# Tag Info

3

I find the explanation given in the first paragraph of Wikipedia article is pretty good. Let me just elaborate some aspects to make it more clear. Megaphone is simply an extension of your vocal tract. Therefore the acoustic impedance of the whole system rises so the pressure and volume flow variations at your vocal chords may grove. A trade-off is ...

3

When one shouts , the sound waves disperse in a semicircle , the power of the voice cords distributed to 180 degrees. A simple megaphone channels the sound in a small angle and thus is directional and stronger. Electric megaphones amplify the sound and still send it in a narrow cone. At the output of the cone the sound wave spreads, but it still is much ...

2

The speed of sound is given by: $$v = \sqrt{\gamma\frac{P}{\rho}} \tag{1}$$ where $P$ is the pressure and $\rho$ is the density of the gas. $\gamma$ is a constant called the adiabatic index. The equation should make intuitive sense. The density is a measure of how heavy the gas is, and heavy things oscillate slower. The pressure is a measure of how stiff ...

0

Sound waves propagate through a medium as the result of collisions between molecules. At higher temperatures, molecules have greater kinetic energy, and as they move faster their collisions occur at greater frequency and they carry sound waves faster. Greater kinetic energy = less inertia = increased speed. However, as sound waves are compressional waves ...

2

Yes sound is a goldstone mode. Consider, for example, an ideal gas with particles at positions $\mathbf{x}_i$. There is a symmetry where we can displace each particle by some displacement $\mathbf{u}$. Of course this symmetry breaks spontaneously. By definition, we only observe $\mathbf{u}=\mathbf{0}$. The goldstone modes corresponding to this symmetry are ...

0

Yes, it's like that. You have described the basic conception of retarded potential in acoustic radiation (which will be one of the typical relativistic topics in electromagnetism). You have not specified the medium between the device and the listener but it actually doesn't matter. For more media you only need to specify more times and subtract them all ...

-4

sound needs to bounce off of air and gravity. so if theirs no gravity sound can not bouse off of sound

0

You need to consider an impedance discontinuities at the ends of the tube. In simplified model the duct has non-zero impedance, ending of the stopped pipe has infinite impedance and open end zero impedance. Therefore the reflection occurs and a standing wave can be created. Number, frequencies and amplitudes of the modes depend on the parameters of the ...

1

Let's review the linearisation and go to the further details. Just the pressure might be not enough. Take the momentum equation: $$-\frac{1}{\rho}\nabla p = \frac{\partial \vec{v}}{\partial t}+\vec{v}\cdot\nabla\vec{v}$$ Here we have to eliminate the convective part $\vec{v}\cdot\nabla\vec{v}$. Usually the argumentation is that changes of the velocity ...

0

When comparing light waves and sound waves in this fashion, we need to consider what is waving. In a sound wave, the position of air molecules are waving. In a light wave, the strength and direction of the electromagnetic field is waving. This does not exert any force on air molecules (actually it does, but that force is so small, and the frequencies are ...

1

Water forms close to perfect spheres in zero gravity due to it's surface tension. There's a variety of videos of water in the space station. Ice, assuming you start with one of those balls of water, you have to ask first, would it freeze outside in (say, the temperature of the station is dropped below 0 C), or would it freeze inside-out, say you stick a ...

0

Sound will behave just like on earth provided it has a medium to travel through (Astronauts on the ISS can communicate normally; watch some videos)

1

The very famous Newton-Laplace equation is a relation between the speed of sound and the pressure of an ideal gas. It can be written as: $$v = \sqrt{\gamma P / \rho}$$ where v is the velocity of sound in the given medium, P is the pressure, γ is the ratio of the heat capacities for the medium and ρ is the density of the medium. The Newton-Laplace was ...

0

The dB scale is logarithmic, so when you have two dB levels, their difference is their ratio. Going from -15dB to -7dB is an 8 dB step. It's that simple. Doing the math more explicitly (and showing where you equations come in): If a signal has a dB value $d$ then its intensity is $10^{d/20}$ (that is just inverting your expression for dB value given sound ...

-4

Sonic booms happen when an object crosses the sound barrier. Light leaves its source at the speed of light, so it never creates the effect, even in a circumstance where the light is strongly interacting with the air, like when there is fog.

0

Yeah, these theories aren't a piece of cake at all. In his well-known article dealing with horn modeling (reading that might be useful!) Dan Mapes-Riordan used the first of your possibilities, i.e. circular piston in an infinite baffle: $$Z = \rho_0 c_0\pi R^2 \left(1 - \frac{J_1(2kR)}{kR} - i\frac{H_1(2kR)}{kR} \right)$$ where J1 is the 1st order 1st ...

