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25

Yes it works. But let's not use it on a massive scale, lest we damage the ecosystem (tip of the hat to @phi1123). A hint to the mechanism can be found in Behroozi et al (Am J Phys, 2007) They state in the abstract: From the attenuation data at frequencies between 251 and 551Hz, we conclude that the calming effect of oil on surface waves is principally ...


10

Before everyone freaks out, no, you don't use petroleum oil. You use vegetable, fish or animal oil. In earlier times, whale oil would be used. The OP's picture looks like a fuel oil leak, not an attempt at wave calming. I have seen references of this technique being used since at least the early 1800s, probably much earlier. Ernest Shackleton made use of ...


5

This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...


4

You can't tell directly. But you can look at a bunch of it and notice that it is at the same temperature here there and everywhere in all directions in every single place where your view isn't blocked by some moon planet star or galaxy. And that temperature is quite cold it is hard to get something that cold. And either there are many things with the same ...


3

There are two different kinds of wave here. The ones that you see on the surface are the (IMO) badly named Gravity Wave (not to be confused with a Gravitational Wave), which is the familiar phenomenon of waves on the ocean surface and arise at the interfaces between dissimilar fluids. Their phase velocity is $\sqrt{\frac{g}{k}}$ and the group velocity ...


3

The frequency of the first harmonic oscillation is the higher the higher the tension in the string. As temperature increases, the length of the string slightly increases. The change of linear mass density is thus negligible, but the corresponding change of tension in the string is not. The tension decreases and thus the speed of waves and frequency of ...


3

First, check your algebra. You actually end up with units $\sqrt{\frac{m\cdot Pa}{kg}}$. Now, one Pascal is one Netwon per meter squared, $Pa=\frac{N}{m^2}$. One Newton is one kilogram-meter per second squared, $N=\frac{kg\cdot m}{s^2}$. Thus one Pascal is $Pa=\frac{kg}{m \cdot s^2}$, and if we plug that into $\sqrt{\frac{m \cdot Pa}{kg}}$ we get units of ...


3

Many people have the misconception that only infrared radiation creates heat. Actually electromagnetic radiation over all frequencies carries energy. The shorter the wavelength, the higher the energy. Different elements and compounds (or molecules and atoms) have different resonant frequencies and so electromagnetic radiation of the same frequency tends to ...


2

One simplification of what gravitational waves are and how they're created, can be understood by visualizing a massive body and its gravitational field when stationary and what happens when the body accelerates. Imagine what happens if such a body were to start accelerate to the right for a while, and then stop. Just like the time-delay with light, it will ...


2

If your phonograph cartridge uses a piezo-electric pickup, the cartridge may have acted as a rudimentary crystal radio receiver. In a piezo-electric pickup, the stylus contacts a crystal and creates an electric current while jiggling as it moves through the grooves of a record. It's likely that while you were jiggling the apparatus, you found the exact ...


2

Actually we can measure optical frequencies at very high accuracy ($\Delta\nu/\nu=10^{-13}$ or better) using frequency combs. John Hall and Ted Haensch were awarded the Noble Prize in Physics in 2005 for this discovery. The idea of the frequency comb is to divide the huge frequencies down to a domain were we can easily count the oscillations electronically. ...


2

It depends what you mean by "function optimally". If you want it to be loud, it must transfer energy to the air and thus it's vibrational energy must decay rapidly. Imagine that the tines are large thin vanes (close together) so they transfer most of their energy to the air in a few vibrations. You will not hear or measure a precise frequency. If "function ...


2

If all you want to know is the voltage at one instant, you could indeed simply add the x-components. If you want to know the amplitude of the resulting sinusoid, you need to do complete vector addition (both x and y coordinates) to get the full amplitude (43.6 V. in your example). This will tell you not just what the voltage is at one instant in time, but ...


2

The fact of a wave doesn't needfully mean there's tension or elasticity in the sense of those phenomena in an acoustic medium. General Relativity describes how spacetime's geometry depends on the distribution of energy within it[1]. But GTR does not tell us anything about the microscopic "machinery" of spacetime that gives rise to this behavior: tension ...


