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The most common approach to model the sound radiation from a vibrating body is generaly the same as in all wave cases: continuity is the key. Let's say that a sphere is vibrating (changing it's volume periodically), then the acoustic velocity of the air particles just on the boundary with the body must be the same as the velocity of the sphere surface. It ...


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May I ask what you think of as sound? If sound is the vibration of air- or in general any material agent- then sound is the sensation you get from the changes in the pressure of the air, it's what reaches your ear and then produces some signals interpreted in your brain. Sound is the vibration, not something produced by the vibration. This vibration which ...


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Take your hand and move it. By moving your air, you moving the air. This is what a vibrating object does - it moves the air. Sound is just the movement of air (or a liquid or solid).


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The relationship is true by dint of the trigonometric identity $\cos(u+v)+cos(u-v)=2\,\cos u\,\cos v$. A compelling experiment is to listen to two tones a few hertz apart (the LHS of the identity) and hear the throbbing beats (the AM wave on the RHS). A pitch fork and a guitar / violin string (the latter readily tunable) is a good way to do this; Given the ...


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To kick things off: you're right. Molecular vibrational excitations are exactly the same as phonon modes. We don't use that language very much because the system is too small (so we can't have things like travelling waves which have momentum, and we need to work only with stationary waves) but the analogy is indeed exact. Now, as to what exactly is ...


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In this context, vibration usually refers to the relative motion of the nuclei. In a periodic solid (crystal), vibrational modes are called phonons. We can say, for example, that a normal vibrational mode of a branch $s$ with wavevector $\mathbf{k}$ is in its $n$th excited state, or equivalently, that there are $n_{\mathbf{k}s}$ phonons of branch $s$ with ...


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By definition you should use total force on the system. In this case total force acting upon system is load, F. What $mx''+kx$ is, a response of the system, not some random force from nowhere, it is just Newton's law. By intuition it looks like that. There is an impedance due to the spring and there is an impedance due to the mass at the end of the spring. ...


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I doubt you can do that. Let's do a back-of-the-envelope calculation. Suppose we are dealing with harmonic oscillation. $$v(t)= 2 \pi \nu A \cos (2\pi\nu t)$$ $$v_{max}=2\pi\nu A$$ where $A$ is the maximum displacement. You want the process to be macroscopic, so we could take $A=1\mathrm{mm}$, pretty small. If $\nu=1\mathrm{THz}$, we get ...


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The period of the natural frequency is $T=2\pi/\omega_0$. Therefore the energy at the end of a cycle (measuring time from the start of the cycle) is:$E(T)=E(0)e^{-\gamma T}$. Hence the fraction of the energy lost per cycle is $F_1=E(T)/E(0)=e^{-\gamma T}$. If the fraction of energy lost in 1 cycle is $F_1$ then in two cycles we lose a fraction $F_1$ on the ...


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Vibrations at 1 THz and over are mostly molecular or macromolecular vibrations. This is the sort of frequency where you get the vibration of individual atoms in molecules or collections of molecules in larger molecular structure. When dealing with molecular vibrations quantum mechanics needs to be used and classical pictures break down - so a simple ...


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Very often, in the audio range people will use speaker coils. Audio speakers are designed to give displacements on the order of a few mm over a range of 20 Hz to 20 kHz - with lower amplitudes at the higher frequencies. Since they are mass produced, they are quite cheap and robust. Take off the membrane, and you have an instant vibrator. Just add AC ...



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