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346

So, I decided to try it out. I used Audacity to record ~5 seconds of sound that resulted when I dropped a penny, nickel, dime and quarter onto my table, each 10 times. I then computed the power spectral density of the sound and obtained the following results: I also recorded 5 seconds of me not dropping a coin 10 times to get a background measurement. ...


131

If you have the dimensions and material of an object, you can compute both the mass and the normal vibration modes. Just the mass is not enough - a large paper "coin" will have a different fundamental frequency than a small tungsten sphere. A summary of everything that comes below - the result of several edits, and including a nice interaction with the ...


28

Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase. If you ...


15

This is not an advertisement. Under the rubric of "do try this at home", I wanted to share one more thing that I discovered after writing my previous answer - but it is so unrelated to that answer that I thought it better to write this as a separate post. I discovered two interesting things. First, when you spin a coin on a hard surface, it "rings" with ...


7

Picture a straight line of balls confined to a line with no gravity, friction or anything else. The balls are free to move around the line apart from the fact that every two neighboring balls are connected with an elastic spring which pushes them apart when they get closer and pulls them together when farther than a distance $a$. When we take one ball on ...


6

My answer will be quite close to that of PhotonicBoom although a bit more graphical. When it comes to light phenomena, there are different ways to comprehend them: we can use a wave picture (Hyugens-Fresnel), we can use the most modern picture we have (QED) or we can use something a bit more intermediate which is the picture of light rays travelling from ...


5

You are right in that there is only one set of physical things going on in diffraction. The reason people talk about two different kinds, is because there are two natural limits in a diffraction problem. The intensity of light you see at any point is the contribution from all of the points at the aperture, where the contribution from any point decreases as ...


4

The answer lies in QED and is quite complicated mathematically. As a simple explanation this is what happens: Photons follow all possible paths from the source to the screen and each path has its own probability amplitude associated with it. Summing up all these paths cancels out most terms in the summation and you end up with your desired final "classical" ...


4

Huygen's principle alone will not answer your question, however the Huygen-Fresnel principle modifies this to include wavelength. It states that every point in an unobstructed beam acts as a secondary source of wavelets with the same wavelength as the primary wave. The amplitude of the optical field at any point is then the superposition of all the wavelets. ...


4

I agree it looks like a transverse wave - like the ripples on a pond. But I believe you are fooled by a simple thing: the waves you are looking at look like "illuminated ripples" but are in reality just changes in temperature (changes in brightness of the sun's surface). If you have a shock wave traveling out across the surface of the sun, what happens? The ...


3

Take the wave equation $$\nabla^2\vec{E} = \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2},$$ and let $\vec{E}(\vec{r},t)$ be a solution. Indeed taking the real part $\Re(\vec{E}(\vec{r},t))$ yields the physical significant values. The initial values at $t = 0$ are $\Re(\vec{E}(\vec{r},0))$ and the problem arises here: this does not give you enough ...


3

I don't mean to take anything away from the previous great answers, but the "simple and to the point" answer is, a very qualified, yes. By qualified, I mean one must know the coin's composition, thickness, diameter(or shape), density distribution, country of manufacture, etc. If we make assumptions and restrictions, then it becomes possible to calculate ...


3

The velocity term here means the speed of a wave traveling through the medium. Take, for instance, a pipe resonating, like in an organ (the equations are basically the same). The v term means the speed of sound in air, even though the wave is standing. The same thing applies here - v means "speed of any wave" not just "speed of this particular wave".


3

I guess you have this image in mind: This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ...


3

Not a complete answer, but this is a classic "sloshing" problem. The interaction between the fluid and the container wall, under the influence of the external (periodic) force sets up a (self-reinforcing) and harmful resonance. This is of immense practical interest: jet-fuel sloshing inside airplane tanks, for instance.


3

There are a lot of good answers already here, but I think I'll add a few pictures that show the ways I have to force myself to think to keep track of the differences between light waves and light rays when thinking about diffraction patterns. Above we have some light coming through the center of the slit. If we draw it like this it seems to make sense with ...


