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11

Because the frequency of a sound wave is defined as "the number of waves per second." If you had a sound source emitting, say, 200 waves per second, and your ear (inside a different medium) received only 150 waves per second, the remaining waves 50 waves per second would have to pile up somewhere — presumably, at the interface between the two media. ...


10

When you pluck a string or hit a drum or sound a not on a flute, the instrument and the air in and around it vibrate and this vibration propagates as sound waves in the air to your hear drum. When you hear an instrument being played, what you recognise as the note is the base frequency. 'C' corresponds to $261.6$ Hz and is the same for a piano or a guitar. ...


8

This has to do with continuity of the wave motion. Imagine you had a change in frequency going from medium A to medium B - say 10 Hz become 20 Hz. How do you make something move at 20 Hz? You need to apply a driving force at 20 Hz of course. But the incoming wave is going at 10 Hz. To add energy to the wave we must be pushing when it it moving away from us ...


7

Update with a more clear answer: Here's a plot of all the velocities involved with shock propagation through a sationary medium: The x axis is the mach number of the shock wave and represents the strength of the shock wave, it could have been velocity or pressure ratio or any other quantity that is monotonic with shock strength. The y-axis is velocity ...


5

The electromagnetic wave is a classical theory while matter waves are quantum mechanical. The wave aspect is a mathematical abstraction which allows us to predict future quantum states of the electron with a known probability.


4

It's not a stupid question. In fact, Quantum Field Theory is the field of physics that seeks to answer exactly this question. In QFT, in addition to the electromagnetic field, there is a single electron field that extends throughout the universe. Stable ripples in the electron field constitute individual electrons. Every fundamental particle has a ...


4

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}) = ...


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

Pitch, in music, is equivalent to frequency. How often the wavefore cycles. This is usually defined by length, i.e. how long the string is, how long the pipe is, etc. It can also be affected by the tension (how tight the string is.) Timbre, the sound of a specific instrument, is defined by the "shape" of the wavefore, whether spikes, round, square, or ...


3

Light will never be completely at rest, but we have succeeded in slowing it down significantly. (See this for example) In a medium, particles can move faster than the speed of light. (The speed of light in that medium) In fact, this is used in some particle accelerators to detect certain particles. When a charged particle travels faster than the speed of ...


2

The first 2-D image you posted is a typical simplification for teaching purposes. In it, they use the height of the sine wave to represent magnitude, and the directions of the sine waves to show how the fields point relative to each other. The light itself however is not itself at all cone-like. You have to imagine this sine wave existing at multiple points ...


2

Frequency, in physics, is the number of crests that pass a fixed point in the medium in unit time. So it should depend on the source not on the medium. If I take a source who vibrates faster than yours then number of crests that my source can create per second (for example) will be more than yours. But speed of the wave depend on the properties of the ...


2

From the famous Double-slit experiment, it is clear that electrons do behave as wave as well as particle. When it is detected by geiger counter, "click" sound appears & no matter how greatly the voltage is decreased along the cathode tube, "click" & never "half click" appears. So, electrons always arrive at lumps like bullets. However, unlike bullets ...


2

Let me say what others are trying to say, hopefully in a clearer fashion: Just because you can relate two variables in an equation does not mean that they are dependant. In this case, you have to constrain intensity $I$ in order to get the relationship. At that point, it is not a general relationship, but only true when $I$ is constrained. An example that ...


2

The speed of mechanical sound waves through the air at 0 degrees C is 331 m/sec. But sound can travel at many different speeds, depending on the medium it propagates through and the temperature and pressure, among other variables. What we call sound is any mechanical wave within the range of human perception that is transmitted to our eardrums via the air. ...


2

You hear the boom when you and the cone overlap. It doesn't matter whether you move "into" the cone, or "out of" it - there will be a sharp transition in pressure. Maybe plane B hears a "moob cinos". It will still be loud.


2

So this is a many-part question. If you look at that diagram, in fact, you will see that the pressure peaks and troughs (the circles) do not "stack up": nowhere are they intersecting. So in fact the only effect in this diagram is that the pressure for the circle of radius $r$ is actually decreasing like $1/r$ as the circle gets bigger. The intensity, which ...


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 ...


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

Generally, in linear systems modes are independent. Energy does not flow from one mode to another. What causes the coupling is a nonlinearity. The nonlinearity reveals itself at higher amplitudes (nonlinear terms are small at small amplitudes). Thus, when you drive the rod just a little bit the energy DOES go to the higher harmonics, but the coupling is weak ...


1

A possible answer for that might be that if you have a rope with the length $L$, you have a frequency $f$ as the first harmonic frequency with $T=\frac{1}{f}$ as the time between two amplitude maxima. This time is determined by the frequency how fast the wave can propagate in the rope, and is therefore bound to the speed of sound in the rope $$\nu = ...


1

First off I think I should sort out a misconception about Huygens Principle. You can apply this principle efficiently if you have a slit, which is equal or smaller than the wavelength you are considering. If on the other hand the slit is substantially larger than the wave length, you should consider multiple Huygens sources. Take a look at this animation ...


1

Unfortunately, I think you are speaking about what people commonly say is "Huygen's Principle", "In order to explain waves diffraction, it says that every point in a wave front behaves as a source, so the next wave front is the sum of all secondary waves produced by these points.", but this is not actually what Huygen's principle says. Huygen's principle ...


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

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!"



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