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## Hot answers tagged frequency

16

Colour is defined by the eye, and only indirectly from physical properties like wavelength and frequency. Since this interaction happens in a medium of fixed index of refraction (the vitreous humour of your eye), the frequency/wavelength relation inside your eye is fixed. Outside your eye, the frequency stays constant, and teh wavelength changes according ...

11

The human ear responds only to the intensity $I$ of the sound it receives (more specifically, to the intensity distribution over the different frequencies) and this goes more or less like the square of the amplitude, $$I\sim A^2.$$ Changing the sign of the waveform changes the sign of $A$, which has no effect on $I$.

10

As FrankH said, it's actually energy that determines color. The reason, in summary, is that color is a psychological phenomenon that the brain constructs based on the signals it receives from cone cells on the eye's retina. Those signals, in turn, are generated when photons interact with proteins called photopsins. The proteins have different energy levels ...

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A human eye may only distinguish thousands or millions of colors – obviously, one can't give a precise figure because colors that are too close may be mistakenly identified, or the same colors may be mistakenly said to be different, and so on. The RGB colors of the generic modern PC monitors written by 24 bits, like #003322, distinguish $2^{24}\sim ... 8 A. All light sources (even lasers) are subject to a diffraction limit, so any light beam will eventually diverge with an angle$\theta$given by $$\theta \approx \frac{\lambda}{A_T}$$ where$\lambda$is the wavelength of the light and$A_T$is the aperture of the light beam source (and "eventually" means for distances much greater than$A_T$). Any beam ... 8 The colors which are perceived by people are defined by the degree to which the light will excite the red,green, and blue photoreceptors in the cone cells of the eye. There are only three discrete colors we can perceive, and they are red, green, and blue. The statistics of the relative and absolute excitations, the amount of red,green, and blue averaged over ... 8 The speed of light in vacuum is constant and does not depend on characteristics of the wave (e.g. its frequency, polarization, etc). In other words, in vacuum blue and red colored light travel at the same speed c. The propagation of light in a medium involves complex interactions between the wave and the material through which it travels. This makes the ... 7 The physics is actually much easier than it seems at first glance. Power generators are engines just like the everyday ones we see all around in our cars, lawnmowers, snowblowers, etc. Except for new power sources like some wind and solar systems with electronic inverters, the vast majority of power is supplied by large rotating AC generators turning in ... 7 Your voice, like any sound, is a combination of many frequencies. Physically, your voice consists of pressure waves. If we plot the pressure as a function of time, we see that it goes up and down in a way that looks somewhat random. You can measure these pressure waves with a microphone, then visualize them with an oscilloscope. Here's a Youtube video ... 7 Re your last question: what you've achieved is essentially the same as if you wire one of your speakers the wrong way round so it moves in antiphase to the other speaker. In principle there will be points equidistant from both speakers where the sound waves cancel and you get a quiet spot. However as soon as you move closer to one speaker than the other you ... 6 I'm getting the impression that a good part of this question (and perhaps also this physics.SE question?) arises from a wrong presumption that time and position should be on equal footing in quantum mechanics. They are not. The position$\hat{\bf r}$is an operator, while time$t$is a parameter. (Notation: In the following boldface denotes a vector ... 6 (This is an intuitive explanation on my part, it may or may not be correct) Symbols used:$\lambda$is wavelength,$\nu$is frequency,$c,v$are speeds of light in vacuum and in the medium. Alright. First, we can look at just frequency and determine if frequency should change on passing through a medium. Frequency can't change Now, let's take a ... 6 It would depend on damping effects being taken into account or not. Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term):$m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right)$Where$m$is the ... 6 If you have some electromagnetic wave e.g. a plane ave: $$E = E_m sin(kx - \omega t)$$ then the energy transport is given by the Poynting vector. For the plane wave above the energy transport works out to be: $$S = \frac{1}{c\mu_0} E_m^2 sin^2(kx - \omega t)$$ To calculate average energy transport we note that the average value of sin$^2$(anything) ... 6 Probably not. A fresnel lens isn't just a rippled surface, it has discontinuities, or straight edges. The area of these edges mostly causes loss of incident power. The optics designer wants a good ratio of its (aspheric) area of use to its unused area at edges. Sound and other vibrations could create sine wave-like ripples on the surface of a liquid, but ... 