2

It depends what you're trying to do. The integral will give you the net current. So if you did the integral for the AC current going to the computer I'm typing this on you'd get the value zero. This is quite correct because no net current flows for kit powered by AC. On the other hand, if you're trying to work out how much power my computer consumes you ...


2

In optics, Fraunhofer diffraction (named after Joseph von Fraunhofer), or far-field diffraction, is a form of wave diffraction that occurs when field waves are passed through an aperture or slit causing only the size of an observed aperture image to change due to the far-field location of observation and the increasingly planar nature of outgoing diffracted ...


2

They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before. Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in ...


2

The other big advantage of $\omega_0 t$ is that it is the time dependence of energy eigenstates. Other time dependencies can be constructed, but they won't have such nice properties as $E=h\nu$, etc. The usual definition of a "photon" has it being an energy eigenstate, so in a sense the frequency must be fixed by definition. Other electromagnetic wave ...


2

Some good points made in different answers. I just want to add my two cents. The short answer is "yes - it can appear louder, and it can be louder". First - appearance. If you have a loud point source far away, and another source closer by, it is possible that the close source clouds the faraway sound. Imagine a faraway train and a nearby radio playing ...


2

The gaps just need to be smaller than the wavelength of the UV radiation. Waves can only pass through gaps larger than their wavelength. In symbols $$ L < \lambda $$ Where $ L $ is your gap length, and $ \lambda $ is the wavelegnth of the UV radiation. Typically this from is $400$ to $10$ nm, with UVA in the $400$ to $320$ nm range, UVB in the $320$ to ...


2

I start my answer with the the second question (similar situations). This is eerily similar to the free-particle dispersion relation (or the relativistic energy-momentum relation) in natural units ($c = 1$ and $\hbar = 1$), $$\omega^2 = k^2 + m_0^2$$ This system of units is commonly adopted in particle physics and quantum field theory. It can be clearly ...


1

A wave is a perturbation in a system that propagates. The wavelength is the typical length along which a wave is coherent, which means that what happens at some position affects the wave behaviour in the vicinity if this point at distances of a few wavelengths. The reason for that is that the medium in which the wave propagates has some rigidity and the ...


1

This is a hard question to answer, in the end. However, be assured that long, long, ago we started looking for nuclear expositions by looking for X and gamma radiation using the Vela satellites http://en.wikipedia.org/wiki/Vela_(satellite) . These did not find much in the way of violations of the nuclear test ban treaties but did discover astronomical ...


1

This is a general property of (pseudo)Riemannian geometry. I do not think there is anything specifically physical about it beyond the geometry. It holds even if $\varphi$ in not Lorentz invariant. In (pseudo)Riemannian geometry the covariant derivative $\nabla_i$ replaces partials $\partial_i$. The Laplace-Beltrami operator $$\Delta \triangleq ...


1

(I am refering only to the updated question) This is a question which has been asked numerous times by physicists after the formulation and popularization of Maxwell's equations. They have previously known the wave equations on strings where the matter of the string was vibrating, or on water, where the water surface was the one carrying the wave. How ...


1

This is a question that was drastically changed and the other answer and the comments to the question and the answer are discordant with the edited question. I already advised in the comment that you read a simplified article in wikipedia on electromagnetic radiation. Classical electromagnetic radiation cannot be simplified easily by analogies. I will try ...


1

It depends on the direction in which you choose the $y$-axis. If a positive $y$ means a displacement to the left, when this figure is accurate. But you right, that choice is at least unconventional and should have been indicated in the figure.


1

First things first: waves that have already reached close vicinity to the beach DO displace water towards the shoreline - just notice how the water moves back and forth at the point it's ankle-deep. This is related to the phenomenon by which they lose their wave form and get a crest. There are many forces acting on a surfer, but two of them are the ...



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