# Tag Info

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It's impossible to say without seeing your experimental setup, but an obvious explanation is that the ultrasound is heating the water and you're seeing plain old thermal convection currents. You say you don't get convection currents in an ultrasonic bath, but my recollection of using (admittedly quite powerful) ultrasonic baths is that you get quite ...

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Sound waves are air vibrations which means that the particles of air are not displaced permanently (in ideal case), while "puff of air" is a kind of "wind" meaning permanent displacement of air particles. In practice both these phenomena often go together, like in the case of hand clapping (or lower quality bass loudspeakers for that matter). EDIT: ...

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Based off of that video, the differences you're pointing out are the nice wavefronts from the speaker at 2:04 and then the clap shown at the beginning and the end. It's true that the wave fronts from the speaker (and even the book) give nice "crests" and "troughs" whereas the clap kinda just... is this blob-y thing. There are several potential reasons why ...

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The collision of the blocks will cause them to vibrate (sound within the blocks). Since the blocks only contacted with each other momentarily (the blocks rebounded), there would have been a very short period of time for the vibrations to transfer from one block to the other (when the 2 blocks are touching). Since the vibrations are travelling fast (the ...

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Actually I have read (although I can't find a reference) that the subjectively perceived psychological notion of pitch itself, although very nearly wholly set by the sound wave's frequencies, is also weakly dependent on the intensity of the sound: that is, a higher intensity sound wave does seem ever so slightly sharper (higher in subjective pitch) than one ...

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A partial answer to where the simple linear no-Doppler model has a gap: in the coupling of the speaker cone to the air. There is a boundary condition imposed on the acoustic wave equation at the location of the speaker cone. However, the speaker cone itself is moving around, meaning that this boundary condition is imposed at different locations at ...