6 Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field$\mathbf{E}$(i.e., light), its movement can be described by the equation$$... 5 I can't claim any experimental experience in this area (fortunately :-) but I thought it was interesting enough to be worth a bit of Googling. The results suggest there is a difference between shells and bombs. There is an extensive collection of eye witness accounts of WW2 at http://www.bbc.co.uk/history/ww2peopleswar/categories/, and searching this ... 5 Power consumption is about linear with frequency. The processor contains millions of complementary FETs as shown. When the input goes low the small capacitance gets charged and it will hold a small amount of energy. A same amount is lost during the charging. When the input goes high again the charge will be drained to ground and be lost. So with each ... 5 Because the frequency and intensity are not related to that extent (by which you got yourself confused). Frequency gives the number of vibrations per second (and so, the unit$s^{-1}$). As @daaxix says, you have used intensity to mean irradiance. Irradiance is the amount of power the wave has delivered per unit area and hence,$W/m^2$. Both are quite similar ... 4 If you are considering human vision there is a definite (and surprisingly small) number of distinguishable colors. This is known as a MacAdam diagram and shows a region around a single colour, on a chromaticity diagram, that is indistinguishable from the color at the centre. The total number of colors would be the number of ellipses needed to completely ... 4 A. Because the total energy is divided over the whole sphere$4\pi R^2$of radius$R$around the source of light which means that the energy density over unit area goes like$1/R^2$. However, this is only true for unfocused light - with chaotic directions. Lasers may create coherent light whose intensity per unit area doesn't drop. The beam may move for ... 4 The main problem with RF energy harvesting is the low power density (<1µW/cm^2, unless near a transmitter). Other approaches are generally more useful, though even picowatts can be enough for some applications. Some links: Overview of Energy Harvesting Systems (for low-power electronics) Energy Scavenging/Harvesting Ambient RF Energy Harvesting 4 In order for the intensity of a light source to stay the same, while each lower frequency photon carries less energy, there must be a greater number (per time, per area) of the lower frequency photons in the beam than the original number of higher frequency photons. As for the second part of your question, I admit that it can be confusing that the power ... 4 I did a bit of discussion on this subject in this thread on Music.SE. The fundamental doesn't necessarily have the strongest amplitude. As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each$\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is ... 4 The oscillating magnetic field associated with an EM wave will induce a voltage in any electrical conductor that it passes through. So in principle all you need to do is stretch a piece of wire across your room then measure the voltage across it. However, as usual, the devil is in the detail. If you've ever listened to a radio in your room then you've done ... 4 For any spatially periodic wave, sinusoidal or not, the wavelength is just the spatial period, by definition. It is true that if you compute the Fourier decomposition of a non-sinusoidal wave, you'll get a potentially infinite number of frequency components, but all of these will have wavelengths that are integer factors of the wavelength of the overall ... 4 To add to the "linear with frequency" point, there is also an additional factor. As that "dynamic power" increases, the temperature of the die will increase and this will also increase the leakage current through the millions of transistors, which will cause more dissipation (termed "static power") There's a long Anandtech thread taking lots of values and ... 4 I would have to see the words in context, but the description "apparent frequency" seems strange to me. The frequency your ear detects is exactly what the most sophisticated scientific instrument would measure. The Doppler shift is real in that the frequency your ear detects is really the sound frequency in your frame. I would guess "apparent frequency" ... 4 The reason for this is mainly operational, rather than to do with the laws of physics. Radio spectrum is a very scarce resource, and is managed independently by each country's regulatory authority. In order to allocate spectrum to a mobile operator, the national regulatory authority has to make sure that spectrum is not being used by any other services. ... 4 The light intensity$I$is the power$P$, that irradiates an area$A$,$I=P/A$, therefore the units$\mathrm{W}\mathrm{m}^{-2}$. This means, that if you have two beams with equal intensities, the area is the same and the power has to be the same as well. You are probably confused by the relation between frequency and energy of a photon,$E_\nu = h\nu\$. But ...

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