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Would it also be true to say that Kolmogorov's hypothesis of isotropy, means that turbulence is considered to be invariant with rotation and reflection, such that in all directions it "looks the same", implying homogeneity. So it doesn't make sense to analyse turbulence in the cubes of physical space, instead analyse it in the spheres of wavenumber space.


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As previously stated, one of your expressions is dimensionally inconsistent. However, it appears that your textbook is defining the vertical displacement of your wave for every point x on the wave, not just at x=0. This is done by noting that since x=vt, you can define a time offset from your origin by subtracting t = x/v from your original time, leading ...


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 ...


0

Yes. Fermat's principle is equivalent to the Eikonal Equation, which can be derived for any system fulfilling the D'Alembert wave equation when a slowly varying envelope approximation is valid. For approximately monotonal sound waves of wavenumber $k$, the Helmholtz equation with position dependent sound velocity $(\nabla^2+k^2(\vec{r}))\psi= 0$ is ...


1

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 The site is not peer reviewed, but it looks as though it might provide helpful explanations.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

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

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 ...


0

Think about how electromagnetic radiation is generated. Electrons and protons emit photons. The wavelengths of this photons depends from the surrounding temperature, for the protons it depends from the chemical elements (H, He, ...., C). Furthermore it depends from the gravitational potential and the velocity of gas streams. Perhaps this are not all ...


0

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 ...


0

The frequency of a signal is hard to change. Some ways that you can change it: reflect it off a moving object change the distance between source and receiver as a function of time (Doppler shift) introduce some non-linear amplification - this will generate harmonics If you just add "noise" (uncorrelated to the signal), you can create constructive or ...


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 ...


1

If the space is truly "open", then reflection should not play (since there is nothing to reflect off), and refraction can only play if there is a change in refractive index (which "normal air" would not have). That seems to leave two things: the fact that the entire body vibrates when you speak - and thus there is some fraction of the sound being directed ...


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 - ...


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 ...


0

Typically not. Changing the frequency requires some non-linear process and background noise is simply additive. There potential exceptions: the noise could be so loud that the air and/or the microphone become non-linear, the noise mechanically couples to the string and/or the instrument in a way that changes the resonant behavior, the noise is somehow ...


0

I may not be understanding the source of your difficulty. There are three facts here. First, the speed, frequency and wavelength are related as $v = \lambda \nu$. Second, the frequency of light remains the same when crossing the interfaces between media. This is a consequence of ensuring that the continuity conditions implied by Maxwell's equations are ...


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


0

The difference in frequency is most likely due to the guitar going out of tune as a characteristic of background noise is that it does not have well-defined, constant frequencies, rather being a collection of all kinds of random frequencies thrown together.


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

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 ...


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 ...


0

So in our results, i found that the frequency recorded in a quiet area was higher than a loud area, varying by around 5Hz. Can this be attributed to the interfering frequencies?


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In addition to the two fundamental frequencies you will also get the beat frequencies as they interfere in the microphone


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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.


0

So, why some objects, even if the velocity before and after the collision seems to be the same, are louder than others? I mean, how do the different material properties enter in the phenomenon? As dmckee stated Sound is longitudinal pressure waves in the air... The impact can set up pressure waves and or ringing in the bodies themselves ...


0

This sounds like ;) it's because the object that the TV is resting on is acting as a sound conductor. Solids conduct sound better then gasses. On the 1st floor, sound goes from the TV, through the platform, through to the floor and to the ceiling of the floor below without losing much energy before being transmitted through the air. On the ground floor, ...


0

In my answer, I assume that "scatter" can be generalized to wave motion. In general, we recognize different kinds of surface waves: gravity waves (main driving force is gravity), and capillary waves (surface tension dominates). Now gravity waves have a velocity that is a function of both wavelength and depth: ...


0

To some approximation -- probably a pretty bad one -- a guitar string is described by simple wave equation for small enough oscillations. If $y(x,t)$ is the displacement of a portion of the guitar string $x$ meters from the fretboard, at time $t$, then the equation governing the motion is: $$c^2\frac{\partial ^2 y}{\partial x^2}= \frac{\partial^2 y}{\partial ...


