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One possibility is that your polarizer interacts with the other parts of your setup (for example, forms a resonant cavity with some other interfaces that enhances transmission). You can test this hypothesis by rotating your polarizer (is the intensity always brighter?). If you include a drawing of your setup, it would be easier to figure out the underlying ...

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This is not possible, as you will see if you derive simple harmonic motion from first principles. The defining equation of simple harmonic motion is: $$\ddot x \propto -x$$ This is linked closely to Hooke's Law, where $F = -kx$, where $F$ is the restoring force, $k$ is the spring constant, and $x$ is the extension from the equilibrium position. So, by ...

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Harmonic motion in physics isn't so much defined by a periodic solution as it is defined by a certain differential equation. The equation for the harmonic oscillator is $$\ddot{x} + kx = 0,$$ where $k$ is some constant. The general solution to this equation can be written $$x(t) = A \sin(\sqrt{k}t) + B \cos(\sqrt{k}t),$$ where $A$ and $B$ depend on the ...

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Let's break this problem into steady state and transient parts (if it works for circuits...). Regarding transient tones, all bets are off. String musicians (I'd love to speak for the rest of music-dom, but I mostly play strings) spend lots of practice time starting a tone with clarity and the correct volume and without any crunch or unwanted harmonics. ...

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It seems that the harmonic (integer multiple) overtones of a sound usually all have the same phase. Is this true...? No, I don't think this is generally true, although it may be true for certain instruments. What led you to believe this? In trumpet tones, for example, the different harmonics come up at different times during the attack, so it seems ...

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This is an excellent question, that deserves a more thoughtful answer (no offense guys). The question that Unknown is asking (I think) is why should there be a node or antinode at each end of a cylinder? When the end is closed, it is fairly easy to see that the air cannot move any further along, so the displacement of the air will be zero - in other ...

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A "sharp" tip typically has a finite curvature; there will be a very small part of the "tip" that is therefore angled at such a way that light will be reflected off it. The sharper the tip, the smaller the radius of curvature, and the smaller the "twinkle" or glint. The second effect is diffraction: Light that passes an object will be diffracted. For ...

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You are basically correct. An air-filled cylinder that's open on both ends will actually resonate at multiple resonant frequencies, given by $$f=\frac{n v}{2L}$$ where $n$ is a positive integer, $L$ is the length of the tube, and $v$ is the speed of sound in air. The fundamental frequency, which generally contains the most energy, is the case when $n=1$, ...

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Depends on the ends of the tube: An open end is a displacement anti-node (unrestricted), a closed end is a displacement node (restricted). Thus, a tube that is open at one end and closed at the other will have natural frequency and harmonics such that there is a node on the closed end and anti-node on the open end (a quarter wavelength). If both ends are ...

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Paper cones were originally chosen for their rigidity and lightness, so they can move air quickly without deforming and couple to a motor easily at the center while also being easy to suspend from the basket by their perimeter with a simple corrugation or foam/rubber surround. Physics only played a major part in the ease of construction and performance was ...

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the waves will obliterate each other but they will still exist, they just won't be moving they would just change form (energy cannot be destroyed it can only change form) so when the waves meet they will cancel each other so sound will change to potential and kinetic will change to sound or whatever

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If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called "photon" or "wave packet". Linking the appropriate formulas from QM and E&M waves, you get the diameter of the wave packet (about λ/2) but not the length. The radius and the direction of propagation do not change as long as the wave packet is not disturbed. ...

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What I am missing in your question, is the dimension of $\lambda$. You can see from your second formula, that the dimension of $\lambda$ is $\text{m}^{-2}$ (because of the Laplacian). Further, to add a bit more information, Normally we would write your Eigenvalue problem as $$\Delta v = - k^2 v.$$ We use $k^2$ here mostly because it will make the answer ...

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The first thing that distinguishes a shock wave from an "ordinary" wave is that the initial disturbance in the medium that causes a shock wave is always traveling at a velocity greater than the phase velocity of sound (or light) in the medium. Notice that I said light - that is because there is also a kind of electromagnetic analogue to a shock wave known as ...

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The phase of a sinusoidal wave is represented as: $$y(x, t) = a \times \sin(\omega t + \phi)$$ So the time evolution (i.e the part that changes with $t$) is only the $\omega t$ term and not the pure constant phase $\phi$ term. $\phi$ can depend on $x$ or other things but not $t$. This is what is meant that the phase is constant. In case you mean sth ...

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Even though the forces started at different times, is there any displacement of the metal box in any of the situations? Or is there any movement at all but is the net displacement zero? Sure. If you think of each force as causing an acceleration, the first one begins an acceleration in one direction, the second an acceleration in the other (or a ...

