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3

I'll piece together some of what's been said in answers and comments in a different light. With acoustics, it pays to be very careful with units. A sound wave has a pressure $p$, and this corresponds to what I'll call it's intensity $I$. Intensity goes as the square of pressure: $$ \frac{I}{I_0} = \left(\frac{p}{p_0}\right)^2. $$ Here $I_0$ is the ...


2

The formula you use does not make any sense - you don't measure intensity of sound in dB, but the logarithm of the intensity, so you cannot multiply by the distance ratio squared.


8

The problem is with your first calculation and also with the somewhat misleading equation that you've found. It's true that $$\frac{I_2}{I_1}=\left(\frac{d_1}{d_2}\right)^2$$ but units are important here. In that formula, $I_1$ and $I_2$ would properly be expressed as power values. To compute with decibels, which are logarithmic quantities, one would ...


0

Calling all finite-element model experts :-) . I can only offer one small tidbit: for wind instruments, aside from the octave hole, which exists primarily to facilitate exciting the higher frequency notes, the tone is primarily defined by the distance from the mouthpiece to the first open hole. As a long-time clarinetist, I'm fully aware that the pitch ...


1

The surface formed by the bubble is such that its energy is minimized. Since increasing the interface between a liquid and air increases its energy due to surface tension, the bubble tends to reduce its radius, which implies that the pressure inside it must be higher than the pressure outside, and following this reasoning you may also get a quantitative ...


0

In a sense yes, because if you disturb a moving object you can always express its motion as the sum of the gross motion and the disturbance. But I think this interference is not important to the nature of the crack. The reason the tip of the whip reaches such high speeds is that as the wave travels down the whip, it is concentrated into a smaller mass of ...


0

Two things against: It's unlikely that the waves will actually all interfere constructively. In a realistic setting the sounds won't be constant, so points of constructive interference are very temporary and don't have time to do anything even if they could. Each of the sound sources is putting out a certain amount of power as sound (in all directions), ...


3

This page quotes the pressure inside a soap bubble as $\frac {4\gamma }R$, where $\gamma$ is the surface tension, about $25\text { dyne}/\text{cm }$ for soapy water, and $R$ is the radius of the bubble. For $R=1$ cm, the pressure is then $100 \text { dyne}/\text{cm}^2 = 10 \text{ Pa}$. This is released when the bubble pops. It doesn't seem like much with ...


16

The air pressure inside the (intact) bubble is larger than in the surrounding. This pressure difference is called Laplace pressure and is caused by the surface tension between the soap film and the air. When the bubble pops the compressed air expands, thus creating a pressure wave, which you ultimately hear as the typical popping sound.


0

Cavitation is the formation of bubbles in a liquid when a sufficiently strong negative pressure is applied. A point in the liquid experiences a “negative pressure” if the local pressure goes below the average pressure in the liquid. This can happen when water in a pipe has a very abrupt turn, near the propellers of ships and submarines, in presence of a ...


22

After much investigation, simulation and a deep literature search, I've figured out the true answer. You perceive a chirp because you are being hit with the echos of the sharp noise that generated the sound. The times between the arrival of those echos is decreasing inversely with time, so it sounds as if it were a tone with a fundamental frequency ...


3

First of all, I would assume that the sounds are originating from the vibrating ball, so the next question I have is whether (1) the chirp results from the way that vibronic harmonics on the surface of the ball evolve over time or (2) it results from the way that the sounds emanating from the ball interact with the room. I've noticed that (A) I only hear ...


0

According to the Wikipedia article on String theory, At sufficiently high energies, the string-like nature of particles would become obvious. There should be heavier copies of all particles, corresponding to higher vibrational harmonics of the string. It is not clear how high these energies are. In most conventional string models, they would be close to ...


0

There are a few things to think about here. First - a cavity will in general have modes. For a simple shape (rectangular etc) these modes can be calculated; as shapes get more complex, this becomes very hard to do. Let's assume your shape is nicely symmetrical, so you can compute the modes. Second, for each mode you will have associated losses: if you ...


0

There is two part in a pipe column : the excitation and the cavity The cavity select a wavelenght depending on its length. But the small open you see in a pipe, it's a menbrane which vibrate with every frequency as you blow (you create something that have every wavelength). It's only the cavity that will choose the good frequency/wavelenght. the stream ...


0

The longer wavelengths aspect is still a little confusing without diagrams. I keep picturing a snake trying to slither between particles in a wall, and you'd think a smaller snake with shorter body ripples would have an easier time. But I do get it, overall. A gut-level angle is that bass has more raw energy than higher frequencies and simply bullies its ...


