# Tag Info

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If there were only one prong (imagine holding a metal rod in your hand), then the oscillation energy of the prong would quickly be dissipated by its contact with your hand. On the other hand, a fork with two prongs oscillates in such a way that the point of contact with your hand does not move much due to the oscillation of the fork. This causes the ...

36

I am by no means an expert in tuning fork design, but here are some physical considerations: Different designs may have different "purities," but don't take this too far. It is certainly possible to tune to something not a pure tone; after all, orchestras usually tune to instruments, not tuning forks. Whatever mode(s) you want to excite, you don't want to ...

12

Q. How do two coupled vibrating prongs isolate a single frequency? howstuffworks.com has an article on How Tuning Forks Work The way a tuning fork's vibrations interact with the surrounding air is what causes sound to form. When a tuning fork's tines are moving away from one another, it pushes surrounding air molecules together, forming small, ...

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The analogy is a very good one, because heat transfer is in fact modelled by phonons, which you could also use to describe sound waves. The crucial difference is that sound waves have a much longer wavelength (at least in the range of some millimetres) than thermal phonons (not more than a few orders of magnitude bigger than the atomic lattice scale). These ...

6

looking at the link you gave, I think it is just a problem of name convention and has nothing to do with oversimplifications. It was indeed ruled out that the reason of the failure was not due to a resonance phenomenon called the Kármán vortex street, a phenomenon arising in fluiddynamics with a certain resonance frequency, called the Strouhal frequency. ...

5

The forces on the screw are not symmetric. Once the screw is no longer turning loosely in the hole tightening the screw compresses the two materials held together (i.e. increases the stress on the material, i.e. stores energy in the material), while loosening reduced the compression (i.e. releases the stress). So a random dislocation will be more likely to ...

3

No, because in a vacuum, there is no way for the two tuning forks (I think you meant this, rather than pendulums) to communicate. The reason a second tuning fork with the same resonance frequency will begin resonating is because, physically, sound waves are hitting it at its natural frequency. Sound waves travel in a medium, so in a vacuum, there's nothing ...

3

The amount of energy lost to vibration in a car engine is typically very small. You can see this easily because the vibration (and the energy associated with it) is dissipated in the engine mounts, and if any significant amount of energy were being dissipated the engine mounts would get hot, which they don't. Most of the inefficiency is because the ...

2

I tried it actually on my bed, a wood floor, a wood table, and a tiled floor. On the tiled floor, I didn't feel the vibration, on the rest it was clear. I believe the reason for the difference is the difference in the elasticity of the materials. For materials with high elasticity, the wave is transmitted more than reflected, on the other hand, for materials ...

2

Two-prong system surely supports more than one mode, consider: squeezing the prongs together (mode that you want) twisting the both prongs relative to stem twisting each prong relative to its base wobble/barrel of both prongs sound wave travelling in the metal from one edge to another etc... If you are a designer of the tuning fork you want one mode to ...

2

It looks like you're asking for some historical view of how the idea of vibrating air was formed. I honestly don't know, but I can tell that the idea of things acting at a distance has been rejected through history (even Newton thought that his gravitation theory had a flaw at supposing the force acted at a distance, potentials partially solved that), so I ...

2

$y(\theta) = A\sin \theta+ B \cos \theta$ is known as the simple harmonic function. All the motions which can be represented by this function are known as simple harmonic motions. Motion of a simple pendulum is approximately a simple harmonic motion for small amplitudes. It stops vibrating after some-time due to drag from air i.e. loss of energy. But, we ...

2

$\dot{z}(t) = \frac{\partial}{\partial t}P(t,s)z(s) + \frac{\partial}{\partial t}\int_s^t d\tau P(t,\tau)b(\tau)$ $= A(t)P(t,s)z(s) + P(t,t)b(t) + \int_s^t A(t)P(t,\tau)b(\tau)$ $= A(t) P(t,s)z(s) +b(t) + A(t) \int_s^t P(t,\tau)b(\tau)$ $= A(t) [ P(t,s)z(s) + \int_s^t P(t,\tau)b(\tau)] + b(t)$ $= A(t) z(t) + b(t)$

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Here are the steps you can take. Degrees of Freedom. There are 3 degrees of freedom, one for the base plate, one for the box and one for the mass. Hence there are 3 variables that you need to track, as well as their derivatives. I will name them $x_0=\gamma(t)$ for the plate, $x_1$ for the box and $x_2$ for the ball. Free Body Diagrams. For the moving ...

