# Tag Info

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This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that do not yet have commentary. Acoustics by L. Beranek An Introduction to ...

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The question could be rephrased as, is it possible to create an analogue of an "image" of a three dimensional object acoustically? The answer to this question is clearly yes, provided that the wavelength of the sound used to do the imaging is sufficiently small compared to the object being imaged. One example is in the energy industry - geophysicists use ...

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The process of auditory masking can partially answer the question. A constant wind can present fairly high intensity middle and high frequency noise (pink noise) to the ear. This will mask these frequencies in the speaking voice. These frequencies (>1 kHz) are generally responsible for the "diction" of speech, so you may be able to tell that someone is ...

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Sound waves in air are a series of high and low pressure areas moving through the elastic medium of the air. If the medium is being compressed and rarefied by other areas of high and low pressure, the integrity of the sound wave may be broken and may become unrecognizable through the background noise of the medium's turbulence. Here is an explanation of ...

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It doesn't have to be a drinking glass. It's the stethescope principle: by placing a rigid object against the wall, and the other end of the rigid object at your ear, the sound energy is contained within the object rather than dispersing into the air. A drinking glass has the added advantage of a nice flat surface to push your ear against, and may ...

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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.

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The reason that the speed of sound is a well-defined quantity is that, for small pertubations, the equations which govern the fluid dynamics can be linearised. In that linearised form, the solution boils down to a simple wave ansatz with linear dispersion relation, i.e. constant velocity.Those are the sound waves. It so happens that in air, this linear ...

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Oddly enough, supercavitation can help increase max speed. According to the not-necessarily-correct PopSci, the Navy has achieved Mach 1: supercavitating torpedo . A link in wikipedia claims the Germans have achieved 800 km/h . Other than the Soviet Shkval, I can't find any updates on such torpedos, so either the Navy programs have gone deep black ...

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Speed of sound in water at 20 degrees Celsius is 1482 m/s., (2881 knots), just for comparison to current claimed achievable speeds. Small related fact: The pistol shrimp can create sonoluminescent cavitation bubbles that reach up to 5,000 K (4,700 °C) which are as loud as 218 decibels, breaking the sound barrier in water. Says Wikipedia YouTube video of ...

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The maximum change of pressure caused by a sound wave is its pressure amplitude. This would be the difference between high and low pressure areas in the sound wave. When sound is measured in pascals, however, for the purpose of computing decibels by comparing with other sounds, it's just the high pressure against the measuring surface, to the extent that ...

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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 ...

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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.

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Actually no because to set up a sound wave you push on some particles and they in turn push adjacent particles.So all you can try is pushing the first set of particles harder,but eventually they will collide with adjacent ones.So you ending up producing oscillations instead of collective motion.

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The answer is: not much. The usual analogy is waves at the beach. You can see a long line of waves rolling towards the land, but the water never actually moves across the land. Waves, whether water or sound, transmit energy by passing it along from one particle to the next. The medium itself oscillates but doesn't really go anywhere. There are ...

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Sound waves are changes in the pressure density of air. In other words, the air is already moving. A fan makes a sound when it strikes air because it's forcing it together and then it rebounds into adjacent air which undergoes yet another compression and expansion.

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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. ...

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Yes this is related to Stoke's law for sound attenuation, which states that a plane wave decreases amplitude exponentially with a factor $\alpha$ given by: $$\alpha = \frac{2\eta\omega^2}{3\rho V^3}$$ where you can see that the dependence on the frequency squared $\omega$ of the sound will yield a higher coefficient of attenuation for higher frequency ...

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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.

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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 ...

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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 ...

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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 ... 0 Another way to view the same relationship which @JohnRennie gave is $$v=\sqrt{\gamma\frac{RT}{M}},$$ where$R$is the ideal gas constant,$T$is the absolute temperature, and$M$is the molecular mass of the gas.$\gamma$is the heat capacity ratio, as mentioned in his answer. Since you were directly given the densities, his form works more directly, but ... 1 The speed of sound in an ideal gas is given by: $$v = \sqrt{\gamma\frac{P}{\rho}}$$ where$P$is the pressure,$\rho$is the density and$\gamma$is the heat capacity ratio. For ideal diatomic gases$\gamma = 1.4$. In fact for air at 20ºC$\gamma$is almost exactly 1.4, while for hydrogen it's 1.41, so pretty close to ideal behaviour. 0 There's no such formula. You can think of pressure as the (negative) spatial derivative of displacement. If the particles at$x = -\epsilon$are displaced a little to the left and the particles at$x = +\epsilon$are displaced a little to the right, then there's low pressure at$x = 0$since less particles are there. With that in mind, ... 1 There isn't a simple formula for fan noise, but the physics can be worked out from the fundamental equations of fluid mechanics and acoustics. It isn't a simple problem however. The noise created by fans is complex and from several fundamental sources, and its amplitude depends on frequency. Here is an example of a fan noise frequency spectrum for a cooling ... 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 ... 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. ...

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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 ...

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Yes, the sound can be reversed. Thanks to JiK, we have this animation (Python source code) of a supersonic jet moving forwards that can illuminate what is going on: The red circle represents the first sound produced by the object, the blue circle the second sound produced by the object and the remaining (black) circles representing the sounds produced ...

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As you stated, the observer would not hear anything as the plane approached him. If fact, he still would not hear anything until the plane had past by since it takes some time for the sound to travel from the plane to the observer. In answer to your question; no, you would not hear the sound backwards since you still hear the sound in the same sequence as it ...

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