# Tag Info

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

I get the same thing reheating some discs of glazed carrots. And there are several videos of folks doing this intentionally with grapes. An article published last year in PNAS says this will happen with almost any pair of similarly-sized object with sufficient water. The shape of the pairs appears to set up a resonance that concentrates the electric field ...

58

In order to properly understand this without any unnecessary "controversy", let's break the whole process of sound generation and perception into 5 important, but completely separate parts. We'll then proceed to explain each part using a few different examples and pieces of derivative logic: Vibration of the vocal folds Transmission of energy from vocal ...

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The reason for having two prongs is that they oscillate in antiphase. That is, instead of both moving to the left, then both moving to the right, and so on, they oscillate "in and out" - they move towards each other then move away from each other, then towards, etc. That means that the bit you hold doesn't vibrate at all, even though the prongs do. You ...

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-...

19

There seem to be a lot of human body mechanical models, such as this one: As for applications, I have heard that sub-audio frequency vibrations have been considered as nonlethal weapons for riot control.

19

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 ...

18

Every resonator amplifies just certain frequencies while it inhibits all others. This is true only for very simple resonators. The shape of the guitar body is such that it has a different size at different angles. This corresponds to different resonant frequencies. In addition, the top has a supporting bracing that is very different on different models 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 ...

15

The car is behaving like a closed pipe, so you get a resonance set up. There's a Wikipedia article here, but for once the Wikipedia article isn't that great, so there's another better article here. I imagine you (like most of us) will at some point have discovered you can make a sound by blowing across the top of an opened bottle, and it's the same thing ...

12

This is an attempt to explain, in a purely intuitive way why sound waves reflect from the end of an open pipe, and therefore can produce a standing wave. Consider a pressure wave travelling up the pipe. I've drawn just a single maximum of the pressure wave to keep the diagram uncluttered: Call the pressure maximum $P_1$ (I haven't marked $P_1$ on the ...

12

A resonance (in the particle physics or related physics sense) and an unstable particle is exactly the same thing. The object has some complex mass and the imaginary part determines the decay width (and decay rate). But these two terms describe different aspects of the same thing. "A particle" refers to the object, the particle species (in your URL's case, ...

11

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)...

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

It would depend on damping effects being taken into account or not. Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term): $m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right)$ Where $m$ is the ...

10

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force ...

10

Mathematical demonstration It's straightforward to see why this happens if you use a bit of linear response theory. Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says ...

10

@BowlOfRed hit on it solidly. The noodles are acting as waveguides, because of their size and shape. Where they meet, a contiguous surface is created, but with a much higher resistivity, as it's a narrow point contact. Based on a table of refractive indexes and a table of frequency/wavelength, this effect would be especially effective when the total length ...

9

The first generation of elementary particles are by observation not composite and therefore not seen to decay. They are shown in this table of the standard model of particle physics in column I. The Standard Model of elementary particles, with the three generations of matter, gauge bosons in the fourth column and the Higgs boson in the fifth. All these ...

9

The answers currently posted are ignoring a few important details so I'm going to give my own. I may rehash some things already said. To make everything absolutely clear I write here a complete derivation of the forced damped oscillator with emphasis on the role of the $Q$ factor. Basic equations Consider the equation of motion of a forced, damped harmonic ...

8

Vibrations begin to resonate together into sound waves we can hear. We can make the sounds loud or soft depending on how much pressure we place on finger. The pitch of the sound can also be changed by adjusting the amount of water in the glass.As you rub your finger on the rim, your finger first sticks to the glass and then slides. This stick and slide ...

8

From here, how do I define the "resonance"? At resonance, the energy flow from the driving source is unidirectional, i.e., the system absorbs power over the entire cycle. When $\Omega = \omega_0$, we have $$\phi(t) = \frac{A}{2\beta \omega_0}\sin\omega_0 t$$ thus $$\dot \phi(t) = \frac{A}{2\beta}\cos\omega_0 t$$ The power $P$ per unit mass delivered ...

8

The power transfer is maximised at resonance because the driving force and the velocity of the oscillator are in phase. If you multiply two sinusoidal terms together (the force and the velocity) with a phase difference between them, then the product has its maximum average value when the phase difference is zero and a minimum value when the phase difference ...

8

We can model the building as a uniform cuboid of density $\rho$ occupying the region $$0 \le x \le L_x$$ $$0 \le y \le L_y$$ $$0 \le z \le L_z$$ with its mass given by $$M = \rho V = \rho A L_z = \rho L_x L_y L_z$$ The building is attached firmly to the ground ($xy$-plane). Ignoring gravity and compressive stress, consider only the effects of the ...

7

Your first question is answered by this Open University article. To summarise, consider a low pressure region travelling along the tube towards the open end. The air outside is at atmospheric pressure, so when the low pressure region hits the end of the tube air from the atmosphere rushes in and creates a compression wave heading back down the tube. The ...

7

In physics, resonance is the tendency of a system to oscillate with greater amplitude at some frequencies than at others. Frequencies at which the response amplitude is a relative maximum are known as the system's resonant frequencies, or resonance frequencies. (Copied from Wikipedia: Resonance.) The Fano resonance and the Feshbach resonance are the same ...

7

I had the same feeling as you when I watched the video again recently. It seemed like one of the ice giants would get ejected after coming too close to Jupiter. It turns out that there's a name for this: the jumping Jupiter scenario. Outside Wikipedia, it's described in Fassett & Minton (2013) (paywall!) and tangentially in Deienno & Nesvorny (2014). ...

7

It depends on what you mean by resonate. Water has three different vibrational modes - there are vibrational frequencies associated with these, but these are not really oscillations like a mass on a spring which we would be familiar with seeing. The webpage you link has some 'vibrational frequencies' of different molcules and notes they are significantly ...

6

You may find by starting from first principles, or by consulting external resources that pressure waves in air (in one dimension) are governed by the wave equation $$\frac{\partial^2 p}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2 p}{\partial t^2} = 0$$ where $x$ is a position and $t$ is the time, and $p$ denotes the pressure difference away from ...

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