# Tag Info

1

In addition to the other answer by Cort Ammon, I have heard of other psychophysical/evolutionary explanation: The frequency distribution of that sound closely corresponds to the frequency of a crying baby, which has been shown to drive people crazy when exposed to it for a short amount of time (we are genetically predisposed to get distressed by such a ...

1

From http://www.livescience.com/16967-fingernails-chalkboard-painful.html: Interestingly, the most painful frequencies were not the highest or lowest, but instead were those that were between 2,000 and 4,000 Hz. The human ear is most sensitive to sounds that fall in this frequency range, said Michael Oehler, professor of media and music management at the ...

63

The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to $$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^... 18 This source shows that for free beams like the bars of the glockenspiel, angular frequency \omega=2\pi f and wavenumber k=\frac{2\pi}{\lambda} are related by$$\omega^2=\frac{YI}{\rho A} k^4$$where I=\frac{1}{12}bh^3 is 2nd moment of cross-sectional area about a horizontal axis through the centre, and A=hb is the cross-sectional area. In the ... 1 If you've got an older web browser kicking around that still runs Java applets you should check out Paul Falstad's Loaded string simulation. You can add harmonics to your heart's content. 2 Basically Oscars answers says it all, but I just want to add a few more things. When a string is plucked its motion need to follow the wave equation$$ \frac{d^2}{dt^2}y(x,t) - c^2 \frac{d^2}{dx^2}y(x,t) = 0 $$with Dirichlet boundary conditions (the ends of the string are fixed). c is the speed of sound of the string's medium. The function y_n(x,t) = \... 2 If a string has multiple waves expressed in it, this is done by adding the waves individually. Each frequency in the harmonic series can be expressed by a wave, a guitar string is the sum of these waves in different proportions. The resulting wave is significantly different than the others. See below for the sum of the first three frequencies in the ... 3 Loudness will not help. You do indeed have a (big) problem with reflection - the geometry is really designed to make the poor rat's task impossible. You need to manage the reflections. Since you state that modifying the surface is not an option, I would recommend tapering the walls - instead of being vertical, you might place them at a 15 degree angle (... 2 You have the correct conclusion, and I think you have the correct analysis, but I don't fully understand your presentation. Think of it this way: If the displacement occurs only between "neutral" and "forward", then the average density (air molecules per volume, or spring coils per length) over the entire system (air chamber or spring) must increase. But ... 1 There is (in theory) no limit to the number of nodes N or anti-nodes A, except that these numbers cannot differ by more than 1. Also, standing waves in the pipe of length L can have other wavelengths \lambda besides those you have stated : for a pipe closed at both ends N=A+1 and \lambda=\frac{2L}{A} for a pipe closed at one end N=A and \... 8 Carrying out the Fourier transform, I get a slightly different result for the frequency spectrum than 'knzouh'. I used u instead of y and c instead of v, so the PDE becomes:$$u_{tt}=c^2u_{xx}-Au_{xxxx}$$Fourier transforming the equation:$$F\{u_{tt}\}=F\{c^2u_{xx}\}-F\{Au_{xxxx}\}$$Transforming x to k:$$\hat{u}(k,t)=\int_{-\infty}^{+\infty}u(...

1

Most sounds (even the sound of a "single note") contain multiple frequencies. For pure sounds, there is the fundamental frequency and its harmonics, but almost any "real" sound contains some additional components - due to the envelope of the sound (e.g. the fact that a string must be plucked, then decays) or due to sampling (your sample is finite - so there ...

20

In plain English, there is stiffness at the ends of the strings where they are fixed in place, which makes the string's frequency of vibration slightly higher (sharper)—effectively shortening the length of the string slightly, for all practical purposes. And the resistance to bending is dependent on the frequency. It behaves more “stiffly” with regard to ...

158

This effect is known as inharmonicity, and it is important for precision piano tuning. Ideally, waves on a string satisfy the wave equation $$v^2 \frac{\partial^2 y}{\partial x^2} = \frac{\partial^2 y}{\partial t^2}.$$ The left-hand side is from the tension in the string acting as a restoring force. The solutions are of the form $\sin(kx - \omega t)$, ...

