# Tag Info

0

Calling something a wave usually carries the connotation that there is some kind of periodic variation. However, this concept is not rigorously defined, as far as I'm aware. A wave can vary over space, over time, or both. For example, the wave $f(t,x)=\sin(x)$ varies in space but not in time, $f(t,x)=\sin(t)$ varies in time but not in space, and $f(t,x)=\sin(... 0 Normally an EM plane wave is taken as sinusoidal vector field in space time. But it is not required to have this form to solve the wave equation. An electric field$E_y = e^{-u^2}$where$u = k(x+/-ct+a)$also solves the wave equation: $$\frac {\partial^2 E_y }{\partial t^2} = c^2k^2(4u^2 - 2)e^{-u^2}$$ $$\frac {\partial^2 E_y }{\partial x^2} = k^2(4u^2 - ... 0 Remember, that when using rotating phasors to represent out of phase voltages in an AC circuit, only one component of the phasor has physical significance. It does not matter which you choose as long as you keep in mind that that is what you are doing. If working with phasors in the complex number plane, I would be more comfortable using the real components ... 0 Basically, the reason we use complex waves in quantum mechanics is because the mathematics is simplest when we do it that way. It is technically possible to formulate quantum mechanics using only real functions, but it is more complicated, and it gives exactly the same results (see the first page of Adler's book Quarternionic Quantum Field Theory for ... 1 There's a misunderstanding what the word "plane" represents in the term "plane wave". A plane wave is a wave in which the surface of constant phase (wavefront) is a plane: (image source) What is shown as a circular thing that rotates for e^{i\omega t} is the phasor that represents the value of the wavefunction at a given (single!) point of space. Phasors ... 0 Sound is really not about air molecules hitting obstacles. It's about the collective motion of the air molecules, which translates to pressure waves in the air. When the wave hits an obstacle, it reflects or is absorbed, but does not continue on its original path. Portions of the wave that bypass the obstacle do continue, but on the far side of the ... 0 The actual question seems unrelated to water: My question is, as the wave packet is superposition of many such waves of various wavelengths and what we actually see is the packet itself moving 'as a whole', modulating the component waves then how can we actually say some waves (smaller k) are hitting the coast earlier than the rest? From your formulation,... 0 Let me start by the meaning of the fast and the slow axes of the quarter half plate, and to do so let me compare it with a regular piece of glass. As usual let the speed of light in vacuum be c. Light travel in glass with a speed v that is less than c, we write v=c/n, with n is the refractive index and n>1 for glass. For a regular piece of ... 0 In that answer for elastic field in a solid, you can see that the density is multiplying the time derivative of the displacement:$$\rho\frac{\partial^2 u_x}{\partial t^2}$$The second derivatives of the displacements with respect to position are in the left side of the equations. So, the differential equations show that the wave velocity is proportional to ... 1 Your waveform would be a square wave with the lowest frequencies (those lower than the dish's cutoff frequency) missing. On an oscilloscope it would resemble a steady train of sharp spikes, each of very short duration. The frequency of those spikes would be equal to the harmonic series required to construct the square wave. You can simulate this scenario ... 1 Normally the carrier frequency is modulated (amplitude or frequency) by a much lower frequency carrying the data. If you decrease the ratio of data frequency relative to carrier frequency, you can reduce data error/incorrect signals. For low fundamental frequency carriers signals, this significantly reduces the data rate below what many find useful. ... 7 In a more realistic way you would draw wave beams instead of rays, like in the image below. And here is the same situation as an animation. (animated image from Wikimedia - File:Internal-reflection.gif) So you are right. The reflected wave and the incident wave indeed interfere with each other. Where the incident and reflected beams overlap, we get kind ... 0 If the incident beam had nonzero diameter, or were a plane wave, then they would interfere to form a standing wave. But rays are an idealization, which assumes a zero-diameter beam. In this approximation, the incident and reflected waves don't overlap (except at the exact point where they meet the surface) so there is no standing wave. 0 How can I solve this integral? You have the right idea but I think you missed the bounds of the integral. If you look at the typical solution to the wave equation, you will note that the limits of the integral are something like r \pm c \ t and the cross section is something like d\sigma = r \ dr \ \sin{\theta} \ d\theta d\phi. If you integrate r^{3} ... 0 The link below shows water diffraction better than the images you found. https://www.quora.com/What-is-single-slit-interference For water we have diffraction but there is no interference for the single slit, as you mentioned. In your images there is a slight pattern but that is likely due to the thickness causing extra reflections. Water waves and light ... 2 That is a lot of questions, but they are all strung together sensibly, so I will try to answer all of them. Yes, two point sources if they are mutually coherent and are stationary will produce an interference pattern where their beams overlap. If you shift the phase of one of the sources (e.g., by interposing a thin sheet of glass to retard one of the ... 