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10

If you have any kind of solid material, it will become a little bit thicker as you compress it, and thinner as your stretch it. This means that a "one dimensional" wave traveling longitudinally down a rod will in fact cause some lateral motion. The ratio of displacements in the perpendicular direction is obtained from the strain (relative displacement of ...


6

The situation you are describing is an example of Fresnel diffraction (or near-field diffraction). In general, when a wave propagates every point of the wave front can be thought of as its own source of waves traveling in all directions (called Huygens construction). It turns out that neighboring point sources along an infinite straight wave front reinforce ...


5

If we look at the sonic boom as a $\delta$-function, where we have a really loud sound for a really short time, then it will be able to excite all frequencies at the same way. You can actually compute this by showing that $$ \delta(t)=\frac{1}{2\pi}\sum_n e^{int},$$ which show how the $\delta$-function is actually composed of all frequencies. Then it's ...


4

A graph should clarify the relationship between two quantities, requiring the least amount of mental effort on the part of your audience. If you are trying to show the change of density as a function of position along a wave, you should plot position along one axis, and density along the other. Whether you use vertical deflection as a measure of density or ...


4

Several things happen with ripples. First - they are comprised of different wavelengths. That may be hard to explain in plain English, but basically the only way that a wave can have a single wavelength is when it goes on for all infinity. A short "wave packet" must contain different wavelengths. Next - the speed of propagation of these different ...


4

You don't need a particular point on the wave. You only have to make sure it's the same point on each successive wave. If you have a microphone, hooked to an oscilloscope, you can measure the time between peaks (or troughs, or zero-crossings), multiply by the speed of sound, and that's your wavelength.


3

You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction. The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = ...


3

That video is very poor in one aspect: particles in the sound field doesn't move "horizontally" nor "vertically". Really, the proper word is "longitudinal motion" and you are in fact asking about "transversal motion". In basic description, the air is considered to be an ideal fluid. Therefore no shear stress is possible and hence no transversal motion as ...


2

Think about what the wave vector represents, and what kind of wave your equation describes. The most general equation for the spatial variation of the electric field for a plane wave is $$\mathbf{E}(\mathbf{r}) = \mathbf{E_{0}} e^{-i\mathbf{k} \bullet \mathbf{r}}$$ where $\mathbf{E_{0}}$ is some vector with no dependence on $\mathbf{r}$. This can be ...


2

As I understand it the derivation of the function $\sin(\theta)$ of angle $\theta$ comes from the triangle. This is purely mathematical. The equation that you quote $$y = A \sin(wt + \theta)$$ can be derived by solving the second order differential equation for simple harmonic motion $$ {d^2x \over dt^2} = -{k\over m} x $$ here for a mass, $m$, on a ...


2

In those notes, it states that the left part of the string pulls the dot with a force proportional to the slope. However, the right side pulls the dot in the other direction with a force proportional to the slope, so that if the slope is constant, there is no net force on the dot. The only way to get a net force is for the slopes to be different on the left ...


2

The standard procedure is the following: starting from $ \langle nlm|\,\partial^2_z\,| n'l'm'\rangle $ insert the identity operator with respect to the position basis $$ 1 = \int d\textbf{r} |\textbf{r}\rangle\otimes\langle \textbf{r}| $$ to have $$ \int d\textbf{r}\, \langle nlm\, |\,\partial^2_z\,|\textbf{r}\rangle\cdot\langle \textbf{r}|n'l'm'\rangle. ...


2

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...


2

Individual photons are not considered rays. Because of the wave and particle nature of photons, they are much more complicated than what they are generally thought of: a projectile of light. In fact, they do not have an exact measurable position, but do travel in straight line trajectories. What we consider rays are lines perpendicular to the wave front of ...


2

A "ray" in geometric optics is a locus of continuous propagation of light. Think of it as mapping where the energy is going in space. In principle there are an arbitrarily large number of them, but we draw a manageable number for visualization purposes. The various [letter]-rays were so named when people didn't know what they were beyond being things that ...