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The Doppler shift for small speeds is $\Delta f/f = \Delta v/c$, where $\Delta v$ is the (signed) speed of the source relative to the detector, and I'm using $c$ as the speed of sound. So let's plug in some numbers. I'm going to use numbers that will produce a large effect to see how larger an effect is plausible. Let's take a woofer operating at $f = 200 ... 1 Another thing that happens that can lead you to think that low frequency sounds attenuate quicker is that if you record yourself one time being close to the microphone and another time being farther away, you'll notice that the farther you are the more the lowest frequencies are picked up. This is due to the proximity effect and not to the low frequency ... 0 You seem to be looking for boundary layer separation, which occurs for sufficiently large Reynolds number. See Batchelor's Introduction to fluid dynamics, sections 5.8 to 5.10. See also: http://en.wikipedia.org/wiki/Flow_separation http://en.wikipedia.org/wiki/Boundary_layer 1 At low Renyolds number, the vortex shedding described in the question does not occur, based upon the following simulation videos: http://www.youtube.com/watch?v=ElmTA0t3bEc http://www.youtube.com/watch?v=8o-JC3R9YBY http://www.youtube.com/watch?v=sN9LP5dNWhc However, for higher Renyolds number see Nakamura et al. "Experiments on vortex shedding from flat ... 2 Chris White's answer using an analogy to incoherent light pretty much answers the question; it's fundamentally a question of the statistics of how wave sources add. Here's a slightly different but equivalent rephrasing of Chris White's answer using matrices: Given$N$wave sources, incoherent waves add "diagonally" ($I\propto N)$, ie, additively. ... 3 Taste: There are 5 basic tastes that the human tongue can detect. They are sweet, savory, salty, sour and bitter. These are detected by taste receptor cells on our tongue, I won't go deep into the biology part. The basic tastes of sweet, salty and sour have different thresholds, or concentration levels, at which they can be detected. In other words, it is ... 4 Taste and smell are mediated by receptors in your body that molecules can attach to. These receptors then give off an electrical signal which is translated in the brain to a certain taste or smell. The details of this are biological and not of importance here. So no, there is no relevant frequency or even wave-like behavior. Touch is a very different thing. ... 4 This is a neat question. Did you know that adding two Sine waves of the same frequency but different phase together always produces another Sine wave? Of course you can imagine two perfectly out-of-phase Sine waves that "cancel" by adding to a line but in that case you can just imagine the result as a Sine wave with 0 amplitude. Using gnuplot with the ... 0 I am going with no. If this was a question of sound spreading with an inverse square law, the answer would be yes. Place a cymbal at a distance where it has the same apparent diameter as the sun. On a quiet day, it would be audible. Place 4 cymbals at twice the distance. It would be just as loud. Repeat this idea at the distance of the sun. If the ... 2 Let me give a more detailed back-of-the-envelope approximation, which might actually be able to decide, given the conditions of the problem, if we would be able to hear the sound of Sun. Assumptions: The space between Earth and Sun is filled with uniform air. This is a non-physical assumption. It basically means we are ignoring the gravitational effects ... 1 Sound, in simple words is vibration of air. So in theory yes, we should hear the Sun if there was a medium like air that could transfer the vibrations. That's just my opinion, of course I can be wrong as this is purely theoretical question and answer. 0 You are right. Like any physical process described by linear equations, there are limits. Think of a sound so intense as to crush the cells in a sound absorbing foam, turning it into a hard surface. A reversible version of that foam is one where the bubbles don't get destroyed, they get flattened to the point that the material they are in starts to play a ... 1 Sound as you hear it is waves of pressure differences in the air, which is interpreted by your ear as sound. So no, you cannot directly hear electromagnetic radiation (EMR). You could, however, take the EMR and convert it into sound waves in the audible range, which you could listen to. This was done in 1990 for Jupiter by Voyager as it passed Jupiter. It ... 0 Electromagnetic radiation is light. Like any other star, you can see the radiation from a pulsar if it is not too far away or too dim. Pulsars and other stars do produce very loud sound. We do not hear it because there is vacuum between stars and us. Sound does not travel through a vacuum. Also stars are so very far away that we wouldn't hear it anyway. 2 The surface of the sun is where local plasma cools enough to recombine and go transparent, the photosphere. You would still be deep within the sun's atmosphere, and it would be LOUD. H-bombs are LOUD at the edge of their fireballs. 3 helioseismology is what you need to learn about. yes, there are sound waves in Sun 1 It was already mentioned by Carl Witthoft, but I think the ocarina does count, as long as you're not too hung up on the resonance chamber being a tube as such. From Wikipedia: The ocarina, unlike other vessel flutes, has the unusual quality of not relying on the pipe length to produce a particular tone. Instead the tone is dependent on the ratio of the ... 1 You don't explicitly state you are looking for a wind instrument so perhaps a drum would count. Perhaps a snare drum since the snare is on the resonant (non-struck) head or a kettle drum maybe qualifies as a pitched instrument. If you are looking for a wind instrument in particular and Carl Witthoft's suggestions of the ocarina or the jug do not fit the ... 7 It can be done, but there's some trade offs. Larger speakers are better at moving longer wavelength (low frequency) waves. When you try to combine a bunch of small surfaces in different locations to recreate a single wave you end up with a some random interference where the wave is stronger or weaker (in 3d-space) (see phased-array antenna for some ... 3 Well, does playing flute-like across the top of a beer bottle count? Or, better put: a jug-player in a country "Jug band" plays his instrument that way. It's also your call whether blocked instruments with holes along the length qualify, such as an ocarina. I sort of guess what you're looking for is an instrument with an air pocket as the resonance, as ... 0 the pitch of our voice cause the vibration of magnet in the microphone, thus causing generation of different voltages of electrical signal. this is not exactly true. the voice doesn't have the pitch. the pitch is a concept from a frequency domain. when you speak you change the air pressure over time. these changes travel in air with speed of sound, they ... 0 Here is an image of the waveform of my voice saying "hello": The blue line corresponds to a vibration in the air (pressure wave) but it's easier to imagine it as the amount a speaker cone needs to be displaced at a given time or the amount your eardrum will be displaced at a given time. The simplest common (digital) way to encode this sound as data is a ... 