21

There are many differences between light and sound waves noted in other answers, such as the impossibility of any object with nonzero rest mass reaching lightspeed. However, there is one likeness that I don't think has been noticed yet and that is the following: a sound wave travelling at the speed of sound does not make a sonic boom! This is because the ...

63

A sonic boom is produced when a macroscopic object (say, roughly: larger than the average spacing between air molecules, $\approx 3\,\mathrm{nm}$) moves so fast that the air has no time to “get out of its way” in the usual way (linearly responding1 to a pressure buildup, which creates a normal sound wave that disperses rather quickly, more or ...

38

I know that when an object exceeds the speed of sound[340 m/s] a asonic boom is produced .Light which travels at 300000000m/s [much more than the speed of sound] doesn't produce a sonic boom right? Why? The answer is already in your own question: just because light is not an object. Sound "is a vibration that propagates as a typically audible ...

3

There are just two requirements, 1) correct frequency, and 2) sufficient amplitude. The correct frequency is, the resonant frequency of the glass cup (pane, cube, etc.). You will know you have sufficient amplitude, when the glass breaks! Both requirements will vary, depending on the material, shape, dimensions of the object, and other variables. If you ...

0

From Wikipedia, the free encyclopedia A sonic black hole (sometimes called a dumb hole) is a phenomenon in which phonons (sound perturbations) are unable to escape from a fluid that is flowing more quickly than the local speed of sound. They are called sonic, or acoustic, black holes because these trapped phonons are analogous to light in astrophysical ...

3

To the point of Is it hard to measure the resonant frequencies directly: it's tricky and careful discussion of the measuring procedures is needed. Some of the main problems: Destruction of the open-end behavior: If you place the speaker and microphone in front of the vocal tract to measure the response, you may have just switched open end behavior of your ...

4

There are two primary factors that allow the cochlea to isolate frequencies. These are generally referred to as passive and active properties: tl;dr version: The passive properties are due to the mechnical properties of one of the membranes in the cochlea, the basilar membrane, primarily the width and stiffness at a given point. The active properties are ...

1

There is a model described in Main's Vibrations and Waves in Physics dealing with the speed of sound variations you might consider useful. Sorry, I would just comment that, but I don't have enough reputation. The other way might be to derive the speed of sound not from the ideal gas laws but from van der Waals equation, but to be honest, I've never tried ...

1

Any structure that leads to a high Q system (the glass) will work and the trick is precisely matching the resonant (natural frequency ). By mounting the glass in a clamp that dissipates energy at a lesser rate than the sound energy that feeds it, the glass is doomed regardless of thickness or lack of imperfections. If the rate of energy input exceeds the ...

2

I think The Physics of Musical Instruments (Springer Science & Business Media, 1998) by Fletcher and Rossing would be a good starting point for you. The general physical description of sound rests on the investigation of the impedance changes on the boundaries. For example: the reflection at the end of the string is caused by the discontinuity between ...

3

The resonances are quite broad: each cavity will amplify a broad range of frequencies, spanning most of or more than an octave. Driving those resonances isn't as simple as choosing a pitch. You have to do some work to efficiently couple the different cavities to your vocal apparatus, and to maintain the resonance while you're singing. The people who are ...

1

The musical property of a guitar, a violin, a cello, and indeed any string instrument, depend to a great extent on the shape of the empty space they contain. Such property also depends on the vibrational properties of the wood that encloses their empty space. Empty space properly enclosed within a thin skin is like an echo chamber that can magnify ...

0

I will second what everyone has said... you will not like what you hear. At the most general level, sound is governed by wave equations, which are differential equations. Feel free to look the details up, I'm just looking to provide an overview. Energy is never created nor destroyed (in known physics). It just changes form. Your subwoofer has changed ...

0

Would anyone want to help me on this journey? Probably but they all will probably give you the answer you don't want to hear. Which is that technology to prevent sound from affecting walls without any construction does not exist.

0

You could use the transfer matrix method. It is commonly used to model propagation of sound in porous materials.

0

was I really feeling the sound in my heart and all over my body? It is definitely possible to feel sound. This occurs when the pressure is high enough and the frequency is low enough for the sense of touch. The heart can definitely produce a sensation of pain, perhaps also that of external pressure albeit with a rather low sensitivity.

0

"Which reading is correct?" That is called measurement uncertainty and that's why we have people called metrologists. Is there a naturally occurring phenomenon with a known sound pressure level to which we can compare a sound pressure level meter's reading? Probably not, at least not any practically feasible one at the current level of technology. ...

