# Tag Info

1

The assumption when linearizing is that the deviations/perturbations are very small compared to the reference (averaged) values. Typically the derivatives of the deviations are of the same order as the deviations themselves. Consider the deviations having this functional form in 1D: $$u'=\Delta u\sin kx\quad \partial_x u'=k\Delta u\cos kx$$ The deviation ...

4

As Thomas has commented, the trick is that we only assume first order terms and this convective acceleration would be small of the second order. In fact, that is one of the first assumptions to drop when you consider more general cases. See e.g. Burger's equation for first generalizations and/or Lighthill's equation for source terms arising in the wave ...

2

Sound is a compression wave not a shear wave, so the viscosity of the liquid has no (direct) effect on it. The speed of sound in a liquid is given by: $$v = \sqrt{\frac{K}{\rho}}$$ where $K$ is the bulk modulus and $\rho$ is the density. The viscosity does not appear in this equation. A quick Google found data on the speed of sound in liquid helium in ...

3

First, some fraction of incident sound power will pass thru any object, human body or not, so your title makes no sense. The real question is how much this body will attenuate the frequencies of interest. Below some attenuation, you either don't care or it's below the noise floor of the sensors to detect it. However, this depends on what you care about, ...

0

Let's use the balloon as an example. When you inflate it the contents are under pressure because the latex wants to shrink back to its natural size. When you pop the balloon that pressure is released. It rushes outward temporarily creating a void. The void then sucks in the surrounding air. The popping sound is the air molecules slamming into each otherwhen ...

1

Nice question and lovely piece of optimization! I have been thinking about that for a while. I can only answer from a point of physical music acoustics. It appears to be interesting feature for drum design and construction but my considerations end with a conclusion that it is not much useful in praxis. Here is why: Case 1 - Directly struck drum: the sound ...

0

It's actually not that complicated for a linear case. Let's derive the 1D wave equation for velocity potential $\Phi$ ($v = \partial_x \Phi$). As usual, $c_0$ denotes the speed of sound. The kinetic energy density is obviously $$\mathcal{T} = \frac{1}{2}\rho_0\mathcal{v}^2 = \frac{1}{2}\rho_0\left(\frac{\partial \Phi}{\partial x}\right)^2$$ Potential ...

0

In case someone else have this question, I finally found that Goodman (Introduction to Fourier Optics, ISBN 9780974707723) explicitly states that the Fresnel approximation is indeed valid in the far-field.

2

Good question. My Problem is that I can't give you a "Mainstream answer". The reason is that Turbulence is considered happening In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations. But with these ...

0

No, certainly not. In linear model there is something like static tension $T_0$ and time dependent tension $T'$. When there is for example a standing wave on a string with fixed ends, there is a minimum of string displacement (zero) but maximum of string tension $T'$ at the fixed ends. (As an effective value throught the time.)

2

Consider a small segment segment under tension From the balance of the horizontal axis you have $$-T \cos \theta + (T+{\rm d}T) \cos (\theta+{\rm d}\theta) = 0$$ $${\rm d}T \approx T \tan(\theta) {\rm d} \theta$$ Integrating by separation of variables $$\int \frac{1}{T}\,{\rm d}T = \int \tan(\theta)\,{\rm d}\theta + K$$ $$T = \frac{{\rm e}^K}{\cos ... 2 Good theoretical answer is that it results from linear acoustical wave equation and its presuppositions. It is therefore good approximation whenever linear acoustics still can describe the wave propagation (that would by e.g. 90% of room acoustics practical examples). Typical examples of problematic models are large-amplitude events (e.g. a shockwave after ... 1 The spectrum of various resonant tube arrangements (half-open, fully-open fully closed) is something that can be measured in a very basic laboratory and gives solid evidence that the claim is true over the kinds of frequencies that are accessible in such a lab. Say a few hundred to a few thousand hertz. 3 It's a bit pat maybe, but if the wavelength-dependence were detectable over human distance scales, the (quality of) sound (not just the volume) of music, speech etc. would depend noticeably on how far one were from the source. 2 If you are cooling your object that you wish to hear, then the exact sound will depend on the exact temperature (as given by yuki96's answer at 17nK). However, any temperature above the nanoKelvin temperature scale will sound the same, but the volume will increase with temperature (according to the Stefan-Boltzmann law). The sound of a warm blackbody (such ... 12 This problem can be solved with noise-shaping. Since the shape of the spectrum is known, it can be used as a base for the power spectral density:$$ P(f,T)=\frac{ 2 h f^3}{c^2} \frac{1}{e^\frac{h f}{k_\mathrm{B}T} - 1} $$where k_\mathrm{B} is the Boltzmann constant, h is the Planck constant, and c is the speed of light. This outputs the relative ... 0 This is not trivial, but is routinely done in laboratories and commercial products (like surround stereo systems). If we consider the strictly physical aspects of sound localization in humans (and probably many other living beings who cannot conscientiously orientate their ears), there are two main phenomena to take into account: 1. Interaural time ... 0 Let's do some rough estimation. Considering: CuriousOne's very good and fitting comment with regards to geometrical estimation given in question and the last but not least the human physiology (10^{-12} \ W can hear only a full healthy man, with perception much less than a whisper and only on a frequency range improbable due to the diffraction) we ... 2 Strictly speaking: never. Outside the idealized models there are always at least a bit of radiation impedance so the pressure on the boundary is not truly zero. But let's focus on idealized models. Usually the zero pressure boundary condition is used for open ends of the waveguides or for the free surfaces of solid bodies. Typical illustrative example ... 1 Decibels are defined as:$$ L_P = 10\log_{10}\left(\frac{P}{P_0}\right) $$This is not unique to sound, any quantity with units of power can be represented using decibels. Decibels are a ratio between two powers, on a logarithmic scale. With regard to sound pressure, sound pressure \rho_s \propto P^{1/2}  so:$$ L_\rho = 10 ...

4

my assumption is that I will hear a single note humming in a constant state. A sound wave is not a thing that you can hear. Assume for a moment that you are just standing in the coffee shop, enjoying the music. What you are hearing is not the waves. What you are hearing is the guitar. The waves carry acoustic energy from the guitar to your ear. The ...

0

You would hear Whoosh... and Boom! The sound of air rushing past your ears traveling at the speed of sound and the shock waves would entirely drown out the sound of any guitar even with the most powerful amps! ;) And if you say well then remove the air and do the experiment in space - you will not hear any sound since it requires the air to propagate. ;) ...

1

I'm pretty sure you will hear nothing from the guitar, because you will be traveling at the same speed as the sound so the sound wave will not be able to make vibrations on your eardrum thus you hear nothing.

-3

They do constructively interfere. Our ear/brain sorts it out. We do not 'hear' sound waves. We sense spectra.

1

The key thing to keep in mind is that the phases of the different signals (including the room reflections) are effectively random. In addition to exact frequency matching, complete destructive interference requires that the the two signals be $180^\circ$ out of phase. Because of the phases of the different signals are uncorrelated with one another, the ...

5

There were certainly no sound waves before the universe began. The article was talking about a period of time after the big bang during which the "stuff" in the universe was so hot that no atoms were stable, so atoms made of protons, electrons, and neutrons broke into their component parts almost immediately even if any happened to be created. No atoms ...

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