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The speed of sound in an ideal gas is given by $$a = \sqrt{\gamma R T}$$ Where $\gamma = \frac{C_p}{C_v}$, $R$ is the specific ideal gas constant and $T$ is the absolute temperature. Taking standard values for air, this makes a graph like this: The linear approximation is plotted by your formula, $a = 331\ \frac{m}{s}\ +\ 0.6 \frac{m}{sK} (T - 273\ ... 5$400$-$700\text{ nm}$corresponds to about$430$-$750\text{ THz}$($10^{12}\text{ Hz}$), not$\text{MHz}$($10^6\text{ Hz}$). To convert from wavelength to frequency, use $$f = \dfrac{c}{\lambda},$$ where$\lambda$is the wavelength,$f$is the frequency and$c$is the speed of light. So, for$400\text{ nm}$, this is: $$f = ... 4 The poster from that link is saying that the work done by the spring (that's Hooke's law there: F=-kx) is equal to the potential energy (PE) at maximum displacement, A; this PE comes from the kinetic energy (KE) and is equal to the integral of Hooke's law over the range 0 (minimum displacement) to A (maximum displacement). Anyway, your professor is ... 4 Internet propagates with radio waves. Radio waves take advantage of a wave guide generated by the charged ionosphere and the ground for long distance propagation. Storm fronts with lightning and charged clouds do interfere with the propagation of a signal. Sudden changes in the atmosphere's vertical moisture content and temperature profiles can on ... 3 I believe it's to do with the fact that the speaker's function is to propagate pressure waves through the medium (air). So, it's mainly a mechanical concern: you want something to push air, and you do not wish to expend much energy. So it has to be light and rigid, which the cone manages to fulfil due to its shape. A plane sheet, for instance would undergo ... 3 Wikipedia gives the formula c_{air}=331.3\sqrt{1+\frac {T(^\circ C)}{273.15}}, valid anywhere the ideal gas law is valid. The expression you quote is given at the first two Taylor series terms. 3 I'm not sure I've understood your question but I think you're asking if a big wave can have wave-features on its large features. If so, sure, why not? You can add waves of different frequencies to achieve results like: 2 Let's consider a 1D cavity with one wall at x=0 and the other wall at x=L. We know we have the wave equation for the electric potential \phi, \nabla^2 \phi - \frac{1}{c^2}\partial^2_t \phi = 0. There would be a similar one for \vec{A} in three dimensions. We additionally have the boundary condition that the potential must be zero on the boundary. ... 2 For a linear system, the superposition principle holds since, be definition, a linear system has the following property: (1) if y_1 is the output for input x_1 and (2) if y_2 is the output for input x_2 then (3) the output is ay_1 + by_2 for input ax_1 + bx_2 In other words, the output for a superposition of inputs is the superposition of the ... 2 A quantity oscillating with frequency equal to zero would simply be static or constant. EDIT When T goes to infinity, it is not possible for an observer to see that the phenomenon is periodic. Think about T=\text{a few times the age of the universe}, for instance. If there is no observable periodicity the concept of frequency is not physically ... 2 I hope you know that intensity (I) of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as$$A^2=I=a_1^2+a_2^2+2a_1a_2\cos{\phi}$$where a_1, a_2 are the amplitudes of the light waves with constant phase difference of \phi, A is the amplitude of the resultant displaement at the point on ... 2 In Mathematica: Refine[-(c^2/(4 \[Pi] c r)) Integrate[ DiracDelta[r - c t] Exp[ I \[Omega] t], {t, -\[Infinity], \[Infinity]}], Assumptions -> {r \[Element] Reals, c > 0}] The output is$$-\frac{e^{\frac{i r \omega }{c}}}{4 \pi r}=-\frac{e^{irk_0}}{4 \pi r}$$The extra factor of$c$is eliminated since the$\delta$function has an argument of ... 2 There's no way that you can add a vector pointing along the$x$axis, and have it cancel a vector pointing in the$y\$ direction. So the amplitude of the light is zero nowhere, and there are no intensity fringes. Instead, the polarization of the wave changes at the screen. As you study the light along the direction that you expect to see fringes, you would ...

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Here is another hypothetical (i.e. extremely impracticable) answer to your question that is rather interesting (althgough Aksakal's Answer is likely to be a bit more practical!). You have to imagine yourself to be a very deft light-catcher with mirrors (I can't help thinking here of Mozart the Light Catcher). You trap light in the box by suddenly (within ...

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The light will die out quickly. Think of playing B tone on a string tuned to A. It's pretty much the same thing. Also, 3m wave is not light, it's VHF used in TV UPDATE: In the sound analogy, if you attach B tone generator to A-tuned string, as @WetSavannaAnimal suggested, there will be a B tone wave on a string, but it will be only at and around the point ...

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You are right that a soundboard adds no energy to the system. However, it does allow the existing energy to be converted to sound better. The greater area of the soundboard causes more air to be pushed than the string alone can, even though the displacement amplitude of the soundboard is less than the string. This exactly the same reason speakers have ...

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The soundboard resonates with the same frequencies as the source. It takes it energy form the vibrating source. As the soundboard distributes this energy over a larger volume of air, the sound is louder, but the energy is depleted quicker, limiting the time you hear the sound. Try this with a tuning fork. Hold it by your ear and time the duration 0of the ...

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The comments above that say the sound is louder because the soundboard itself begins to vibrate are correct. This is called resonance. It sounds louder because the motion of the board is mechanically more efficient at converting the energy of the system into sound waves than the string alone. The board is an effective radiator of sound energy. A louder sound ...

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Here's my visual approach to answering your question: the "spherical electrical waves" that you're referring to are just a way to represent the wave crests of light. When considering the oscillating electric field component of the electromagnetic wave, the individual waves propagate through space symmetrically in all directions when being emitted from a ...

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The generation of waves by wind is still an open question. Jeffrey's (1925) made a prediction based on wave shadowing, that is, he proposed that wind over waves would lead to higher pressure over the troughs and lower pressure over the crests, leading to wave growth. It turns out the theoretical growth rates for waves, based on this mechanism, are much too ...

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