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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 ...

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Actually, it can be theoretically derived from D'Alembert equation (that is satisfied by each component of ${\bf E}$ and ${\bf B}$ in absence of sources in view of free Maxwell equations). The idea is to compute the field (any component of ${\bf E}$ or ${\bf B}$) in $p$, when it is generated by a spherical point source localized in $q$ emitting a spherical ...

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I like Brandon's very physically intuitive answer: mine is a little drier. It is simply that three waves $E_j(t);\,j=1,2,3$ mix through $n^{th}$ order nonlinearity by way of $n^{th}$ power term $\left(\sum_{j=1}^3 E_j(t) e^{-i\,\omega_j\,t} + E_j(t)^* e^{i\,\omega_j\,t}\right)^n$ in the Taylor series for the input to output transfer function. So in the ...

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Re question 1: when you learn this stuff in school you usually simplify the system by modelling it as a simple harmonic oscillator so the amplitude of the system will be given by some equation like: $$A(t) = A_0 e^{i\omega_0 t}$$ where $\omega_0$ is the natural frequency of oscillation. Typically you study what happens if you apply a force that also ...

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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 + ... 3 The reason is not quite as intuitively put as for ropes, but it is essentially to make the fields consistent with the electromagnetic boundary conditions, which in turn can be traced to (1) Kirchoff's voltage law and (2) no conduction currents can flow in a dielectric. Consider a tiny, thin rectangular loop running parallel to the interface with one half ... 3 Imagine a wave that advances 10 meters per second. if every second a wave crashes on the shore, the frequence is one per second (1 herz) the wavelength is then 10 meters since each second the wave advanced 10 meters and therefore the waves are 10 meters apart. If we halve the wavelength (the wave still advances at 10 meters per second) then the waves are 5 ... 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 ... 1 It depends on "how monochromatic" a source you need for your current use. Further, you can have multiple modes of a single wavelength. Using a Fabry-Perot etalon can clean up things a bit. But if your question is not how to achieve, but rather how to evaluate, your source, then you will be limited by the resolution of your spectrometer, or the peak-spacing ... 1 The reason simple harmonic oscillators, and hence their sinusoidal solutions, are so ubiquitous, is that any generic system will look like SHM close to a stable equilibrium. Roughly speaking, this is because smooth potentials look like polynomials (this is Taylor's theorem, in essence), and the quadratic term is the first interesting piece. Concretely, with ... 1 If you're at the molecular level, it's not really QM. Further, consider the calculus of Fourier sine expansions. You can deconstruct any periodic waveform into a collection (possibly infinite) of sine waves. As John Rennie mentioned, sinusoidal behaviour is a direct consequence of quadratic functions, and for better or worse, the forces in our universe ... 1 In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, ... 1 I might add a few commas to that Wikipedia sentence, as "A perfectly collimated beam*,* with no divergence*,* cannot..." to show informative rather than additional parameters. To answer your question about "collimated" vs. "plane wave" , consider two point sources at th plane of focus of a lens. Each point source gives off spherical waves; the lens ... 1 Let us start with 1D continuity and Euler equations written in terms of p and u: \begin{gather} \partial_t p + u \partial_x p + \rho a^2 \partial _ x u=0,\\ \partial_t u + u \partial_x u + \frac1\rho \partial _ x p=0.\\ \end{gather} Here we used an equation d \rho = a^{-2} d p, derived from definition of speed of sound. Dividing the first equation by ... 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 ... 1 A boundary is defined as the line or surface between two different types of materials. For any given boundary, that reflects and/or transmits, we can give it transmission T and reflection R constants to find the amount of reflected or transmitted stuff(like light).$$\text{Stuff} \cdot R = \text{Reflected Stuff}\text{Stuff} \cdot T = ...

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It would seem from this question, that I am incorrect, as it has both waveforms reaching their zero simultaneously. From that question, it seems that the sums of the amplitudes ranges from -2 to 2; essentially just a doubled amplitude of a single wave (as far as sums go). Sad, as I rather enjoyed the symmetry of my interpretation. Oh well.

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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. ...

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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. ...

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