# Tag Info

1

Actually, the reason is anthropic! The original unit for sound intensity is the bel, named after Alexander Graham Bell. But the bel is "inconveniently large" for most purposes1, so we use the decibel (literally 1/10 of the bel or 10 dB = 1B), hence the factor 10. Ordinary conversation is $\sim$65 dB, this would be 6.5 B

3

I believe there is an acoustic analogy to evanescent waves, therefore a spring and mass lattice model must approximate evanescent waves in the same way that a finite difference model of the continuous medium example below must simulate evanescent waves. In homogeneous mediums scalar acoustic fields fulfil the Helmholtz equation: $$\left(\nabla^2 + ... 4 No, it has nothing to do with sound quality. In fact, the grid or covering is carefully chosen to interfere with the sound as little is possible. Speaker cones must be light weight, so are made from paper or other thin and delicate material. The grill is to physically protect the delicate speaker cone from getting dinged, a curious cat, or some moron with ... 5 When you snap your fingers there are multiple sound waves, but the speed of sound is so fast you can't distinguish individual waves. The frequency of sound waves is around 100Hz to 10kHz so each wave completes one oscillation in 0.01 to 0.0001 seconds. What you're hearing when you snap your fingers is the envelope i.e. the overall amplitude of the sound ... 1 How should I model this situation for a closed cavity? Just solve the wave equation with boundary conditions matching the geometry and material properties of the cavity. What does resonance in a closed cavity even look like? (as opposed to a "whistling" kind of cavity) It is about the same in both cases - a standing wave, with energy ... 0 From your description I deduce that approximation of geometrical acoustics should be enough. For its applicability we need to ensure that The sound could be described as small perturbation (so, no nonlinear effects). Wavelengths of sound are much smaller than the dimensions of structures with which the sound interacts. The main equation for geometrical ... 0 I will give you mathematical background that explains where sound waves, as well as shear, and other waves come from in continuum approximations and why viscosity has no influence on such waves: In Eulerian form, the vector equations of motion for a fluid or solid continuum can be written as$$\frac{\partial \mathbf{q}}{\partial t} + ...

0

The exact answer can easily be computed by solving the elastic wave equation using the finite-difference time-domain method. You mentioned the image and ray-tracing methods for room acoustics. Theses are geometric methods, which would not be useful for solids since there would need to be separate rays to describe the compressional and shear waves, and the ...

4

Do low frequencies carry farther than high frequencies? Yes. The reason has to do with what's stopping the sound. If it weren't for attenuation (absorption) sound would follow an inverse square law. Remember, sound is a pressure wave vibration of molecules. Whenever you give molecules a "push" you're going to lose some energy to heat. Because of this, ...

-1

Wave always carry energy infinite distance. The energy of sound may dissipate along distance by many cause at the same rate in any frequency Low frequency wave not carry in "longer distance" than high frequency. But it may have less problem in propagate to things. Lower frequency wave tend to pass though bigger object with lesser absorb or reflect. So in ...

-1

I think it has nothing to do with the frequency, but usual sounds consist of many fundamental frequencies and their harmonics, and usually you would have sounds built with such a formula like $$S(t)=\sum_{N=1}^\infty \frac{1}{N}\cos{N\omega t}.$$ So you see, lower frequencies would usually have higher amplitudes, and live longer. You can prove to yourself ...

1

Your question is very similar to a question my mother asked me and since the answer is the same, allow me to start with a proxy question: How can a paper speaker reproduce the intricate timbre of such a wide variety of musical instruments? A wood flute sounds different than a metal flute, a steel string guitar different than a nylon string guitar, etc. ...

1

In the future, please search for these answers first by e.g. looking up "how speakers work". A quick search will bring up what you want. An electromagnet in the speaker cone is usually fixed next to a permanent magnet. Electricity is channeled through it to create a magnetic field which, by changing polarity, can flip the poles of the magnetic ...

1

No, but it can sound like it. If there are two sounds that have similar frequencies, we will only hear the loudest one (this phenomenon is used to help compress music files). This is caused auditory masking. Now, if you have a loud sound far away, and a less loud sound close to you, the one close to you will sound louder, and will mask the farther away one. ...

0

Maybe possible if you can construct "Air Lens" It is in my mind for such times. Sound wave can refract like light wave. So it possible that if you can shape the pressure of air in an area in a fit form. It could be guide spherical sound wave into point cone. Make it louder at focal point This is the explanation that my idea occur in reality ...

3

The only way that I can envision a sound getting louder as you walk away is if you happen to initially be located at a node where destructive interference causes the waves to be near zero amplitude. When the waves are exactly out of phase at the point you stand, you will not hear anything. As you move in any direction away from said point, it will get ...

2

I think that in general, as you move away from a sound it gets softer due to the dissipation of energy. However, I can see possibilities using exotic configurations of air density where the sound does get louder. For example, imagine that the air density increases while retaining the same bulk modulus. Then, as you move away from the source, the sound ...

0

Let me add one more possibility that I've only just thought of: The two tuning forks behave as point sources for the sound, so they will generate an interference pattern just like a Young's slits experiment. This means that as you change the separation of the tuning forces the intensity of the sound at some fixed point in space will oscillate up and down, ...

5

What you hear in this experiment is the combination of the Doppler effect and the beat. As John Rennie points out, the frequency change due to the Doppler effect would be hardly audible. However, the frequency between the two tuning forks will now be slightly different, which results in a intensity modulation, called the "beat".

1

Imagine instead of pressure fluctuations, that the tuning fork was throwing balls at you at a rate of 1 balls per second, and that the balls were moving at 1 m/s. Now imagine that the tuning fork is moved away from you at 1 m/s, but the balls are still moving at the same speed. What do you think would happen to the rate at which the balls hit you in this ...

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