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No. For one, you can have frequency lower than one. Think of Sun's frequency in the skies, it crosses the sky once per day, not one per second. On the other hand, it crosses the sky in different place every day during the year, until it gets back to where it was today in one year. So in this regard the fundamental frequency is really once per year.

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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 ...

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This is what the situation looks like: The two radio transmitters behave like a Young's slits experiment, so they will produce a diffraction pattern in the plane of the car's motion. As the car drives north it drives through the diffraction pattern. You have to work out what the diffraction pattern will look like given the separation of the transmitters, ...

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Since it's the momentum of the wave that you are after, here's a good way to estimate its mass: On the open ocean: Waves are approximately sinusoidal in shape. Take a trough - the lowest point for a wave - as the base. Estimate the height of the wave and perform an integral to estimate the cross-sectional area of a wave. Then multiply this by the ...

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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 ...

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The collision of the blocks will cause them to vibrate (sound within the blocks). Since the blocks only contacted with each other momentarily (the blocks rebounded), there would have been a very short period of time for the vibrations to transfer from one block to the other (when the 2 blocks are touching). Since the vibrations are travelling fast (the ...

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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 ...

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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 ...

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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

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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 ...

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If you look at a light wave as a rotating x and y axis which propagates forward in the z direction, the equation which might result takes the appearance of a screw [or helix]. The equation of the wave is not only a function of time, but also in z. y = A e^ (i( B*z + w*t )) , i is the square root of -1 Note an equation of a helix ...

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For larger objects the radio wave gets reflected. Compare this to a water wave hitting a wall. For smaller objects the radio gets diffracted. Compare this to a stick placed in the path of water wave. This stick bends the water wave which is similar to diffraction A light wave consists of larger number of smaller waves. A mountain reflects most amount of ...

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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 ...

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I was just thinking some spherical light source. Like, a light bulb? – Anthony Jul 22 '13 at 3:32 For a light bulb there cannot be infinite strength because the power is given by I*V and the bulb has resistance, so the current is limited. The resistance that is heated to give off light will not be giving off light at r=0, for whatever topology you ...

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The magnetic field polarizes orthogonal to the electric field in free space. We generally only talk about the electric field because Maxwell's equations define a one to one relationship between the two. It would make just as much sense to only talk about the magnetic field. We choose the electric field because, in general, when light interacts with matter ...

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The magnetic field does not vanish when light is polarized. A changing electric field induces a magnetic field, and a changing magnetic field induces an electric field. This is why, in the propagation of an electromagnetic wave, there is always an oscillating electric field coupled with a magnetic field oscillating perpendicular to this electric field. You ...

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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 ...

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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. ...

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The fact that the wave equation is ubiquitous in physics does not mean that the derivation of it is the same for each physical situation. I'll show you how to derive the wave equation of electrodynamics since it is pretty elegant and point you to some places to look at the derivation for other physical situations. For electromagnetic waves in vacuum, start ...

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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 ...

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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 ...

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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. ...

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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 ...

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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 ...

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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 ...

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Wow, this is a very detailed question. Thanks for your effort. Lets ignore diffraction effects, which will scatter some small amount of extra power out of the laser beam. The loss at elements 1a to 1d will not simply sum up. This is because the power will only be lost at one of the mirrors, and will not be there to be lost at the other mirrors. So, to ...

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Just to add to Anna v's answer and Aanel's answer which are both admirably pithy and correct. Polarisers making use of the Brewster angle deflect the light of the "blocked" polarisation, rather than absorbing it. The "blocked" light actually passes into a refracting, the unblocked light is reflected off and redirected to the output. Polarising beamsplitters ...

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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 ...

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The wavevector does not always point to the direction of propagation. Read this from wikipedia to understand it better. The wavevector is basically parallel to the direction of propagation of the wave if the medium is isotropic (i.e uniform in every direction).

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Mathematica gives $$z=\gamma\frac{\sqrt{\gamma^2-a^2-4(y^2+L^2)}}{2\sqrt{\gamma^2-a^2}},$$ where $$\gamma=\frac{(2n+1)\pi}{k}.$$

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It is fringe width ,the distance between centers of adjacent secondary maxima's or minima's and it is $\lambda d/D$.This is what question expects from you here to use.But instead I think you are confused because actual width of fringe looks only half of above.But note it is varying intensity of light there so you choose only lines as fringe centers and just ...

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Light that is not transmitted is either absorbed or reflected. Wire grid polarizers tend to reflect. Polarization beam splitters separate the two polarizations in different directions. Polymer based ones absorb it i believe...

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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.

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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'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 ...

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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 ... 0 In phase simply means that the particle are moving at the same direction at all times. For a standing wave, this is achieved only if they have the same amplitude at all times. Using parthvader's animation below, if you take any of the two black dots, that represent a particle, you can see that all the points that move in parallel to that point are all half ... 0 A standing wave is formed when two identical waves travelling in opposite directions meet and superimpose on each other. This usually happens when one wave is the reflection of the other. Now, to better understand this, see this animation, It is pretty apparent that the black particle that doesn't appear to be moving will be in phase with the particle ... 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 ... 0 Air resistance acts only on macroscopic objects travelling through air. Electric fields are not affected by this at all. Electromagnetic waves do not even require a medium to propagate, so there is no point in even discussing how they can propagate through air. I think you are confusing Permittivity of air with air resistance, as anna mentioned in the ... 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 ... 0 A is incorrect because a node, rather than an antinode, would exist at the reflection surface. C is incorrect as the oscillations would be out of phase. D is incorrect because sound is generally a longitudinal wave, although I would say the diagram is confusing on this point. 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 be$$I_{\textrm{out}} = I_{\textrm{in}}\cos(\theta)^2$$where \theta is the angle between the polarizations of the ingoing light and the pane, ... 0 To answer the "then what" part of the question, note that d (2\pi/\lambda)=\frac{-2\pi}{\lambda^2} d \lambda. Then (2π/λ)*dv/d(2π/λ) = (2π/λ) (\frac{-2\pi}{\lambda^2})^{-1} dv/d \lambda = -\lambda dv/d \lambda . This gives you your answer. 0 Start with:$$ u = k \frac{dv}{dk} + v $$and use the chain rule:$$ \frac{dv}{dk} = \frac{dv}{d\lambda} \frac{d\lambda}{dk} \tag{1} $$And \lambda = 2\pi/k, so:$$ \frac{d\lambda}{dk} = -\frac{2\pi}{k^2} $$Substituting this into equation (1) and multiplying by k we get:$$ k \frac{dv}{dk} = - \frac{dv}{d\lambda} \frac{2\pi}{k} = - ...

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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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The answers so far have been very good, but I will attempt to answer my own question, building on what has already been said. Lighthill and Ffowcs Williams make use of the term pseudo-sound to denote sound which can physically be observed by say a microphone, but which will not propagate into the homogeneous quiescent region. The example of a turbulent jet ...

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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 ...

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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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