0 If you just want a "rough" idea of how far away the lightning struck, then the answer is yes. You can use the loudness of the thunder as an indicator of the relative distance (a loud strike will be closer than a faint strike). More accuracy could be be gained with a loudness measuring device calibrated with known average strength of strikes and how far ... 2 Think about the definition of pressure: $$P=\frac{F}{A}$$ Now, let's consider the definition of a force. $$F=\frac{dp}{dt}=m\frac{dv}{dt}$$ Hence, for a given area and particle mass, the pressure is a function of the velocity: $$P=\frac{m}{A}\frac{dv}{dt}$$ 0 The answers so far have been very good, but I will attempt to answer my own question, building on what has already been said. Lighthill and Ffowcs Williams make use of the term pseudo-sound to denote sound which can physically be observed by say a microphone, but which will not propagate into the homogeneous quiescent region. The example of a turbulent jet ... 2 Hydrodynamic perturbations = change in pressure due to a flow velocity (particles don't return to equilibrium positions). Acoustic perturbations = change in pressure due to the fact the particles undergo an elastic restoring force (for a compressible fluid) which causes perturbations to travel at the speed of sound. Any change in the pressure/velocity ... 1 That's an interesting link, explaining how if fluid contacts a plate, and if there is a vibration pattern in the plate, what vibration pattern you get in the fluid. As I read it, if the speed of sound in the plate is very high compared to the fluid, the fluid sees a plate that vibrates into and away from the fluid, creating a sound wave that propagates ... 3 It isn't possible to create an audio source in mid-air using the method you've described. This is because the two ultrasonic waves would create an audible source if the listener were standing at that spot, but those waves would continue to propagate in the same direction afterwards. You would need, as I point out below, some sort of medium which scattered ... 2 It's certainly possible to produce a sound in the sonic range by intersecting two ultrasonic signals, however, at least as the effect is generally modeled in physics, this interference pattern is only detectable at points where the original waves are intersecting and each of the original waves will continue propagating in its original direction unmodified by ... 1 Yes, a sound could become louder as you move away. One way for this to occur is if the sound waves are created in such as way as to focus to a point. As the sound is focusing the loudness will increase and after it passes through the focus the loudness will decrease with distance. @Thaina had the right idea by suggesting that an air lens could focus the ... -1 There is no way unless you see the Lightning. You need a frame of reference. Even radar detectors have a frame of reference. You need a visual approach. Or einstein to work out some new formula In anycase, Without seeing the lightning how do you know the sound came from one. It could be a Scud missile up above breaking the sound barrier. When lightning ... -1 There are a few different reasons. Let's only face the digital channel. Only a band limited signal is being used. G.711 uses a sampling rate of 8kHz resulting in a usable bandwidth of 4kHz which is left for the voice. It's OK for voice telephony but almost unusable for music. Other codecs use different bandwidths, for example G.722 (Wideband telephony) ... 0 This is due to signal processing, not physics. Telephone carriers apply aggressive compression optimized for recording only speech well. The AMR codec, still in use, dates to 1999 and achieves up to about 13 kbit/s. Any other codec would not record music well at that bitrate, either. Even MIDI consumes more data. 13 I think yes, you can get some distance information from analyzing the thunder sound. I am basing this mostly on having personally heard many thunderclaps, both far and near, of reasonably known distance to the lightning. Anyone that has ever experienced a near (100 m or less) thunderclap can tell you it sounds a lot more "sharp" than a distant one, not to ... 40 The limitation you're hearing has been part of the phone network since long before digital sampling had any part in the telephone system. It is related to the fact that the connection from a land-line phone in your house or office back to the "central office" of the phone company is essentially a continuous connection through a pair of wires. There's ... 24 This is an interesting question. Unfortunately, the answer is probably no for two reasons. There is a nice way of telling how far away a lightning strike was by counting the seconds before the thunder reaches you though. First let me tell you why your method probably won't work, then I'll tell you how to calculate the distance based on the time of arrival ... 6 I presume you did not see the flash, so cannot use that as a timing mark. From Wikipedia "The dependence on frequency and pressure are normally insignificant in practical applications. In dry air, the speed of sound increases by about 0.1 m/s as the frequency rises from 10 Hz to 100 Hz. For audible frequencies above 100 Hz it is relatively constant. " You ... 3 The simple way to do this is counting seconds after seeing a lightning flash. The light from the flash travels at$299792458\$ meters per second; for all practical purposes one can approximate it as arriving instantaneously at your location from where the lightning strikes. The speed of sound is quite a bit lower, however. It depends on temperature a bit, ...

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According to Wikipedia the frequency range of the plain old telephone service is 300Hz to 3.4kHz. So any music you listen to will be missing the low frequencies and missing the high frequencies. If you remember back to the last time you heard hold music on the phone you'll probably remember that it sounded a bit muffled, but I have to say that it's still ...

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The brain is quite good at filling in for a few missing harmonics. For example music still sounds reasonable on a smartphone speaker even though that speaker is incapable of creating low frequencies. This Wikipedia article explains the phenomenon (thanks to Glen for the link). So it's possible that if you heard the 2nd, 3rd and 4th overtones of 250Hz your ...

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Phone companies only built the phone to carry voice frequencies. Bass and tweeter frequencies are generally out of the range of what the phones were built to do. I used to listen to a radio show that when a caller called in with some lame joke, they would play crickets chirping to the person on the phone. It took them a long while and several awkward ...

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Have a look into the Nyquist theorem. The sampling frequency needs to be at least double the rate of the sampled frequency. I.e. that's why the human ear can hear up to ca. 20kHz and the CD samples at 44.1kHz. Wikipedia Nyquist-Shannon Theorem What do we hear instead if we do listen to (originally) 5 Hz to 20 kHz music through the phone? Is everything ...

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