1

I think the sound makes the particles move more because there's more interaction when the sound particles are included, but I'm probably wrong because I don't know much about physics...

1

The speed of sound is not constant. It is a function of the elastic modulus (or its closest analog in the material) and the density of the material. In gases, it noticeably depends on temperature. In solids, the speed of sound depends on the kind of wave; different kinds of wave have different elastic moduli. For purposes of estimating the speed of sound,...

1

Consider a box full of air in stady state (closed system). If the volume of that box is $22.4~\mathrm l$ there are $6.022\cdot10^{23}$ particles inside that have maxwellian distribution of speeds. Those speeds are randomly oriented and we can defocus from single-particle description to whole body desctiption supposing it is continuous and homogeneous. Then ...

26

It does, but the effects are negligible in the regions we think about. If you think about a volume of air as a box of atoms bouncing around, you can apply an oscillating pressure gradient across that box and show that it behaves close enough to an ideal wave propagation medium that you can get away with using such an ideal model. The variations you are ...

0

I think the sound wave DOES spread out and dissipates over enough distance. Consider this: A small stick of dynamite explodes in a flat empty field. The compression wave of the air (which is what cuases sound) moves away from the source of the explosion at about 340m/s and propagates in a hemispherical manner outward. One second after the explosion, ...

1

I think this may be correct. The individual air molecules are indeed moving about randomly, and oscillating. When a large amount of energy is generated, these random motions are superposed by an alomost uniform behavior, making the propagation of sound possible. This is quite like the free electrons in a metal. Without a potrntial difference, the motion of ...

16

They do, its just usually negligible in practice. There's also scattering because the particles are not all the same (H2O, N2, O2, etc.)--but that, too, is usually negligible. Its mainly because there are so many particles in a single wave. Consider that the wave must be extremely short before it becomes noticeable (megahertz).

5

Interesting. In your title you say "background noise." So I was going to suggest the sea shell effect - putting a sea shell to your ear you think you hear the sea, even if it is far away. If you did think you heard music, that could have been an illusion. Your brain is wired to look for patterns - see animal shapes in the clouds, hear people talking (or ...

0

Intensity is energy per unit area; over short distances, the intensity can be considered constant. In a time period dt, the sound wave travels a distance c.dt, so the total energy passing through an area A will be equal to the sound energy present in a volume cAdt. (think of water flowing through a pipe, if it travels at 10 m/s through a pipe with a 0.1 m ...

3

In this type of problem one has to take great care in defining intensities. In this case there are 4 different intensities: 1. $I_{ss}$, the intensity received by the static observer as perceived by himself, 2. $I_{ms}$, the intensity received by the moving observer as perceived by a static observer. 3 $I_sm$, the intensity received by the static observer as ...

3

The maximum and minimum are "local" values. As you move closer to A (at 0.2 m you are MUCH closer to A than to B) the amplitude of A is much larger - so although there may be destructive interference between A and B at that point, this is by no means perfect interference, and the resulting amplitude is still quite large (lot of A minus a little of B). ...

1

With respect to the air the wavelength will be $\lambda = 343/f$ everywhere. With respect to the ground the wavelength will be $\lambda = v_{sound}/f$ but $v_{sound}$ will depend on position with respect to the source - picture your classic Doppler shift diagram with a set of not quite concentric circles. Downwind from the source the wavelength, measured ...

2

To add to the existing answer, I think there is a nomenclature issue. When you say "bass" people understand "low frequencies" but what you probably mean is "beat". Rapid changes in amplitude, like a beat, carry a lot of high frequencies. You do hear mostly the beat from other peoples' headphones, ans it's annoying. You can think about the extreme case: the ...

0

There is a difference between pressure and displacement: pressure is "absolute", displacement has a direction. With speakers on the left and right, pointing to each other, the pressure wave from the right will be the mirror image of the one from the left, the displacement wave is not only mirrored but also changes sign (mirrored by the x-axis). Taking that ...

2

First, I am going to provide a little background on equivalent pressures at different altitudes from Earth's surface. Layers of Earth's Atmosphere Troposphere to Mesosphere At sea level, the neutral atmosphere of Earth has a pressure of ~$10^{5}$ Pa (or ~1000 mbars). The below image from https://en.wikipedia.org/wiki/File:...

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