# Tag Info

2

Because Light Wavelength is actually less than a sound wave. And Diffraction is more in longer wavelength waves, as is less in wider slits.

2

My fluid mechanics is not strong, but a steep increase in the pressure from a beam that you are heading towards as you approach the speed of light does indeed happen. You can reason this from several perspectives. Suppose you measure a plane wave in one frame and find out that the electric and magnetic field are $E_0\,\cos(\omega_0\,t)\,\hat{X}$ and ...

0

Maybe your confusion comes from thinking that oscillations must be perpendicular to the direction of propagation, but it doesn't. The fluid does oscillate, just as a sound wave in air oscillates. Hence the name baryonic acoustic oscillations (BAOs). Particles are compressed, and this compression moves away with the speed of sound (which is something like ...

3

You are not required to. Functions can be decomposed into a wide array of orthogonal basis functions, including the Bessel functions (in the Hankel transform) and the Legendre functions. The sine function just happens to be the overall simplest to deal with in the general case.

1

You refer to Fourier Series. The brilliance of Fourier was to use sin to express a function.You know that you can create any vector from the sum of some unit vectors.Exactly the same think happens here. The number you multiply the unit vectors is the coefficients in F.S. To answer to your question of why we use sin and cos is that they have (mathematicly) ...

2

The Poynting vector is useful not because we say so, but because of Poynting's theorem, which in essence states that the Poynting vector can usefully model how electromagnetic energy is moved around a system of changing electromagnetic fields. More precisely, you can define a quantity $$u=\frac12\left(\varepsilon_0\mathbf E^2+\frac1{\mu_0}\mathbf ... 0 If you have access to a class room setting, find a long, small diameter, stiff spring (e.g., 1" dia, 10' long) and attach one end to the handle of an immovable object. Take the other end of the spring and step back several steps so the spring is stretched. Now, hold the end tightly with one hand, and use your other hand to displace the spring several ... 2 Take a look at this Desmos animation. Either animate it by clicking the play button next ot the time (t) variable, or drag the slider around to watch the behavior of the waves. Watch carefully what the standing wave (the black trace) looks like when the waves constructively interfere (I.E. at times when they both look identical) and when the waves ... 0 Yes, there is no reason for the supersonic aircrafts making a significantly larger amplitude of sound as compared to the ones that go at the speed of sound. (But yes, the aircrafts going with a speed power than that of the sound create the sound waves in such a way that they do not interfere and the enhanced amplitude is not generated whereas when the ... 0 The sonic boom is not just the combination of the engine noise and ordinary sounds of the plane, smashed together because of the speed. It is the propagating effect of the air being smashed into by the plane, faster than the air can get out of the way. The amount it compresses the air depends on how fast the plane is going, and the loudness of the sound you ... 2 Short answer: Yes Slightly longer answer: If you scatter the wavefunction of a propagating electron from a potential (surface of a material for example), it generally splits into two parts - a transmitted part and a reflected part. As the names indicate, the reflected part represents a 'reflected' electron, the transmitted part a transmitted one. However, ... 0 I'm pretty sure that we can't do any better with trig identities; it looks like any such expression would have a term vaguely like \sqrt{1 + r \cos \theta}, and then the square root ruins the nice intuition. In lieu of that I'll offer a derivation of the group velocity that uses no fancy math at all! Hopefully that's what you wanted, even if you didn't ... 1 I wonder why Poynting vector can be used to describe the intensity of standing EM wave. It's because a standing wave isn't really standing. Hence the photon in the cavity is off like a shot when you "open the box". It didn't accelerate from zero to c in an instant, it was always propagating at c. The Poynting vector denotes the wave motion even when it ... 1 The standing wave solution you quote is a solution to Maxwell's equations - as is any linear superposition of travelling waves. The relationship that E = cB is really only applicable to travelling transverse electromagnetic waves in vacuum, however I note that in this case E_0 = cB_0. The Poynting vector (in vacuum) is defined by$$ S = \frac{\vec{E} ...

-1

What do light wave oscillations look like? You know what ocean waves look like. Imagine you're in a gin-clear ocean, and there is no surface. Now imagine an ocean wave passes you by. A light wave looks like that. By which I mean it doesn't look like anything. I know that light waves oscillate, but I don't know how. They don't really oscillate. ...