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Imagine you standing some distance from me, and you move a charge back and forth along the line joining us. Waldir, you are quite right that the electric field I observe will fluctuate, and that these fluctuations will not reach me instantly - they will travel at the speed of light. However, this is not electromagnetic radiation. Why?- The electric field ...

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Electrical amplification is about using an input signal to modulate a larger amount of power that comes from a separate power supply of some sort. And yes, there is such a thing as a magnetic amplifier that works on a very similar principle (even though the inputs and outputs are usually electrical). But you can't get an output value that's greater than the ...

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There are several ways to amplify the magnetic field, though the mechanism is not same as for electrical signal amplification, but still they are fruitful. compression:- since a magnetic flux through a surface remains conserved, if we compress the field lines or stretch (or fold) the field line then we can increase the energy by working against the field ...

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In QM the Schrödinger equation, is the equivalent of Newton's law in Classical Mechanics. The Schrödinger equation describes the state of a quantum system (i.e. atoms, subatomic particles etc.), and how the quantum system changes over time. I think you are getting confused because there are two main places where the term wave appears. (1) The Double Slit ...

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"Bright light can never hurt your eyes" seems false to me… enough energy focused on the retina will cause damage, regardless of the wavelength. Otherwise you would not need to wear laser goggles… That aside, materials typically have certain ranges where they absorb light more strongly than others. There is no hard and fast rule for this, but if you google ...

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Uv goes thru glass, I thought that comment strange when I watched it. Laminated glass (which it could have been) would shield about 95% of the uv, I believe due to the resin interlayer.

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The following picture (from http://hyperphysics.phy-astr.gsu.edu/hbase/waves/imgwav/circonwave.gif) gives you a better sense of how to reconcile your observation with "circular motion": As you can see - there is circular motion for particles at the surface: they don't have to go under water to do it though. Incidentally this also shows that in the trough ...

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Imagine the horn has a variable speed $v(t)$ relatively to the observer, the position of the horn relatively to the observer is given by $x(t)=\int_0^t v(u) du$, supposing $x(0)=0$ Suppose we are interested only at the (periodic) maximums of the sound, corresponding to a period of the emitted sound, at $t_0$, $t_1=t_0 + T$, where $T$ is the sound period, ...

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The question, answers and explanation are poorly worded. Since the observer's velocity changes, the nature of that change, his/her acceleration, is significant. If the observer begins to accelerate away from the source, and continues to accelerate, then the perceived frequency will continue to decrease (as long as the observer stays sub-sonic!) At any ...

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If the observer moves away with a constant velocity, they will hear a different frequency, but $f'$ will remain the same. Perhaps the problem meant that the observer accelerated away, or it is possible that the textbook editors made a mistake (which happens more frequently than most people realize). As for your second question - if the observer ...

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The energy flux of an acoustic wave is $$\vec J = \vec v p \;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$ The relevant energy density to be used in these calculation is actually $p+1/2 \rho v^2$, but since we are discussing a small amplitude wave (= no shock wave), $v$ is an infinitesimal quantity; thus $1/2\rho v^2$ is lower order than $p$ (second vs. first), thus it ...

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There is one limit in which this computation is easy to do. Let us consider a massive, perfectly rigid ball striking a a perfectly rigid floor. In this case, there is nothing oscillating, so that we can neglect sound generation by oscillations in the ball or in the floor. Yet there will be sound, because the ball displaces air in its fall, and the air ...

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The confusion you face is a historical one. Originally the interactions of different bodies was thought to happen at a distance more or less instantly, such as the case in the time of Newton and his gravitational theory. But when we discovered electromagnetism, and in particular, when Maxwell completed his formulation of Electromagnetism as contained in ...

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A wavefront (your signal) has a fixed amount of energy given to it by the transmitter. Whatever happens to the wave once it leaves the transmitter is independent of the transmitter, thus receiving a signal does not drain any additional energy from the transmitter (though it can drain energy from the wavefront itself). EDIT: As pointed out by @Alfred ...

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A wave can propagate in any medium that is: a) elastic b) less than critically damped Neither homogeneity nor isotropy are necessary. Any elastic system will return to it's original state when deformed, the question is just whether the deformation can propagate, and this is down to how quickly the energy of the deformation is dissipated. If the damping ...

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A wave is generated by a disturbance in a medium. For a wave to propagate, do not necessarily need a medium. For example, an electromagnetic wave can propagate in vacuum, while a sound wave requires an elastic medium to travel. The requirements for the propagation of a wave, are dependent on the nature of the wave.

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There is no such thing as “conduction of electric wave in conductor” (and I am unsure about where “electric waves” can be observed). There is a conduction of electric current in a conductor. One can say that electric potential in a piece of conductor is always the same (so the electric field is zero inside it), although it is not always so due to resistance, ...