3

This page gives a chart of Young's modulus over temperature for various metals. Taking the top line of the table, the modulus drops from 31.4 Msi at -325F (-200C) to 24.2 Msi at 800F (427C). Due to thermal expansion, you will have more square inches. Using a linear expansion of $12E-6 K^{-1}$ the area of a bar will increase $1.5\%$ The longitudinal ...


4

Have a look at this black body radiation spectrum, which is approximately the spectrum of "light", electromagnetic radiation, that a body radiates because of the intrinsic kinetic degrees of freedom of the molecules. Look at the frequency spectrum for 300K, about room temperature. Acoustic frequencies are of the order of a few thousand Hertz, infrared is ...


-1

Because the width of the typical internal wall cavity corresponds to the wave-length of low audio frequencies and their constituent harmonics.


2

Decibels are a unit of measurement expressing a logarithmic ratio between the intensity of sound and a given fixed intensity. When you see a negative value in decibels does not mean of course that you have negative sound: such a concept has no physical meaning. What actually has physical meaning is the power or the intensity of sound and those are never ...


16

Zero decibels isn't soundless. The decibel scale is a logarithmic one. For sound each 20 decibel step changes the air pressure associated with the sound changes by a factor of ten. So if you take 20dB as a reference, 0dB is a factor of ten quieter and -20dB is a factor of 100 quieter. Completely soundless would be $-\infty$dB. Zero decibels corresponds to ...


0

Google didn't immediately come up with anything significant for "Ludvigsen's methodology", but let me give this a shot nonetheless. Sound is a propagating pressure wave. So as it goes by, the pressure increases, then decreases, then increases again, etc. Pressure increasing means the particles in the material (typically air) are closer together for some ...


15

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 ...


3

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 ...


0

You could start with a pack of cards and ask how long does it takes for the whole pack to free fall after the bottom card supporting the rest of the pack is released all cards are individually held from the sides, and then released at the same time. From this, I think the answer to your original question is that it depends upon the shape of the ...


0

From what you have; $$k=\frac{\omega}{\nu}=\frac{\omega}{\sqrt{\frac B{\rho}}}=\frac{\omega\sqrt{\rho}}{\sqrt{B}}$$ So, $$I=\frac12 B\omega kA^2=\frac12B\omega\frac{\omega\sqrt{\rho}}{\sqrt{B}}A^2=\frac12\sqrt {B\rho}\omega^2A^2$$ Dimensional analysis can asist you in checking the truth of the above formula.


3

It should be obvious that the linear speed is inversely proportional to the distance from the center to that point on that groove. Yes, speed varies over a record, so the wavelength of the wiggles in the groove gets shorter for the same frequency as you get further into the record. However, the master record was created with this same phenomenon in ...


0

Your question is a little misstated, since it's not so much the "groove resolution" as the high linear frequency limit to which the groove can be cut. The stylus (and piezo or magnetic components which convert motion to electrical signal) doesn't really care about linear speed in the along-track direction, just the transverse speed. So long as the groove ...


0

Organ pipe, whistles, and the like are not fed with pulses of air. Think about it. You blow into a whistle steadily, but yet what comes out is 1000 pulses per second or so. You certainly aren't huffing 1000 times/second into the whistle. Whistles and organ pipes are their own oscillators. You put in power in the form or moving air, and that gets turned ...


3

With $t\to-it$ and $P(t)=\delta(t-t_0)$ and some constants redefinition, this is an equation for the Green's function of the Schrödinger equation $-i \hbar \partial_t \psi = \left(-\frac{\hbar^2}{2m} \Delta + V\right) \psi$ Where $V \sim cm$ is a constant. In most contexts, one needs either $V=0$ or $V=V(r)$. The solutions however become oscillatory which ...


44

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 ...


1

It depends on the direction in which you choose the $y$-axis. If a positive $y$ means a displacement to the left, when this figure is accurate. But you right, that choice is at least unconventional and should have been indicated in the figure.


2

By going to the Analyze Menu and selecting Plot Spectrum: You can generate the power spectral density with different windowing functions and settings: As well as export it.


0

A sonic boom is a continuous high pressure sound wave following the aircraft. If a plane is flying over a long path say 100 mile, an observer is standing at (say) 10th mile will hear this boom as the plane will pass by and the observer standing at 100 mile will hear this boom after some time as the plane will pass by him.


0

A scenario that conceivably could work in some situations: Think of how passive sonar behaves--specifically, convergence zones. Given the right gradient in the sound transmitting media you get a situation where sounds radiated over a range of angles are all focused back at the same point. As you get close to that distance from the source the sound energy ...


347

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. ...


2

Some good points made in different answers. I just want to add my two cents. The short answer is "yes - it can appear louder, and it can be louder". First - appearance. If you have a loud point source far away, and another source closer by, it is possible that the close source clouds the faraway sound. Imagine a faraway train and a nearby radio playing ...


132

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 ...



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