0

If you look at one specific point in space of your light wave, the electric and the magnetic field which is perpendicular to it are both in their turn perpendicular to the direction of propagation. So, both fields lie in a plane perpendicular to the direction of propagation. Both fields are varying with time in a sinusoidal way. Different polarizations are ...

0

This is possibly not a full answer, but the Wikipedia article Polarisation of Light below seems to suggest to me that, because of the chance of partially polarised light occurring, there may not be a well defined answer to your question: So what do light wave oscillations look like? because the mathematical appearance of the light wave oscillations ...

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.

0

Somehow you have to impose rigidity on a flexible object. I think this could be done by turning the string into a long thin magnet. The ends would then repel each other. You could put the string in orbit and then perturb it to look at the waves. I think it is intuitively obvious that standing waves would be possible. I don't have the maths to prove it.

1

A "standing wave" is not a real wave - it is simply our observing the superposition of two waves - one traveling to the left and one traveling to the right. If they have the same amplitude and propagate at the same velocity there will be stationary points on the string. This is true regardless of whether the ends of the string are "open" or "closed". ...

1

Yes, it most certainly can. It's much easier to visualize if you consider a length of flexible steel or plastic. You can shake it a bit, then toss it in the air so it's not constrained, and it will (if properly initiated) vibrate at a resonant wavelength. I think the confusion most people will get from your question is that everyday string is "floppy," ...

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 ...

1

Well you are going to get a reflected and transmitted wave, the two properties that change from the original input wave are the amplitude and wavelength of the two waves. I am going to use 1 and 2 to denote the input side of the string and transmission side respectively. Amplitude The amplitude is determined by these two formulae: $$A_r=A_i \frac{Z_1 - ... 1 Personally I find statements like "differently polarized beams don't interfere" to be imprecise, but I can understand where the writer is coming from. The interference of waves of like polarization (or of longitudinal waves) is characterized by differing average power at different points, including places where the average power is reduced from the value ... 0 Consider the typical powered speaker; a paper cone on a solenoid. When current is applied to the solenoid, the cone moves (forward or backward, depending on direction of current). Let's say you apply a steady DC current to the solenoid and the cone pushes forward. It will push a high-pressure wave ahead of it, and then pressure against the cone will ... 0 Yes, based on your comment. There is a low pressure area which follows the high pressure but it also travels at the speed of sound, so you wouldn't catch up to it at 50 m/s. 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. ... 0 Taking the simplest example of a sound wave, the component that determines its frequency, is the "up - down" motion of the wave (perpendicular to motion direction), whereas its propagation speed, (in the motion direction) is determined by the "resistance" of medium. Therefore, if the medium changes, only the wavelength changes. 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 ... 0 You have the intensity:$$I=\frac{1}{2}\rho\omega^2s_m^2$$which is a relationship between displacement amplitude s_m, angular frequency \omega and intensity I. What this is telling you is how the intensity is related to the angular frequency and displacement amplitude. There is nothing remarkable about this, all it is saying is the intensity is ... 0 I don't know where you got this formula but I think it's wrong. See https://en.wikipedia.org/wiki/Sound_intensity Sound Intensity is given by I = p \cdot v where p is the sound pressure; v is the particle velocity. Sufficiently far away from any source or diffraction object the relation ship between particle velocity and pressure is v=\rho \cdot c ... 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 ... 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 ... 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. 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 ... 0 As light spreads out the density of photons per area goes down. However the intensity can never drop below one photon. Hence we can see galaxies from billions of light years away. 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 ... 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 ... 0 Yes. Materials that absorb electromagnetic radiation and emit it in a different frequency are known as fluorescent. You probably see them as the coating on the inside of fluorescent tubes, where they absorb ultra-violet light and emit a lower frequency visible light. 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. ... 0 There is a system in which the frequency will change when the medium changes: a string fixed at both ends, such as a guitar string. If you pluck a guitar string, then change the medium status by changing the tension, the pitch you hear will change. This is because the wavelengths are fixed (2L, L, L/2, L/3, etc) but the speed of the wave is changing. ... 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 ... 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 ... 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 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 ... 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. ... 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 = ...

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 ...

0

As per my answer here. The air on the inside of the explosion is also moving "faster then the speed of sound" so relative to that air, the shock wave is traveling at subsonic speeds. Thus the shock wave travels at a weighted average of the velocity that sound would travel in the mediums on either side of it.

Top 50 recent answers are included