2

A transverse wave popagating in a medium requires that the medium has a non-zero shear modulus. The velocity of a transverse wave is given by: $$ v = \sqrt{\frac{G}{\rho}} $$ where $G$ is the shear modulus of the medium and $\rho$ is the density of the medium. Gases and (Newtonian) liquids have a zero shear modulus so they cannot propagate transverse ...


2

When we say two sources have the "same" frequency, there is often a limit on what "same" means. Specifically, we usually recognize coherence as an essential attribute for creating interference. Let's look at the example of a laser. To create interference, you can split a laser into two beams, then let these beams interfere with each other. With the right ...


2

Most light sources do not emit a constant phase, unlike radar or audio sources. So while all the light may be of a given wavelength, the random (in time) phase shifts can kill interference.


2

The FourierTransform.com is a website maintained by an enthusiast. The site is not peer reviewed, but it looks as though it might provide helpful explanations. Here's a link which provides some basic introduction to the Fourier transform. And here is another link to class notes provided by Prof. Carlton M. Caves for an introduction to the Fourier ...


2

SUMMARY: This is a very good question. In a lossless medium, fundamentally the answer to your question is "no, an individual ray does not lose energy in propagating" because it represents a plane wave (in photon language, a momentum eigenstate), whose intensity does not vary as it propagates. Intensity information is encoded in the flux density of rays ...


1

The inverse square law is "simply" a statement of the fact that a diverging cone of "energy"/particles/stuff will have a cross sectional area that increases with the square of the distance. This is a result of basic geometry and the fact that area is proportional to length squared. As the stuff-beam impacts over an increasing area as distance ttravelled ...


1

For a solid, the speed of sound is given by $$c = \sqrt\frac{B}{\rho}$$ where $B$ is the bulk modulus, and $\rho$ is the density. So contrary to your assertion that "wave speed is fastest in denser media", the same type of relationship holds as for string: speed goes up if the force constant (restoring force for displacement) is greater, and it goes down ...


1

Don't confuse theory with reality. Electromagnetic theory (i.e., the theory of Maxwell) talks about waves of electric and magnetic field strength. The standard model of particle physics talks about photons---discrete particles/packets of energy---whose appearance are governed by probability densities that obey wave-like laws. As far as we know, gamma rays ...


1

Light doesn't "vanish" - it "finds another way". When you have a thin bubble you see interference fringes in the reflection - when the bubble gets sufficiently thin, the reflection disappears completely. This means that the light was transmitted. In general, if you have an interference pattern, for every minimum (fewer photons) there is a corresponding ...


1

Sure, this definition is useful for any kind of cyclic phenomenon, why not? Think for example of 'phases of the moon'. In the case of fourier analysis however, 'phase' usually means: phase of a sinusoidal component, not the phase of the waveform that is being analyzed.


1

The most mathematically general way to write a Fourier expansion is to use complex waves with complex amplitudes. In this case the phase of the waves is represented by the complex phase of the amplitude. You can see this if you write the amplitude in polar form. Here's how it looks for your two-component wave example: $$ f(\mathbf x,t) = z_1 e^{\mathbf k_1 ...


1

Greatly rewritten based on feedback in comments In order to understand this issue, it is worth considering what a telescope (or any optical / radio imaging system) really does. Taking a simple parabolic mirror, the shape is chosen such that the total path length for all rays "from infinity" to the focal point is the same. By making the path lengths the ...


1

Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with ...


1

The velocity of the standing wave is the velocity of the incoming and reflecting wave that formed this standing wave. See http://www.physicsclassroom.com/class/waves/Lesson-4/Formation-of-Standing-Waves


1

The Laplace equation $\nabla^2 \psi = 0$ is a linear differential equation. Now note that if $\phi$ is real, then so is $\nabla^2 \phi$. Moreover, by the linearity of the equation, if $\phi$ is real, then $i\phi$ is pure imaginary, and so is $\nabla^2(i\phi) = i \nabla^2(\phi)$. Okay, back to your situation. Let's say the solution is $\phi_1 + i\phi_2$ for ...


1

Generally that would be diffraction - the spreading out of a wave. Reflection is bouncing off objects - e.g. an echo of someone's voice. Refraction is about a wave changing direction e.g. water looks shallower than it is when viewed from above in air because the light from the bottom is refracted (bent) at the water/air surface. In this case - diffraction - ...



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