1

It is effectively defined that way because it's the simplest form that satisfies the relevant conservation of energy equation. As discussed by Feynman, "There are, in fact, an infinite number of different possibilities for [energy density] and [flux], and so far no one has thought of an experimental way to tell which one is right!"


1

As a bus passes you at speed, it displaces the air that occupied the volume which the bus fills. The displaced air flows past the bus, creating a high pressure area near the bus which pushes you away. After the midpoint of the bus has passed you, the displaced air starts to flow into the area vacated by the bus as it leaves. This creates an area of low ...


1

According to "Fundamentals of Physics 9ed" by Halliday Resnick (p. 496), "there will not be a steady wave pattern with nodes and antinodes, and the total wave will not be a standing wave." The result will be somewhat like waves reflecting off the irregular shores of a tiny pond - unrecognizable patterns, or like confused seas on the ocean, where the waves ...


1

Although the part of the rod that is most displaced "wants" to get home, it is not because it somehow remembers its original location, but because its neighbors are pulling and pushing it back. Thus we say that the potential energy is most concentrated in those parts that are the most stretched or condensed (they pull/push the most). The measure of ...


1

Suppose you shine a linearly polarized laser at the wall. Let's call the direction of laser propogation $\hat{z}$ and the direction of the electric field polarization $\hat{x}$. Then if you plot the $x$-component of the electric field vs. $z$, you will get a sine wave. The wavelength of the light is the wave length of the sine wave. So if one peak was at ...


1

The wavelength is not defined as the length after which the waves repeats itself: that is only a pictorial representation that works in one dimension for simple one component waves but it is not valid in general. Instead, given any solution of a wave equation represented as Fourier transform $$ \psi(\textbf{x},t)=\int ...


1

What is the explanation for this phenomena? "Water Hammer" https://en.wikipedia.org/wiki/Water_hammer There is a study from "Hasson and Peck", 1964, which explains "Thickness distribution in a sheet formed by impinging jets." It's all basically Bernoulli's equation; "Velocity -> pressure -> velocity" and then simply continuity. Anouther good source is the ...


1

No. The particles of the medium in which the wave is travelling (assuming that this is a mechanical wave) aren't actually displaced in the direction of the wave, but are merely disturbed. In the case of a longitudinal wave (e.g. sound), a particle will for a time be moving in the direction of propagation the wave, for a time be at rest and for a time be ...


1

No. A sonic boom is an acoustic disturbance caused by supersonic flow over an aircraft's surface. Supersonic flow creates a discontinuous shock boundary that emanates from the aircraft surface and the shock wave propagates behind the aircraft with a large amount of energy, however dispersive as it travels through the atmosphere. Resonance does require an ...


1

The vibration of the particles isn't only up and down. In the other question Wendy Krieger said ocean waves are transverse waves. Take a look at the Wikipedia Wind wave article where you can see this gif: GNUFDL image by Kraaiennest, see Wikipedia See the red test particles? They don't just go up and down, they go round and round. They have an angular ...


1

Yes, I think your description is pretty reasonable. Suppose you represent your medium as made up of lots of thin strips: Energy is transfered along the medium as one strip rubs against another: This diagram is supposed to show friction (the red line) pulling the right strip upwards as the left strip moves up - possibly not one of my more effective ...


1

You will benefit by finding some tutorials on wave theory. In brief, assuming a spherical wavefront from the emitter, you are correct there's no direct path to the receiver. However, the edge of yourabsorber there causes diffraction (Huygen's principle), so thatsome of the sound wave (energy) will make its way to the receiver. You can see a demo of this, ...


1

In the equation: $$y(x,t)=Asin ~k(x-vt)$$ $A$ can be varied independently of $k$ and $v$ and hence of $f$. That is what is meant by saying that the amplitude doesn't depend on the frequency. Now, when you write the equation as: $$A = \frac{y(x,t)}{sin ~k(x-vt)}$$ it means that the ratio of the height of the string from the mean position at some point to ...


1

No, quantum/particle theory came later. What Einstein realized was that, travelling at the same speed as a light wave, the forward component of the fields "freeze", leaving only the lateral components able to vary. So the interplay between electric and magnetic fields which characterize electromagnetic radiation would be grossly affected. All of this takes ...



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