# Tag Info

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The human voice box produces a fundamental frequency and its harmonics because the mechanism is like that of a relaxation oscillator. However, we have limited control over the relative amplitude of the harmonics (we do have some - that is how we change the "color" of a tone we sing, and the sound of vowels). In order to produce the Shepard scale, you need ...

13

i've programmed some shepard tones and even a voice generator. The human voice can't make that sound for the same reason that a single or even 3 trumbones couldn't make it. if you had 12 trumbones you could conceivably put them on a wheel system so that the pitch of each is increased and when the top one reaches to top is muted and send down to the lowest ...

10

The transitions between two frequencies will not move with the phase velocity! Nor will the beginning of transmission! Only in the steady plane wave parts in the middle, the wave crests will move faster than light. And that's just "appearance" in the sense that bewteen any two times, you can find a maximum of the electric field and pretend the wave moved ...

7

Point a laser to the sky, track where it lands very far away from the Earth, and then move your arm. The point you are tracking in space is moving faster tha the speed of light, even though light itself isn't. This isn't the same as group velocity, but it is very similar. There is no problem in having arbitrary points you imagine going faster than light, ...

5

Sound could be considered a renewable resource if taken from a source that was created by continual physical processes - such as the sound of waves crashing against rocks. Although those sound waves contain energy (which is the kinetic energy of moving/vibrating air particles), their energy density is very low. Therefore they are not useful for generating ...

5

The comments are on the right track. First thing: the R,G,B filters in a standard digital camera each pass some IR light, i.e. they're not low-frequency blocking. Usually cameras add an IR-blocking filter on top of the color grid because of this. An example RGB filter from FujiFilm is shown in this picture: which is from somwhere near here You ...

4

The $\omega t$ term encodes the time-dependence of the wave form while $kx$ expresses its spatial dependence. Think of a wave form frozen in time, i.e. $y(t=t_0, x)=A\sin(kx+\omega t_0+\phi)$. The $\omega t_0$ is just an additional constant phase and can be absorbed into $\phi$. The result is an oscillation in space with a wave length (distance between ...

4

I think you might be a little confused. The phrases 'renewable energy' and 'un-renewable energy' are used to refer to industrial sources of energy. These industrial sources include Wind, Solar, Wave, and Nuclear power, and traditional fossil fuels (coal, oil, natural gas etc.). If a source of power is renewable, it is not depleted (used up) when utilised ...

3

This is an interesting question and the answers which have been given show that the $v$ in your equation should be called the magnitude of the velocity or just the speed of the wave. The mixing of the terms speed and velocity happens all the time. Now there is an equation for wave velocity but in comes about in a somewhat convoluted way. Suppose that you ...

3

I will answer this question in two parts: first comparing plane waves, spherical waves and cylindrical waves (which are really the same thing as I will explain). how this relates to ray optics (which is completely different). Plane waves, spherical waves and cylindrical waves are 3 different examples of doing a decomposition of a wavefield. The method is ...

3

A permanent magnet has a fixed north/south polarity - in this example, lets say north is facing up and south is facing down. This magnet has a membrane of some kind attached to its north face. An electromagnet beneath the permanent magnet can switch the direction of its north/south polarities by changing the direction of the electric current running ...

3

Here is a series of images from two point objects (stars) to illustrate the idea of resolving power. The objective lens of the telescope produces the diffraction. If the distant stars are closer together the diffraction patterns come closer to one another. Diagram (a) shows to diffraction patterns (the intermediate image in a telescope) well spaced ...

3

The first problem you might have with the wave equation is to visualise the graph. This is because the displacement of a particle in the medium through which the wave is travelling $y$ depends on both position $x$ and time $t$. So you need to draw a 2D graph on a 2D computer screen and then this not even take care of a 3d wave with position $(x,y,z)$ and ...

3

Huygens's Wave Theory is what you call a first order scalar diffraction theory of light. So what does it describe and what does it fail to describe? First order means that electromagnetic effects like induced currents in surfaces etc. are ignored. These can be described by solving Maxwell's equations for the same system instead of working with the wave ...

3

This is a common misconception. The function above can be interpreted as follows. Sound of frequency $\dfrac{\omega_1+\omega_2}{2}$ with amplitude modulated by the cos function of frequency $\dfrac{\omega_1-\omega_2}{2}$. The cosine function becomes zero twice every cycle as well as reaching a maximum magnitude twice every cycle. So the intensity of the ...