0

Is there an object I can put on my body that would allow me to feel the bass of the music more, without picking up the vibrations from the ground? Anything rigid and lightweight will convert the sound pressure more effectively into force. That's why loudspeakers have cones. I know sound waves are redirected if they hit an object, but is there ...

0

Excitation is due to interaction (exchange of momentum, changes of volume and force) of the droplets with the liquid surface. The air volume in the bottle has its own characteristic response which emphasizes particular frequencies of the excitation. During filling, the air volume decreases and small things tend to vibrate more at higher frequencies. A small ...

1

I know very little about sound and the ways to measure intensity and power, so please correct me if I confuse terms and concepts. You probably do not want to use intensity and power at all. Intensity is the product of particle velocity and sound pressure. To me it seems that pressure is the quantity you should use. Sound power is as power as any power ...

6

First: what frequency should you hit? There are many, many different factors at play in determining the natural frequency of an object I know from experience. These are (not limited to): Thickness, density, elasticity modulus (you'll need two of those, e.g. Young's Modulus and Poisson Ratio), and of course shape. I'm not aware of any papers publishing a ...

1

well, that's good but not perfect. because in this model you can not have volume 0 unless you infinitely far from the source. the volume zero for us human should be when we can not hear any more sound. In physics it is called "threshold of hearing". so we want our function Volume to be some how that when we get pass the threshold of hearing, the volume ...

0

It sometimes occurs. It is called sonochromism. Here is my source.

1

I am reasoning my way to an answer here - I have never seen this problem before so I could be completely wrong. I think the issue is that you need to make sure the normal force is reduced as the tension on the "signal" side of the capstan is reduced. This is presumably why the slightly stiffer wire (as opposed to the rope) gave more predictable results at ...

1

Yes. a wave created with a certain frequency in the water remains the same even if the medium is changed. The only thing that is important in here is the amplitude of our wave. When the sound wave from inside of water hit the surface, some of the wave reflect back into the water and a lower amplitude wave is continued in the air. so the energy of the wave ...

-1

i don't think it has anything to do with quiteness, a typical surburban area, the noise difference during day and night ain't that great to mask train horn (which is pretty loud). It's the air temperature at work. In the daytime the ground air is warm, bends airwaves upwards so for certain distance, you ain't gonna hear any sound at all, but during the ...

1

There are distance measurement devices commercially available, but if none will do, then I recommend making your own using a pulsed laser, a detector, and accompanying electronics. Another method would be a small buoy with a radio transmitter and a sound generator inside. The radio transmitter sends a signal once per minute and the sound generator emits a ...

0

The interior walls of the cochlea are covered in tiny hairs called cilia. The longest hairs are in the vicinity of the window, and the hairs get shorter as one moves deeper into the cochlea. The hairs respond to different characteristic frequencies - shorter hairs to higher ones, longer hairs to lower ones. The brain interprets signals from nerves connected ...

2

UPDATE - With a reference to: http://www.researchgate.net/publication/48323925_Applying_physics_makes_auditory_sense__a_new_paradigm_in_hearing OP, user263399, COMMENT: Can you explain the phase wave and its cause? On reading the linked paper I'm confused on how ther explanation involving the change in liquid volume velocity would create a ...

0

Yes it is customary when talking about penetration length to mean the length when the intensity has dropped to 36.8% of its surface value. This is motivated by the fact that the intensity decays as $I(z)=I_o e^{-z/d}$. See you need a constant of dimension [Lenght] in the exponent denominator, so you call that $d$ the penetration length then its immediate ...

0

I'm surprised to see no mention of turbofan jet engines in the answers so far. In fact, the blade tips of most modern turbofan engines do reach supersonic speeds. As predicted by a comment above by tpg, this does produce shockwaves from each blade tip and what you hear is a 'buzzing' sound, which is commonly described as sounding like a buzzsaw. If you ...

0

you can simply use the formula as Velocity = 331.4 + 0.6*Temperature + 0.0124*Relative_Humidity Temperature is in Celsius Degrees Relative Humidity can be measured by sensors in %age

1

When you pluck the string, you impart energy into it that's slowly radiated as sound. There are ways to radiate the energy faster, in which case the string loses energy faster. You're increasing the power and decreasing the time, so energy stays constant.

1

Obviously, the sound waves, which cause a hearing sensation in our ears, cause physical movement of the eardrum and the like. A ‘wavy movement’ is observed on the basilar membrane, with a short wavelength, with which in combination with the accompanying vibration frequency by means of the equation [propagation velocity = frequency × wavelength] an extreme ...

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