0

So the shape of the string (of length $\ell$) is defined with a general series equation that depends on constants $A_1,A_2,\ldots$ and $B_1,B_2,\ldots$. $$ y(x,t) = \sum_{i=1}^{\infty} \sin \left(\frac{i\pi x}{\ell} \right) \left( A_i \sin \left( \frac{i \pi c t}{\ell} \right)+ B_i \cos \left( \frac{i \pi c t}{\ell} \right) \right) $$ The wave speed ...


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 ...


9

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 ...


0

They are three dimensional! Two of the dimensions tell us where to locate a point on the surface, and the third tells us how high the wave is above the surface. Or to put it another way, the value of the wave is the "vertical disturbance".


0

The waves are of 1D,2D,3D and if their nature is described within those dimensions of space. 1D Waves : The direction of propagation of wave and the oscillation of the particles of wave are in same direction.(Spring Waves) 2D Waves : The direction of propagation of wave and the oscillation of the particles of wave are perpendicular to each other in the same ...


0

By measuring the shape of the transient output envelope it is possible to infer the phase of a pulse of known frequency and length if the pulse is passed through a bandpass filter. You just have to make sure that the filter is not energized before the pulse hits it.


0

We (humans) exist in and so perceive our world as one of 3 dimensions in space. Although some physical phenomena we observe may appear to be 2 dimensional in nature, if observed more closely we will see that behavior poking its way into the third dimension somewhere. Strictly speaking 2 dimensional behavior cannot exist in a three dimensional world. While ...


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 ...


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 ...


-1

One has explored radio waves experimentally. It was found, that outside the emitter (antenna rod) exist both a swelling magnetic field and a swelling electric field. The electric field was parallel to the rod, the magnetic field perpendicular to the rod. This radio waves are electromagnetic radiation, but they are a special case of EM radiation. The most ...


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 ...


1

Distance is equivalent to time. The time at which the cosmic microwave background was emittied was the time when the universe made a phase transition from being a plasma to being atomic matter. During this time, the universe finally became transparent. Before this time, the universe was so hot that all matter was opaque. The CMB is the wall that ...


1

First, when you rewrote the first formula using the "explicit notation", it is very hard to see what you gained. On one hand, you indicated that $y$ is a function of $x,t$ which would be fine (although it's clearly redundant to write this long expression instead of $y$ all the time) but I can't understand why you wrote $x_0,t_0$ instead of $x,t$. Even more ...


1

The term 'amplitude' is often used somewhat ambiguously. The most rigorous definition is that amplitude is simply $|A|$ (the modulus of $A$). In your case (unmodulated wave or oscillation) there's only one amplitude. But others count the number of peaks and troughs as 'amplitudes' like you are doing. You count only three but that's because Fig.14.2 only ...


0

here is an animated version of image for more in formation refer to http://www.physicsclassroom.com/class/waves/Lesson-3/Boundary-Behavior and a good tool for playing with https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html


1

My intuition is that the frequency should stay the same because the waves in the light rope are caused by the waves in the heavy rope. The point where the ropes attach will oscillate with a common frequency. So, for $(b)$, the frequency would be the same. For $(c)$, use the equation $v= f\lambda$. You already correctly determined that the velocity ...


1

I'll start with the second of your questions. Yes, light from very distant galaxies gets redshifted to such long wavelengths that there practically isn't any light to see. The lower limit on frequency is zero. Obviously. Technically one could say there is no signal at $0\,Hz$, but that still put a lower boundary on the frequency. Objects on the edge of our ...


1

Your understanding of resonance seems about right on a qualitative level. If one were to ignore losses like friction, drag, or the like, "driving" a system at its resonance frequency would indeed result in feeding it more and more energy which is stored in the form of a large amplitude of the oscillation. For a completely lossless system, the amplitude would ...


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 ...


0

Suppose in zero gravity you squirt a smooth stream of water from a nozzle. A cylindrical tube of water with surface tension is not a minimum energy configuration. Spherical drops are minimum energy. The cylindrical tube is a metastable state, like a ball perfectly balanced on a point. Any slight variation from a perfect cylinder will be increasingly ...


1

Well the imaginative experiment you have put forth is a fairly satisfactory method I would say. The wavelength of a guitar string is simply the distance between two consecutive peaks or troughs of the oscillating string (for simplicity let the frequency be constant). The wavelength of the electromagnetic field, or light, is just the corresponding value for ...



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