5 TL;DR The "leading-order" correction to the ODE is$$\frac{d^2y}{dx^2}=\frac{\mu}{T_0}g(1+2k^2x^2)^3$$where T_0 is the tension in the middle of the rope, and k=\frac{4y_{max}}{L^2} for an estimated maximum displacement y_{max} and for a rope hanging between supports separated by a length L. The solution to this ODE is:$$y(x)=y_0+\frac{\mu g}{... -2 I believe your problem is that you want it to make sense. Pictures of it happening don't make sense of it. Mathematical descriptions of it happening don't make it make sense. You are asking for an explanation of something in terms of waves, and for the last 100+ years there has been no emphasis on looking for explanations. Still, you ask the question. You ... 1 One of the Maxwell equations in the vacuum has the magnetic and electric constants: $$\nabla \times \mathbf B = \mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t}$$ so that in the wave equation, derived from the above and the other three: $$\frac{\partial^2 \mathbf E}{\partial t^2} = \frac{1}{\mu_0 \epsilon_0}\frac{\partial^2 \mathbf E}{\partial x^... 1 Let's consider a more general version of the problem first. Suppose the distance between the emitter and receiver is d(t); we'll allow this to be an arbitrary function of time. We'll also suppose that the amplitude of the signal emitted as a function of time is S(t), again allowing it to be an arbitrary function of time. Suppose the signal has a speed c... 0 The answer is yes, as there is only one photon involved. 1 Depends how deep you want to understand it. Mathematicaly, even a vertically linearly polarized light can be described by 2 diagonally lineraly polarized light beams. In that sense, it was always there, just "cancelled out". The retarder retards one diagonal component but not the other. Quantumly... i struggle with it a bit but this video from 3blue1brown ... 1 An answer from an experimental physics physicist, who used theories to analyze data of high energy physics experiments: Fields are similar to a coordinate system on which the behavior of elementary particle interactions can be modeled with mathematical functions. In essence they replace the luminiferus aether with a Lorenz invariant "substance" that ... 2 Mathematically, you can define any dynamic law you'd like on a field. A dynamic law is simply a map \Phi^t(C) mapping a system configuration C - no matter what it is, could be a field, position/velocity pair, state of a game grid, etc. - to an "evolved" one after a lapse interval t that satisfies the composition principle \Phi^s \circ \Phi^t = \Phi^{s ... 0 I think that wave equations are related to pure fields. If we allow non equilibium situations in theory of elasticity, the general solution is an elastic wave. The static solutions (as a catenary for example) stay there as special cases. But, by general solution it doesn't means taking in consideration the air vibration and its effect on the string. This ... 2 Fields are not, in general, solutions of a wave equation. They are solutions of the equations of motion derived from a particular Lagrangian density. The dynamics of a field are described by a Lagrangian density which is specified as part of the foundations of the field theory. For the free scalar field \phi, this Lagrangian density is:$$\mathcal{L}=\... 0 I am not sure what your main issue here is. Any function describing the deflection of a string for$x = 0 \ldots L$can be assumed to be periodic as it can repeat itself for all other intervals, such as$x = 2L \ldots 3L$. Specifically, a plucked string at$x_p$has the form $$y(x,0) = Y_0 \begin{cases} \tfrac{x}{x_p} & 0 \leq x < x_p \\ 1-\... 1 Actually, you've gotten the issue backwards! You complain that the Fourier series is illegitimate because we don't have an "actual" periodic function, so we repeat the function by "brute force". But that's not the right way to look at it. The Fourier series properly represents functions defined on the circle, i.e. functions f(x) for x \in [0, a] with ... 0 We have the general solution$$y(x,t) = \sum_{n=0}^\infty sin(\frac{n\pi x}{a})(b_ncos(\frac{n\pi ct}{a})+c_nsin(\frac{n\pi ct}{a}))$$For a string constrained to y(0,t) = y(a,t) = 0. We can then take the derivative of this with respect to time;$$y_t(x,t) = \sum_{n=0}^\infty \frac{n\pi c}{a}sin(\frac{n\pi x}{a})(-b_nsin(\frac{n\pi ct}{a})+c_ncos(\frac{n\... 0 For any aperiodic function$f(x)$defined on a finite interval$[0,a]$, we can compute the Fourier series of its periodic extension over the real numbers: $$f_p(x)=f(x)\mod{a}$$ Unlike$f$,$f_p$is periodic, and therefore there's no problem in constructing the Fourier transform. 0 Snell's law is always valid for a translation invarant interface. A short pulse contains many frequencies and a narrow pulse many directions. This can be considered uncetainty. As materials are usually dispersive, refraction will change the pulse shape. 0 I assume you are referring to the fact that the refractive index depends on wavelength (this is called dispersion). The answer is that the spot would spread out, with different wavelengths going in different directions. 1$f$is meant to be any arbitrary function$f:\mathbb{R} \to \mathbb{R}$. This can be understood from the wave equation $$\frac{\partial^2 y(x,t)}{\partial t^2}=v^2\frac{\partial^2 y(x,t)}{\partial x^2}$$ which describes the underlying physics of a wave. The most general solution of this differential equation is $$y(x,t)=f_1(x+vt)+f_2(x-vt)$$ where$f_1$... 1 Writing your second equation using the same symbols of the first equation$(s\rightarrow d)$: $$w = \frac{m \lambda D}{d}.$$ Then, from geometry,$w = D \tan\Theta$, so you end up with: $$d\tan\Theta = m\lambda$$ for the double slit diffraction pattern. Which, as you are saying, is different from the general diffraction grating formula: $$d\sin\Theta = ... 