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No the time taken does not depend of the velocity attained by the first ball(if they are ideally rigid) it rather depends on the elasticity or rigidity of the balls. So for ideally rigid bodies, the time taken to transfer approaches 0. Nothing would happen with an increase in distance between the two balls. See: Is the reaction force for a stone hitting a ...

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I think there is a simpler approach: When you have pulse with velocity $v$, the value of the function has to be constant along lines given by equation $x = x_0 + vt$. In other words: $$f(x_0 + vt,t) = \rm{const}$$ Before we proceed further, we notice that the argument of exp can be simplified like this: $$f(x,t) = A \exp \biggl[-\biggl( \frac{ax-bt}{c} ... 1 Let the wavefunction be$$y(x,t)= A\sin\left[\frac{2\pi}{\lambda}(x- vt)\right].$$Now,$$\frac{\partial y}{\partial x}= \text{Rate of change of wavefunction when time is constant}\;,\\ \frac{\partial y}{\partial t}=\text{Rate of change of wavefunction when position is constant or transverse velocity}\; .$$Now differentiate y w.r.t. x keeping t ... 0 https://www.youtube.com/watch?v=YfYPJZCSI-E&index=9&list=PLq1e8D3Q-o-W-QrxTK4UIiJGyMKEsaBjo this should help you, as the cavity collapse the 2 fronts collides and form a jet. depending on the length of the cavity form, they might be 2 jets formed. The only question now is how does the speed relates to the height of the jet? 0 You can apply the Huygen's construction to any distance over which the local environment is optically homogeneous. Encountering a reflective or refractive surface is encountering a non-homogeneity, so you can't draw big circles that include a mirror or a lens. You can modify the Huygen's principle so that you can use near such boundaries, but the ... 0 It sounds to me like you already understand how a standing wave is formed. For each of the contributing plane waves, the relationship that the perpendicular components of the E-field and B-fields are in phase and of equal magnitude (for CGS units and vacuum) is true. Maxwell's equations are linear. Any solutions to this equation can be superposed to make ... 2 Sound is a compression wave not a shear wave, so the viscosity of the liquid has no (direct) effect on it. The speed of sound in a liquid is given by:$$ v = \sqrt{\frac{K}{\rho}} $$where K is the bulk modulus and \rho is the density. The viscosity does not appear in this equation. A quick Google found data on the speed of sound in liquid helium in ... 1 Huygens' principle of taking the convex hull of the spherical waves emitted by all points on a wavefront indeed gives you two new wavefronts: One "forward" and one "backward". Those are both meaningful: The "forward" wavefront corresponds to the retarded solution of the wave equation, the "backward" wavefront corresponds to the advanced solution of the wave ... 1 By denser, I assume you mean with a larger refractive index. The answer is established from the Fresnel equations giving the ratio of the electric field amplitudes. The reflection amplitudes for (for example) light travelling from glass to air can be either positive, negative or complex depending on the angle of incidence and the polarisation state of the ... 3 First, some fraction of incident sound power will pass thru any object, human body or not, so your title makes no sense. The real question is how much this body will attenuate the frequencies of interest. Below some attenuation, you either don't care or it's below the noise floor of the sensors to detect it. However, this depends on what you care about, ... 0 Why doesn't a backward wavefront exist? It exists. See the attached picture. Source -1 Both diffraction and interference occur in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs. This is what give the diffraction pattern. I ... 0 Diffraction is the phenomenon of the change of the movement from the straight line (in a flat, not curved space) in the cases, that it is not a reflection. For the expression "change of the movement from the straight line" it would be better to say "deflection", but this seems not to be so ok because, due to Wikipedia it could be misunderstood in every day ... 0 The lengths of the openings do not need to be equal or less than the wavelength for there to be a fringe pattern. For instance in a double slit experiment with 500nm wavelength light and slits that are 100,000nm wide, separated center to center by 200,000nm you will get (interference) or a fringe pattern with 2,500,000nm spacing's. If you change the ... 2 If the WiFi antenna is emitting at 2.4 GHz, you could detect a slight improvement of the signal, but unless the door is solid and very thick, I doubt it will make much difference. If it is emitting at 5 GHz the improvement could be bigger, as the wavelength is reduced and the door appears "bigger" to the electromagnetic wave. Finally, if your antenna uses ... 1 Interference and diffraction are the same thing. In fact so is refraction. The propagation of light is conveniently described using the Huygens-Fresnel principle. The amplitude of the EM wave at a point is calculated by summing up the amplitudes of all the EM waves reaching that point, taking the relative phases into account. This describes the phenomena we ... 0 The open boundary condition means, as stated in the question, that at the boundary no force acts on the end of the string in the direction of elongation. As the tip of the string has infinitesimal mass, we can argue as if we were considering conditions for static equilibrium (if the forces caused by the string would differ from the forces caused by the wall ... 0 It's actually not that complicated for a linear case. Let's derive the 1D wave equation for velocity potential \Phi (v = \partial_x \Phi). As usual, c_0 denotes the speed of sound. The kinetic energy density is obviously$$ \mathcal{T} = \frac{1}{2}\rho_0\mathcal{v}^2 = \frac{1}{2}\rho_0\left(\frac{\partial \Phi}{\partial x}\right)^2 $$Potential ... 1 The energy is reflected due to the discontinuity of the string mechanical impedance. Therefore it "can't be used as a part of the pulse" because it never gets to the denser rope. It's actually a special case of very general principle: whenever there is a discontinuity in propagation medium, energy reflection occurs. That's the same in optics, when you are ... 3 The main issue is that one cannot use a non-relativistic dispersion formula E=\frac{p^2}{2m} to deduce what happens at relativistic velocities. This Phys.SE post deals with the same theme. 2 You typically have one position and one velocity variable per oscillator. The equation of motion of mass i is m_ix_i''=k_ix_i+coupling terms. If the forces from the coupling terms are small, the frequencies do not change much. If the coupling is large, you will have as many modes as oscillators, but the frequencies can be anything. You wind up finding ... 1 Try to think the rope made up of small solid balls connected with springs. When you make the bump as shown in your picture the springs are expanded. Now you let go of it. The rising ball applies force upward to the ball on its right. The already expanded springs soon tend to decompress again. In doing so the the falling ball applies a downward force to ball ... 1 Each wave source has an initial phase.When a wave is incident on a surface some or all part of it is is reflected off the surface.It so happens that the medium changes the phase difference by some quantity. For instance if you add any quantity to the graph of a particular function(you may want to google it) the graph tends to shift backward or forward as per ... 1 The short answer is that it doesn't matter. Whichever direction you choose to define the reflected E-field initially you will always gets its direction with respect to the incident E-field coming out the same. In the case you have in your question (where you have reversed the sign of E_R and defined it to be in the -x direction and opposite to the ... 0 In case someone else have this question, I finally found that Goodman (Introduction to Fourier Optics, ISBN 9780974707723) explicitly states that the Fresnel approximation is indeed valid in the far-field. 0 Your laser cavity is a Fabry-Pérot interferometer. The free spectral range tells you how close two neighboring laser modes can get: \Delta \nu=\frac{c}{2nl} (for a linear resonator, length l, refractive index n). The resolution of your spectrometer needs to be smaller than this free spectral range. You can increase the free spectral range by either ... 0 The sentence can be proved by Fermat's principle and it is not relevant to wave nature of light. Waves have some properties like velocity (and could be different in different media), phase and the important property called interaction. When we say light is a wave, we must check the above properties in it. And the first experiment that showed that, light is ... 0 If the impedances are complex then it means you have dissipative terms. For example, if the problem was normal incidence from vacuum into a conductor, then energy conservation is not as simple as saying the (magnitude of the) Poynting vector of the incident wave equals the sum of the Poynting vectors in the transmitted and reflected waves. In a conductor, ... 0 You only expect transmitted power to be equal to the incident power minus the reflected power when no energy is supplied to charges (which can then lose energy to heat). The Poynting vector does not have a divergence free energy flow, it can gain or lose energy by getting or supplying energy to charges. And the charges can give or get energy from the fields ... 2 This is the wave equation in Euclidean space (or more simply, the Laplace equation). It shows up when you perform a transformation into Euclidean space, the Wick rotation $$t \rightarrow it$$ It is often used in quantum theory to resolve problems of convergence. While you are not guaranteed that you will find the same results ... 1 I have never seen an application of the equation before, but an informal way to see what it represents is to make a change of variables: x=ix', where i is the imaginary unity. Then your equations becomes (\partial_t^2 - \nabla^2) \psi'(x') = 0. This is a wave equations. A simple solution is the planar wave \exp(i(kx'-wt)=\exp(-kx) \exp(-iwt), which is a ... 0 A pulse refers to a disturbance that travel from one location to another location through a medium. While, A wave refers to the disturbance or variation that travels through the medium. 2 Good theoretical answer is that it results from linear acoustical wave equation and its presuppositions. It is therefore good approximation whenever linear acoustics still can describe the wave propagation (that would by e.g. 90% of room acoustics practical examples). Typical examples of problematic models are large-amplitude events (e.g. a shockwave after ... 1 The spectrum of various resonant tube arrangements (half-open, fully-open fully closed) is something that can be measured in a very basic laboratory and gives solid evidence that the claim is true over the kinds of frequencies that are accessible in such a lab. Say a few hundred to a few thousand hertz. 3 It's a bit pat maybe, but if the wavelength-dependence were detectable over human distance scales, the (quality of) sound (not just the volume) of music, speech etc. would depend noticeably on how far one were from the source. 0 Actually the pendulums swing independently, but one can consider all of them together as a Hamiltonian system. The movie illustrates the Poincare recurrence theorem. It also illustrates quasi-periodic motion on the Liouville torus in a completely integrable Hamiltonian system 0 I too had the same problem, so I started constructing EM wave just as Maxwell did.To get a EM wave he combined two simple laws. When an electron moves with velocity v, some magnetic field will be associated around it. As an electron moves, the electric field associated with it also moves; this results in a change in the electric field in space around the ... 1 The pure simple harmonic motion is in real life very very rare. There are some cases which are really close (e.g. for engineering purposes). That might be: Small-amplitude oscillation of a mass on a spring (small enough for spring nonlinearities not to be pronounced) or other kinds of these simple or moreless model oscillators. Tuning fork. Strictly ... 2 For your example of a violin string, you can immediately determine that it is not simple harmonic motion by listening to it. Simple harmonic motion is a pure tone of a single frequency. Violins don't sound like that so you immediately know there are harmonics and it therefore is not a simple harmonic oscillator. As some other people have mentioned, a tuning ... 4 This is, unfortunately, not a simple task in general. My experience on non-reflecting boundary conditions is for the Navier-Stokes equations, but you should be able to do a similar approach for your system. As you noted, a fixed boundary (u=0) will lead to one type of wave while a free boundary (\partial u/\partial x = 0) leads to another wave. What you ... 1 As Kevin Zhou pointed out in his comment, behind every edge light will be distributed in fringes. As long as one expose curved edges or with light from a point like source or with light from parallel rays there will appear an detectable intensity distribution behind edges. Using monochromatic light and a point like source will give the best results. The ... -3 Let's take this one step at a time. Bear with me. Every wave is characterised by some periodically changing disturbance. For example, that entity is air pressure for sound waves It's air. When an ocean wave moves through the sea, the sea waves. When a seismic wave moves through the ground, the ground waves. When a sound wave moves through the air, the ... 1 The oscillating quantity in matter wave is probability-amplitude, a complex number . Suppose the electron is in state |\psi\rangle. The wavefunction of finding the electron at any coordinate x is given by \langle x|\psi\rangle= Ae^{-iEt/\hbar}\cdot e^{ipx/\hbar}. Probability of finding the electron around x is given by$$P(x,x+dx)= |\psi(x)|^2 dx= ...