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In classical mechanics, you can describe a crystal (in some approximation) by a Hamiltonian that is a quadratic form in coordinates and momenta of atoms. After you diagonalize this quadratic form, you obtain a Hamiltonian of a set of independent, rather than coupled, oscillators (modes). Then you can quantize this system and you do get zero-point energy for ...

1

Your original question was : Afaik, the particles vibrate according to their energy level. Is this vibration in 3D space? One has to state whether you are talking classical particles or quantum mechanical elementary particles or quantum mechanical atoms and molecules. Classical particles move in three dimensions, any motion. It might be constrained ...

1

Imagine that the oscillator is a swing and you are the force pushing it. The phase shift is nothing more than the statement that you have to act differently than the swing. Obviously, you shouldn't push in the exact opposite direction (which rules out a phase shift of $\pi$). Imagine the red line being the amplitude of the swing, and the green line is ...

1

Physical sound is a mechanical wave caused by the motion (vibration) of air molecules. The source of a sound wave does not necessarily need to persist in 'vibrating' itself (like a violin string) to initiate a sound wave. A single very sudden (i.e. step function) displacement of an object in air could generate a sound wave. Lightning is the source of ...

1

Sound doesn't require vibration, sound is vibration. If you take something and make it vibrate, say by plucking a guitar string, you see it vibrate and you hear it vibrate. The vibrating string makes the air vibrate, and that air vibration travels outward like ripples in a pond. The part that hits your ear is what you hear. You can see why if you pluck a ...

1

Consider a pendulum, such as a playground swing. It is a 2nd-order system, because it swings in a repetitive motion at a certain frequency like, say, 30 swings per minute. It also is damped, because if you set it swinging and then leave it alone, it rubs against the air and its swings becomes smaller and smaller until it seems to have stopped. Now, if you ...

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It means that they cannot "interfere" basically, since two orthogonal modes are not at all "similar", aka the if one mode is projected to an orthogonal one, the result is "zero". It also means, that for any other arbitrary function in the span of the orthogonal modes, you can write that arbitrary function as a weighted sum of those orthogonal modes ...

1

The classical string equation that you are referring to, is formulated by making a number of assumptions, which include that the vibration of the string does not affect its tension. This makes Young's modulus irrelevant for results calculated from the idealized equation. In the real world, materials with low moduli of elasticity will follow the ideal ...

1

how was it known or deduced that vibration is necessary for any form of sound wave? It is likely that theories involving wave propagation have been around for thousands of years. The speculation that sound is a wave phenomenon grew out of observations of water waves. The rudimentary notion of a wave is an oscillatory disturbance that moves away ...

1

I won't address the details of the fluid dynamics that cause the cage to produce noise because it is well out of my area of expertise. But I have performed an experiment in my office. The small desk fan I bought was too noisy. Since it lives on the top of a tall bookcase and well out of reach, I removed the cage completely. It is a lot quieter now. My guess ...

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Regarding the general question posed in the title, yes, it is possible in general to have photon emission (or absorption) without electrons changing energy levels. For example, in nuclear magnetic resonance (NMR), EM radiation (photons) are absorbed and re-emitted with changes in nuclear spin state. NMR spectroscopy relies on splitting between nuclear spin ...

1

In the case of the IR photons the system (i.e. molecule) does change it's state (by changing its vibrational or rotational mode). You can also imagine such a thing happening in a purely atomic context. This would be a direct analogy of the Mössbauer Effect, only happening in an atomic context rather than nuclear one. By the totalitarian rule I have to ...

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leftaroundabout gave an excellent explanation for the thermal conduction of insulators. However, in the case of metals, a significant amount of energy is carried by the excitations of electrons (the width of their Fermi-Dirac distribution). The thermal conductivity is then related to how far an excited electron can travel before being scattered, and is ...

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