0 For the level you seem to be aiming for, my advice is to stick with binary observables instead of continuous ones. So here's a good exercise for your students: Let A=(1,0) and let B=(0,1) Take an observable O_1 with eigenstates A,B. Take an observable O_2 with eigenstates U,V where U is proportional to A+B and V is proportional to -A+... 3 I think the Heisenberg uncertainty principle, or at least the simple wave mechanics version of it (which may be the only one your students are equipped to understand at this point), really stems from the idea that most wavepackets don't have a definite frequency and wavelength. That is, in p = h/\lambda, there is not just one \lambda. Maybe this concept ... 0 There is a nice paper on this topic: Newton et al., "Predicting the playing frequencies of brass instruments," September 2014, Conference: Forum Acusticum, Krakow. Brass instruments such as trumpets and trombones are known in acoustics as "lip reeds." Helmholtz defined two types of reeds, inward-striking and outward-striking. In the idealized case, one ... 1 The note produced by vibrating air in the instrument. Blowing air over the player's lips is what sets up the vibration. You can do this without an instrument. The instrument has a resonance frequency. Vibrations at that frequency get reinforced. The oscillating pressure acts on the lips and encourages them to vibrate at the resonance frequency. This makes ... 1 The wave from an impulsive point source is the Green's function G(\vec{r},t) for the inhomogenous wave equation. In other words, if we wish to solve the equation$$ - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi = \rho(\vec{r},t), $$we can do so by solving the related equation$$ - \frac{1}{c^2} \frac{\partial^2 G}{\partial t^2} + \... 0 The width of the slits has nothing to do with the separation of the fringes. Doubling the width of each of the two slits leaves the separation of the fringes the same but each fringe will now have twice a much light forming it. So the answer is$2I. 0 I would proceed as follows \begin{align} \hat{k} =& k_x\hat{x} + k_y\hat{y} + k_z\hat{z}\\ \hat{e} =& e_x\hat{x} + e_y\hat{y} + e_z\hat{z} \end{align} \begin{align} \hat{e}\cdot\hat{e} =& e_x^2+e_y^2+e_z^2 = 1 \\ \hat{k}\cdot\hat{e} =& k_xe_x+k_y e_y+k_z e_z = 0 \end{align} We are always free to arbitrarily choosee_y=0$for one of the ... 0 We have, for a free space plane wave, that$\hat{\boldsymbol{e}} \cdot \hat{\boldsymbol{k}} = 0$. This tells us that$\hat{\boldsymbol{e}}$lies in the plane perpendicular to$\hat{\boldsymbol{k}}$. You are correct that in a fully general situation there is nothing else which determines the direction for$\hat{\boldsymbol{e}}$within this plane. However, ... 1 Now if we think that sound wave is like an object and use relative motion than sound will approach wall with speed$v + v_{s}$... No, I think you are misunderstanding something here. A linear sound wave will always propagate at the speed of sound once emitted in a homogeneous, uniform medium. If your expression were correct, how could a shock wave form? ... 0 As M Enns, said white light is a mixture of individual photons with different frequencies. There is no overall frequency or color for white light. The only reason we see white light is because our eyes have evolved to receive individual photons of different frequencies. We have different receptors which interact with different frequencies. As this video ... 2 It seems from the way you have phrased your question that you think the frequency of the light corresponds to the number of photons per second. That's not the way it works. The number of photons per seconds would correspond to the brightness of the light for example you could have relatively few blue photons per second in the case of dim blue light or many ... 0 As you already mentioned, the polarization is perpendicular to the wavevector, this is expressed mathematically as$\mathbf k \cdot \mathbf{\hat e} = 0$and specifies a plane that your polarization vector can lie in. To start, you can assume that$\mathbf{k} = k\mathbf{\hat z}$so that the polarization lies in the$x,y$plane. Then on can parameterize the ... 2 The first boxed equation refers specifically to the ratio of the magnitudes of electric to magnetic fields in a electromagnetic wave propagating through a vacuum. The second equation refers to the velocity at which a particle must travel for it to experience no force given a specific electric to magnetic field ratio. If you consider uniform fields (ignoring ... 1 Every point on the wavefront may be considered a source of secondary spherical wavelets which spread out in the forward direction at the speed of light. The new wavefront is the tangential surface to all these secondary wavelets. The word new is the key to your question. Huygens' Principle explains wave propagation in steps: 1) decompose to original ... 0 I don't think that it will be a transverse wave. A wave is considered transverse if the displacement vector is orthogonal to the wave vector. If we consider the vertical ($z$) direction to be the direction for deflection and the horizontal ($x\$) direction the direction for extension, we have the following for Euler-Bernoulli Beams. \begin{align} &u_x = ...

1

Radio waves and microwaves are used for long range communication. Micro waves are used for the satellites which are in distant orbits and radio waves are used for low orbital satellites. They are generally regarded as low frequency waves. The low frequency waves are used for such communication because they scatter less in atmosphere as compared to high ...

Top 50 recent answers are included