# Tag Info

## New answers tagged frequency

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For example, when you are looking at the photon emitted from an electron going from energy level n=2 to n=1, the photon has a discrete energy and all photons that do this transition will have the same energy. However, if the photon passes from different energy levels (say n=3 to n=1), it will have a different energy. Both photons have discrete energies; ...

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This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that do not yet have commentary. Acoustics by L. Beranek An Introduction to ...

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Active motion sensors emit any one of three types of energy: (1) Infrared light, (2) microwaves, (3) sound waves. If your motion detector does not emit any energy but infrared, and if it has no passive capabilities, it will not be activated by sound waves. Are you certain that your motion detector employs only active infrared technology? Some motion ...

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Yes this is related to Stoke's law for sound attenuation, which states that a plane wave decreases amplitude exponentially with a factor $\alpha$ given by: $$\alpha = \frac{2\eta\omega^2}{3\rho V^3}$$ where you can see that the dependence on the frequency squared $\omega$ of the sound will yield a higher coefficient of attenuation for higher frequency ...

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Taking the simplest example of a sound wave, the component that determines its frequency, is the "up - down" motion of the wave (perpendicular to motion direction), whereas its propagation speed, (in the motion direction) is determined by the "resistance" of medium. Therefore, if the medium changes, only the wavelength changes.

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Let me say what others are trying to say, hopefully in a clearer fashion: Just because you can relate two variables in an equation does not mean that they are dependant. In this case, you have to constrain intensity $I$ in order to get the relationship. At that point, it is not a general relationship, but only true when $I$ is constrained. An example that ...

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You have the intensity:$$I=\frac{1}{2}\rho\omega^2s_m^2$$ which is a relationship between displacement amplitude $s_m$, angular frequency $\omega$ and intensity $I$. What this is telling you is how the intensity is related to the angular frequency and displacement amplitude. There is nothing remarkable about this, all it is saying is the intensity is ...

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The problem is how waves interact with matter. Thickness of walls is not to be taken into account, since basically a wall is (partially) not metallic, therefore attenuates e.m. waves but doesn't reflect them. You cannot generalize on the material too, because air is in fact not very different from concrete. What changes? