# Tag Info

## New answers tagged waves

7

What you have calculated is the phase velocity, $v_p$, of the de Broglie wave associated with the particle. The phase velocity can be greater than $c$, and indeed it is always greater than $c$. The velocity of the particle is the group velocity, $v_g$, and as you have demonstrated the two are linked by: $$v_p v_g = c^2$$ The group velocity must always ...

0

Let $U$ be the emf.Then $$V=U(1-e^{-t/ \tau})$$ With $\tau=RC$. Rewriting this expression with factor $e^{-t/\tau}$ we obtain: $$V=U e^{-t/\tau}(e^{t/\tau}-1)$$ Let us show that the term $U e^{-t/\tau}$ is indeed of the form $Ri_{d}$. $$i_{d}=A \epsilon_{0} \frac{\partial E}{\partial t}$$ Now $$E= q/ (\epsilon_{0}A) =\frac{UC}{\epsilon_{0}A}(1- e^{-t/\tau}... 0 With coherent light from a laser one can assume that the path lengths of a Michelson interferometer are equal (for the purpose of your question) and due to the narrow bandwidth of the laser light one can also assume low dispersion (equal path lengths through the glass). With everything assumed to be equal the only difference between the two paths that the ... 1 The antenna in a radio transmitter or receiver is basically just a piece of wire. OK, the wire is usually made into a special shape, but that is just to make it more efficient. Every piece of wire in a physical circuit acts like an antenna and picks up some of the EM field radiated from every other piece of wire. At low frequencies this effect is too ... 3 At 300 MHz, the mouth of the waveguide would be of order ~1 meter across. At 30 MHz, ~10 meters across. At 3 MHz, 100 meters. At 300kHz, 1000 meters. At all of these frequencies, coaxial wires are perfectly adequate for conducting signals- and very cheap compared to the cost of making tubular waveguides with the dimensions listed above. 0 Space is almost vacuum so the sound wave will not propagate in the space. Instead, if we talk about the radiations, the impact of photons will impart pressure on the asteroid surface and might help in changing the course of the asteroid. The sun exerts solar pressure on cosmic objects. If we devise a technique of making the asteroid more/less reflecting then ... 2 No, sound waves can't traverse vacuum of space. They would only work in the atmosphere when it's already too late. 1 The solution to the homogenous problem L(u)=0, initial displacement f(x), and zero initial velocity, is for instance readily solved as :$$\frac{1}{2}(f(x-ct) + f(x+ct))$$which could be written in a fancy way as$$f \circledast \frac{1}{2} ( \delta(x-ct) + \delta(x+ct))$$But this isn't really useful. On the other hand, the Green function is the ... 2 Anything happening on a guitar string can be always written as a superposition of standing waves of different harmonic frequencies. Even the initial motion, which is clearly bouncing back and forth, can be written in this way. (This should not be surprising. After all, standing waves themselves are written by combining solutions that only propagate in one ... 14 Since sound travels as longitudinal waves, sound waves should only be able to propagate in a medium through compressions and rarefactions. However, water, as a liquid, is generally treated as an incompressible fluid. The underwater environment is characterized as an inhomogeneous medium, such inhomogeneity is due to point-to-point changes in underwater ... 1 Well basically the incompressibility of the water is just an approximation. You can always compress the water molecules; in fact, let's show how the phenomenon of propagation happens. We know that compression of the molecules by giving them a potential energy, so if we make them vibrate this will create a difference of pressure. The energy will be ... 1 One complication is that a sound wave can be described as a pressure wave or a displacement wave and they are 90^\circ out of phase with one another. The function at the top is the input sound wave I suspect it is the voltage which is applied to the loudspeaker which is producing the sound waves, so which description of a sound wave is this voltage ... 54 If for your purposes of your task you can treat water as uncompressible, then you can also assume that sound propagates instantly in it. Indeed there will be no sound waves in this case, the movement will be just propagated instantly from source to the observer. 0 All is OK. Remember that the mic responds to pressure and finding a signal (pressure) maximum at the closed end is exactly what you would expect. 116 Water is compressible (nothing can be completely incompressible). Treating water as incompressible is just a (usually very good) approximation. Therefore, longitudinal waves are possible. Wikipedia reports the bulk modulus to be about 2.2\ \mathrm{GPa}. This puts the speed of sound in water at about$$v=\sqrt{\frac{\beta}{\rho}}=\sqrt{\frac{2.2\ \mathrm{...

3

By the relation $c=\lambda\nu,$ for an infinite wavelength $\lambda$ you would have zero frequency $\nu$. Consider a plane wave with the usual phase factor $$\cos\left(2\pi\left(\frac{x}{\lambda}-\nu t\right)+\phi_0\right).$$ For $\lambda=\infty$ and $\nu=0$ this simply reduces to $\cos(\phi_0)$, i.e. a constant independent of position and time. For light ...

0

By the Planck-Einstein relation, $$E = hc/ \lambda$$ For an infinite wavelength, you would have zero energy, which is problematic. Also, you would need to measure this "infinite wavelength". What kind of mathematical shape would this look like? A constant? I guess if you consider the constant state to be "infinite wavelength", then yes; but noone would ...

0

The boundary condition is that the sound pressure is zero there.

1

Because of the boundary condition implicit in the open end. The pressure at the end must be the same as the pressure outside the tube (which is represented here by the straight gray line), so the red and blue must cancel out at the ends.

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$\vec{p}$ ... electric dipole moment $e$ ... Euler's constant $i$ ... imaginary unit As others have already mentioned: don't confuse $\vec{p}$ with $\rho$

0

Flat metal surfaces reflect radio waves the same way that mirrors reflect light beams. for the case of a flat metal surface on a plane being hit with a radar beam from an antenna on the ground a distance away, almost none of the incident beam bounces off backwards in the direction of the receiving antenna and so a plane built of such flat surfaces yields a ...

1

The dispersion phenomenon (the variation of velocity with frequency) typically occurs when the wave is guided by a medium having at least one of its dimensions of the order of the wavelength (elastic waves in rods, in plates, acoustical waves in pipes, ...) or when it propagates along an interface (Rayleigh surface waves, for instance).

1

In laymen terms, sound propagation depends on air particles colliding. The closer the particles to each other the quicker they can collide. In a less dense environment, the particles must travel farther to collide. Particle size has an effect as well as can be heard in glove boxes containing Helium vs Argon. Temperature also has an effect in the amount ...

1

You cannot, really. Your question is based on an incorrect assumption that it is the energy of the sound waves that cause attenuation. Instead, let's have a closer look at the three main mechanisms of attenuation in free air: Attenuation due to viscosity (proportional to $f^2$). Simply put, this is higher at high frequencies because the fluid is then ...

6

Sound is simply a compression wave. The velocity of the wave is inversely related to the square root of fluid pressure, and directly proportional to the materiel’s Young modulus. A lower density means a lower pressure, which increases the wave velocity as you noticed. Traveling through a compressible medium such as air, the simple equation for the velocity (...

0

A single frequency has a distinct mathematical shape function, sin(wt-kx). This is a smooth function. These are also solutions to the wave equations so any solution, e.g. phenomenon, can be described by a suitable expansion or superposition in terms of sine functions with different frequency, w, and/or wave number, k. The lower curve "looks" closer to ...

14

The speed of sound in a gas is given by $\sqrt{ \dfrac {\gamma \,P}{\rho}}= \sqrt{\gamma \, R \, T}$ where the temperature, $T$, is in kelvin, $\gamma$ is the ratio of the specific heat capacities of a gas at constant pressure and constant volume and $R$ is the specific gas constant. With increasing altitude there is a decrease in the density but also a ...

37

Wikipedia gives a pretty much straightforward answer. In an ideal gas, the speed of sound depends only on the temperature: $$v = \sqrt{\frac{\gamma \cdot k \cdot T}{m}}$$ So it neither decreases, nor increases with altitude, but just follows air temperature as can be seen in this graph:

4

I'm not sure what bothers you but I guess you simply need to name your displacement variable with a different name from the coordinate names. You could for instance name $v$ the displacement component in the $y$ direction and $u$ the displacement component in the $x$ direction. Then $u \cos (\omega t -kx)$ would be a longitudinal wave, and $v \cos (\omega ... 0 The theory of vibrating plates is a bit more complicated than that used for the vibration of membranes, as commented from a previous answer. And yet, this is assuming a thin plate (shear is neglected just as in the Euler Bernouilli beam theory). The theory was developped by Sophie Germain, Poisson, and ultimately by Kirchhoff. So Kirchhoff plate theory ... 0 When laymen say they can move a laser spot across the surface moon faster than light with just a flick of the wrist, in violation of special relativity, physicists point out that nothing real is moving across the surface of the moon. The apparent position some energy is moving, but not the energy itself. Perhaps a better example is the intersection point of ... 1 I know (radio frequency) waveguides, so I'll answer for waveguides. I suspect that you can extrapolate from there with decent accuracy. In a waveguide, you grind through all the math, and you find that the actual phase velocity in the guide is faster than light. "Woo hoo!" you say, "that Einstein guy was off base!". But if you take a waveguide and apply ... 0 If I understood you correctly, you see radio waves and visiblel light as two different phenomena, because they are genererated and described in different ways. However, both are in fact the same phenomenon and, as anna v already pointed out, only differ by their frequency. When talking about visible light, you seem think the emission of a photon in an atom ... 2 This is the accepted electromagnetic spectrum: Radio waves are identified as the low frequency part of the spectrum, frequency is the only difference. It is the interaction with matter that makes the difference, as matter responds to different frequencies, that is what makes the radio waves behavior separate from the rest of the spectrum. All ... 1 The lower plot would have almost all of its energy concentrated at one frequency and its sound would be a pure clean tone at that frequency. Whereas the upper plots would have higher frequencies in addition to the 'center' frequency and its sound would not be pure but would be slightly distorted from a pure tone. These harmonics could have a variety of ... 0 When we say that electromagnetic waves propagate through space and time. What is the quantity that is actually traveling? Electromagnetic radiation consists of photons, emitted by excited subatomic particles. After being emitted, photons propagate through empty space as indivisible quanta at the speed of light. Photons can carry very different energy ... 1 When applying Newton's IInd law, you're relating the acceleration of that particular particle of the string with the forces applied on it at a particular instant in time. The acceleration, at that time, is proportional to the sum of the forces on the particle, at that time. In this particular case, the tension force is proportional to the separation between ... 0 You're interested in the net force on an element of string at a particular time t. That is determined by the curvature of the string at time t. How that curvature will change with time is irrelevant. 1 In general they are simply solutions to different differential equations. The simple harmonic oscillator obeys the following differential equation: $$\frac{d^2 y}{d t^2} = -ky,$$ where$k$is a constant. The solution to this equation can indeed be written as a sinusoid with a given frequency and phase. The wave equation is a differential equation given by:... 2 Hawking considers a massless scalar field in the Schwarzschild spacetime. This satisfies the massless Klein-Gordon equation$\square \phi \equiv \nabla^a \nabla_a \phi = 0$. The expanded equation is $$-\left(1-\frac{2M}{r}\right)^{-1} \partial_t^2 \phi +\frac{1}{r^2}\partial_r \left[\left(1-\frac{2M}{r}\right) r^2 \partial_r \phi\right]+\frac{1}{r^2}\Delta_{... 1 Instead of the string imagine that there is river and on the water there are some water waves which are moving relative to the shore. Here is an animated gif of surfers riding a tidal bore with the water waves travelling from left to right. You will see that the surfers do not move relative to the crest of the water wave just behind them, so they ... 1 If you are driving alongside a tight string that has no wave, you will see the straight string moving past you. Now if the same string has a wave in it, and you are driving alongside it at the same speed as the wave is moving down the string, you will still see the string moving past you, but you will no longer see it as straight. You will see it curved by ... 4 The phenomenon where waves with different frequencies have slightly different speeds is known as "dispersion," because an impulse which begins with lots of different frequencies traveling together will "disperse" and spread out as the faster frequencies move ahead of the slower ones. Acoustic dispersion is pretty easy to demonstrate. Here's a great video ... 0 We solve the wave equation in a pipe with a sharp expansion, with the boundary conditions of non-reflection from the right boundary. A periodic signal is set on the left border. The initial data is zero. The animation frames show the passage of a wave through a pipe over 6 periods. We used FEM and Mathematics 12. -1 In general, the speed of sound is the same, regardless of frequency. The speed of a wave is given by the following formula:$$v=f\lambda$$Where$v$is the speed,$f$is the frequency and$\lambda\$ is the wavelength. If the frequency is higher, the wavelength will be shorter, but the speed is still the same. Regarding your diagrams, the lower one is ...

0

In general you could say that the speed of sound is the same in a certain medium (like air or water) under certain conditions. As Wikipedia (Speed of Sound) states, describing the speed of sound in the medium air: In fact, assuming an ideal gas, the speed of sound c depends on temperature only, not on the pressure or density (since these change in ...

0

Your analogy between adding waves and adding numbers is not accurate, not a faithful representation of the same idea. There are many ways to get 5 from smaller numbers. The decomposition of a function into sine and cosine waves is unique. This is because these functions form an orthonormal basis for a function space. The proper analogy would be getting ...

0

Nothing travels from one point to another in any wave, based on my interpretation of your statements. Electric and Magnetic fields are as real as the test particles they act on in the limit as those test particles are made to vanish. E = lim(F/q) as q-->0. In the field paradigm, these vectors exist in all of space when some source is present. In a static ...

2

The waveforms that define the shapes of the electric and magnetic fields travel through space at the propagation speed, c, along with electromagnetic energy. An analogy--Think of waves traveling in water: the water molecules only move up and down but the wave shape travels along with its peaks and troughs moving at the propagation velocity. In the process ...

0

A good place to read up this argument is in Sir James Lighthill's book "Waves in Fluids" I took the course from him that turned into this book, and I recall him discussing whether this claim was generally true, or only a rule of thumb. I think the discussion made it into the book. Certainly it has many non-trivial applications of the group-velocity.

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