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The continuous stream of air that you are blowing in, it doesn't enter the pipe continuously. When the stream of air hits the hard edge in an organ pipe, it flaps in and out of it due to the difference in the density of the air outside and inside the pipe. This oscillation of the air in and out, it will be a periodic energy supply for the standing wave in ...

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I am answering your question here, but please provide more information about your goals/experience, as specified by the comments. Primarily, I would like to say that I was planning on answering your question much less in depth than I ended up doing. However, while brushing up, I got carried away and figured out some very interesting calculations concerning ...

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The trouble is that your table, or whatever object it is, will act as a waveguide. That's because the sound waves will (partially) reflect of the wood/air surface then travel back into the table and interfere with other waves. The result is going to be hideously complicated to calculate. As Luboš says in a comment, if the thickness of the table is much less ...

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Yes. Higher frequencies are attenuated more over distance than lower frequencies are, which has a rounding effect on the square wave as the upper harmonics are reduced. Reference Do low frequency sounds really carry longer distances?

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There is a good explanation of this in Matter and Interactions vol II by Sherwood and Chabay. I no longer have the text; I will try to summarize its explanation as I remember it. The electrons in a substance are analogous to charged masses on springs. The electrons in insulators are relatively tightly bound; those in conductors are loosely bound or unbound. ...

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An overview in layman's terms: First, it is important to note that not any electric field will induce current in a conductor, because other than the fact the intensity of the field defines the speed of each charge (bigger difference of potential), the oscillation frequency of the $\mathbf{E}$ also plays a very important role, if the frequency is too high, ...

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The reasoning has to be the other way around: Light acts on the metal and makes the electrons move. This, however, results in an energy loss, as the electrons feel a resistance and thus the radiation loses energy. This can be formulated more precisely with counteracting electric fields. That's why all good conductors are opaque. In insulators this can not ...

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Since cables carry electricity moving at the speed of light, why aren't computer networks much faster? Perhaps I can address your confusion with a rhetorical question: Since air carries sound moving at the speed of sound, why can't I talk to you much faster? The speed of sound is much slower than light, but at 340 m/s in air, it's still pretty damn ...

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"Surely this is a bottleneck" - No, it's really not. Any real-life network connection is not speed-limited by the propagation speed of the signal in the cable, but by the processing delays in the various routers, switches, and network interface processing at each end.

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Two reasons: 1) The speed of light in a "medium" is (almost*) always slower than the speed of light in a vacuum. 2) Electricity propagating in a wire is subject to inductive and capacitive effects which slow it's progress. And even if wires were infinitely fast, integrated circuits are not. Again, inductive (a little) and capacitive (a lot) effects limit ...

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Why only 64% What does propagation speed mean? I know there are other variables effecting the latency and perceived speed of computer network connections, but surely this is a bottle neck. Speed of signal propagation is distance the signal (packet) travels in one second. It is usually lower than $c$ because EM waves that carry the information travel in ...

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A transmission line is made of a pair of conductors which have some resistance, inductance, capacitance, and leakage conductance. We can take all of these per unit length: The wave equation for signals in this line, in the limit of a lossless cable with $R=0$, $G=0$, is $$\frac{\partial^2 V(x)}{\partial x^2} + \omega^2 LC \cdot V(x) = 0$$ You have to be ...

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As you've probably guessed the speed of light isn't the limitation. Photons in a vacuum travel at the speed of light ($c_o$). Photons in anything else travel slower, like in your cable ($0.64c_o$). The amount the speed is reduced by depends on the material by the permittivity. Information itself is slower still. One photon doesn't carry much ...

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The speed of electrons that flows in the cable, i.e. the current, is only a few m/s. The EM wave propagates much faster. Anyway, the speed of a computer no depends intrinsically of the speed of electrons, but the speed of energy transfers between electronics components.

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How sure are you that electricity travels at the speed of light? Although electricity propagation moves at the speed of an E/M wave, and not electrons, its speed depends on the dielectric constant of the material. Only in a vacuum, I think, would it travel at the speed of light.

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If you consider a mechanical wave in a string, that is possible as long as you keep $\omega A$ fixed. That is because the energy of a mechanical wave is given by $I = \frac{1}{2}\rho v\omega^2A^2$ Many textbooks would contain a proof, and http://cnx.org/content/m12793/latest/ may help as well.

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You should look at the form of the advanced fundamental solution of D'Alembert equation, built up in geodesically convex open sets including the source localized at the event $y$ and the test point localized at the enent $x$ receiving the wave generating by the source. The construction, at least for analytic manifolds with analytic metrics, is obtained by ...

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