lines. They can bend when they encounter a changing wavespeed at non-normal incidence, this is how a lens works. Mechanical waves act in the same way. They travel in a straight line until they encounter a change in the material parameters (such as a change in density) which changes the speed of the ... 2 The question is not very clear. Here is the experimental confirmation of the electron-probability-wave. Electron build up over time The concept of probability can be measured only statistically. This is true for classical probabilities as well as quantum mechanical ones. The sequantial experiments, one electron at a time, show that an electron ... 2 The deformation wave will travel in both directions - there's no way for it to "know" the shortest path. And the resulting set of vibrations will interfere with each other in interesting ways, causing complicated resonances. So, let's look at a simpler example: just a thin, large torus. We'll look at two points 90° apart from one another; one at \theta = ... 1 It is true that the deformation wave travels at the speed of sound, but you have to get away from thinking of objects as rigid if you ask about deformation waves. One good image is striking a foam ball with your fist. The overall shape of the ball will change as the ball wraps around your fist. Some of the deformation will travel across the diameter of ... 1 So, my best understanding: The basic solution to the wave equation is$$\Psi(x,t)=Ae^{ikx-i\omega t}$$Where the signs are arbitrary. If you combine this with the good old Euler Formula this expands to$$\Psi(x,t)=A\cos(kx-\omega t)+B\sin(kx-\omega t)$$Where the imaginary part is absorbed into that B 1 In the case of a pebble falling from some height into water, I believe surface tension will be altogether negligible. You should calculate the Weber number to check this, http://en.wikipedia.org/wiki/Weber_number The dominant effect will be the pressure generated by the displacement of water by the pebble entering. Again, I am not sure that the viscous drag ... 1 I think your equation is slightly incorrect, should be:$$\frac{dv}{dk} = \frac{1}{k}\frac{d\omega}{dk} - \frac{\omega}{k^2} This is because $\omega = \omega(k)$, $\omega$ is a function of $k$, so you need to apply the chain rule to evaluate the derivative. The dispersion relation may also be a topic of interest.

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That's an interesting link, explaining how if fluid contacts a plate, and if there is a vibration pattern in the plate, what vibration pattern you get in the fluid. As I read it, if the speed of sound in the plate is very high compared to the fluid, the fluid sees a plate that vibrates into and away from the fluid, creating a sound wave that propagates ...

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Try differentiating twice just as you said. Then to combine them, you can do a few different things: Visually compare the two expressions, and note that they're identical except for a multiplicative constant, or divide the two expressions and see if you actually do get $v^2$ or $1/v^2$, or multiply your expression for $\partial^2 y/\partial t^2$ by ...

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