3

You have to be careful to interpret your statements 1 and 2 correctly. In statement 1 you are talking about an infinitely long wave train which is made up of only one single frequency. As soon as you have a wave train of finite length then the wave train is made up of the sum many frequencies. So in statement 2 what you think of as a pulse of a wave with a ...

3

A radio wave may "bend" due to diffraction or scattering off objects, but it will not "bend" just because a receiving antenna exists.

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The basic phenomenon is that high frequency sound is more strongly attenuated than low frequency sound. The mechanism for sound attenuation is viscous damping. The absorption coefficient is $$\gamma= \frac{\omega^2}{2\rho c^3}\left[ \frac{4}{3}\eta + \zeta + \kappa\left(\frac{1}{c_v}-\frac{1}{c_p}\right) \right],$$ where $\omega$ is the frequency, $\rho$ ...

3

A few observations. First - if you record sound for a short time, the bandwidth of the sample will result in a smearing of the peaks. This only really matters if the sample is very short - with a 1 second sample you would have 1 Hz resolution, but if you sample for 0.01 second, the bandwidth is 100 Hz. Second, you are using a scale that is quite compressed ...

2

Based on the results, the pipe is clearly open at one end but closed on the other. Therefore $\lambda_n = \frac{(2n+1)L}{4}$ Your formalism is a bit unusual. If I can advise you, try to use something like this for harmonics: $$f_n=\frac{c}{\lambda_n} = and \ so \ on$$ Show explicitly the dependency on $n$.

2

You will get destructive interference when the difference in the distances from you to the two speakers is $n + \tfrac{1}{2}$ wavelengths. In your case that's 0.68m, 2.04m, 3.4m, and so on. You get constructive interference when the difference in the distances is an integral number of wavelengths. However the experiment is hard to do in a living room ...

2

There are three points to be noticed: If you just blow without closing the lips, you would change the boundary condition. The trumpet waveguide is not "nicely predictible", the approximation of an open tube does not work cause the bore variations $S(x)$. You need to solve this kind of beasts for reasonable 1D propagating pressure approximation: $$... 2 Since this is clearly homework and excercises question, I will provide just hints. This kind of treatment is the same how beats are descripted. Study this article, it will help you get that. Since k=\frac{\omega}{c} where c is constant and the distance is the same for both the signals, it will not cause any more uncertain phase shifts. 2 The principle that every point on a wavefront can be thought of as an emitter of spherical (or, in 2D, circular) waves is applicable to any waves - see any introductory high school course on waves, where the demonstrations are typically done with water surface waves. Note that, as @ignacio pointed out, the construction is only exact in odd dimensions (in ... 2 When a receiving antenna picks up a signal, a current flows. This current acts as a secondary "transmission", and this will partially cancel out the electromagnetic field that is incident - this is how you get power from the EM field into the antenna. If you look "slightly downstream" from the antenna, you will see a reduction in the EM field (assuming for a ... 2 Edit: I think you'll find all the details you need at this question. As Asher commented, when a wave is described as sinusoidal, or triangular, or square, that's its amplitude profile. When a wave is described as plane or spherical, that's the spatial profile perpendicular to the direction of propagation. For example, a plane wave of sinusoidal ... 2 Light waves are emitted spherically, however electromagnetic waves nevertheless have polarization. The Maxwell equations that the electromagnetic field satisfies are$$\begin{array}{rlcrl} \nabla\cdot \vec E ~=& c\rho &~~& \nabla\times \vec E ~=& -\dot {\vec B}\\ \nabla\cdot \vec B ~=& 0 &~~& \nabla\times \vec B ~=& \vec J + ...

2

I find Pulliam's notes for the Euler equations to be a pretty good introduction to this topic using the equations of fluid motion. The idea is that you start with a conservation law: $$\frac{\partial \vec{Q}}{\partial t} + \frac{\partial \vec{F}\left(\vec{Q}\right)}{\partial x} = 0$$ where $Q$ is your variable vector and $F$ is your flux function. You can ...

2

No. The general relation is given by $$v = \lambda\nu$$ Where $v$ is the velocity of the considered wave and $\lambda$ and $\nu$ its wavelength and frequency. Of course in the case of an electromagnetic wave which is traveling at the speed of light you gain $$c = \lambda\nu$$ If you're treating instead some massive particle, then thou have p = ...

2

EM waves are formed when an electric field couples with a magnetic field. The magnetic & electric fields of an EM wave are perpendicular to each other & to the direction of the wave. The wavelength is just that--the length of the wave through one frequency cycle.

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