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There are no matter waves. In quantum mechanics the wave equation that describes the measurable observables gives wave functions , i.e. complex sinusoidally varying mathematical functions; the complex conjugate square of these functions gives the probability of finding a particle of mass m in the location $(x,y,z)$ at time $t.$ When the experiment is done, ...

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The answer is yes. There are hundreds of facilities all over the world called synchrotron radiation sources where electromagnetic radiation with different wavelengths (ranging from IR to hard X-rays) is produced by circulating electric currents. However, I would not call them antennas.

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So yes, if you compute the Poynting vector (energy flux density), $\vec E \times \vec H$, for an exponentially decaying evanescent wave, you indeed find zero time-averaged energy flux perpendicular to the reflecting plane. Ask you say, this leads to a conundrum --- how do evanescent waves transfer energy across barriers? For sure, we know they can transfer ...

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Yes, any object will have a wave-like nature, this is very interesting and perfectly real. The thing and the reason why we don't see the macroscopic world around us acts a little like the quantum world is that nothing as a small enough momentum. $h$ you know is very very small about $10^{-33} kg \space m^2 /s$ so to be able to observe this wave-like nature ...

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as usual for this kind of question: an infinite number, but their fading intensity will make them un-noticable after a while. see http://www.wikiwaves.org/Ship_Kelvin_Wake and https://en.wikipedia.org/wiki/Wake . As illustrated in these images, the pattern of emission is a repeated curvy triangle, which front corresponds to long wavelengths on the rear and ...

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