# Tag Info

## Hot answers tagged vibrations

727

So, I decided to try it out. I used Audacity to record ~5 seconds of sound that resulted when I dropped a penny, nickel, dime and quarter onto my table, each 10 times. I then computed the power spectral density of the sound and obtained the following results: I also recorded 5 seconds of me not dropping a coin 10 times to get a background measurement. In ...

292

If you have the dimensions and material of an object, you can compute both the mass and the normal vibration modes. Just the mass is not enough - a large paper "coin" will have a different fundamental frequency than a small tungsten sphere. A summary of everything that comes below - the result of several edits, and including a nice interaction with the ...

72

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 ...

59

This is not an advertisement. Under the rubric of "do try this at home", I wanted to share one more thing that I discovered after writing my previous answer - but it is so unrelated to that answer that I thought it better to write this as a separate post. I discovered two interesting things. First, when you spin a coin on a hard surface, it "rings" with ...

35

Summary Yes they can. False positives arising from the acoustic sources you name are ruled out by seismological analysis and the examination of correlation between the separate gravitational wave detection stations. Pretty obviously, the LIGO detectors are probably the most sensitive microphones ever built. So how do we know they are detecting gravitational ...

27

There's a nice article on exactly this subject in this PDF file. Summarising from the article: the vibration arises because the track is not completely smooth and the train wheels are not perfectly circular. As the train moves along thetrack, the result is an oscillating force at each wheel/track contact, and this is transmitted to the ground at each ...

26

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, high-...

25

The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$B = \frac{\hbar^2}{2I}$$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This ...

20

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern ...

18

This is a subtle issue! Your intuition is correct (a driving at $f_0/2$ should be very effective) even though the graph seems to contradict this. The reason is that the graph displays the response to a sinusoidal driving force. If you indeed drove the mass sinusoidally at frequency $f_0/2$, it would indeed be ineffective -- you'd be holding onto the mass and ...

17

What you are seeing on the square plate are the resonant modes of the structure. Each of these modes has a particular frequency associated with it, and is rung up when the plate is driven at that frequency. These resonant modes act like standing waves on a string: where some parts of the plate are moving a lot while other parts are standing still. The sand ...

16

Is it that the wooden block vibrates with lesser frequency than the metal block? If so, why is that? 'Yes', to the first question. Metal is stiffer than wood and produces higher frequencies (higher pitch). This follows from the wave equation (here in one dimension): $$u_{tt}=\frac{E}{\rho}u_{xx}$$ $E$ is Young's Modulus and $\rho$ the material's ...

15

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like: $$A(t,x) = \sum_{i=0}^\infty A_i \sin(n\omega_i t - k_i x)$$ i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with ...

13

I don't mean to take anything away from the previous great answers, but the "simple and to the point" answer is, a very qualified, yes. By qualified, I mean one must know the coin's composition, thickness, diameter(or shape), density distribution, country of manufacture, etc. If we make assumptions and restrictions, then it becomes possible to calculate ...

13

Yes, plucking a guitar string does create standing waves, but... No, plucking a guitar string does not create a standing wave, as the sum of standing waves is in general not a standing wave (thanks for Ben Crowell for pointing this out), since a standing wave must have a stationary spatial dependence and a well-defined frequency: $$y(x,t) \propto \sin(2\pi ... 13 The metal block has a relatively low level of internal damping, however the wooden block has a high level of internal damping: Much of the energy imparted to the wooden block is dissipated internally as heat and deformation, also the higher frequencies are damped more than the lower frequencies (it acts like a low pass filter). So the wooden block will ... 13 The first image shows a string oscillating at its fundamental frequency f. The second image shows a string oscillating at 2f. The third image shows a string (or the wooden soundbox of a guitar or the air in and around a guitar) oscillating at both frequencies at the same time. Finally, here is a graph showing the height of a point a short way along ... 11 When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey$$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n}, where $\lambda_0$ is twice the ...

11

Just an addition to John Rennie's answer. The equipartition theorem can only be derived in classical statistical physics. In quantum statistics it is not correct. For each degree of freedom there is a characteristic temperature below which the quantum effects are significant. This temperature is very high for rotation around the axis of the molecule; I guess ...

11

Damping implies a loss mechanism. In liquids, where molecules move freely in close proximity, this loss mechanism is a transfer of momentum from one molecule to another. In pure crystalline metals, the position of atoms in the lattice is fixed, and the forces between them are elastic. That is, if an atom is displaced, it will experience a force that puts it ...

11

The string oscillations are mainly transverse (a standing wave). The string motion causes the tension to oscillate thus applying a varying force on the guitar top through the bridge and saddle. The string engage the air very little (as is evident on an electric guitar without amplification). This is because the acoustic wave impedance of the air does not ...

10

First I'll try to explain why the amplitude vs frequency diagram only has one maximum, then I'll go back to why this seems to contradict your intuition. Let's take the simplest forced oscillator formula, with no damping (this won't affect our conclusion), for instance that of a spring undergoing a force $F$ : x''(t) + \omega_0^2 x(t) = F(t)...

9

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 ...

9

Sound waves do generate changes in temperature because the propagation of sound is an approximately isentropic process. Keep in mind though that changes in static temperature can very well occur without the generation of heat. Moreover, the pressure changes associated with sound waves are of such a small magnitude that the observable temperature changes are ...

9

Put more simply: sound waves are attenuated as they propagate through air (this is more easily measured for very short wavelengths, e.g. ultrasound). This means they lose energy - which is turned into heat of the air. The amount of heating, however, is very very small. Let's do the math. A sound wave of 120 dB (really loud) has energy of only $1 \frac{W}{m^... 9 I'd like to point out the example of cast iron. It is renowned for its excellent vibration-damping properties. It is wrong to reach a blanket conclusion saying that metals are bad for vibration damping. The properties of any solid depends part on the material it is made up of and part on the micro-structure of the material. By micro-structure, I mean the ... 9 For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency. For an ideal string, the harmonics have frequency${{f}_{1}}$,$2{{f}_{1}}$,$3{{f}_{1}}$.....and${{A}_{1}}\cos (2\pi {{f}_{1}}...

9

This exceprt doesn't quite get what a black hole is. Once you're inside the Schwarzschild radius, it's not that "you can't go fast enough too get out", it's that the idea of "going outside of the black hole" is a concept that doesn't make sense. Once you are inside, all timelike paths (which are the allowable physical paths of objects that move from the ...

8

What do you think a $\Delta^+$ or $\Delta^0$ is, if not an excited nucleon? (To be sure the $^{++}$ and $^{-}$ states do not correspond directly to a nucleon, because there is no allowed lower spin state with that valence content.) The thing I am not sure about is how closely these excitation match the ones you are envisioning. They match nuclear excitation ...

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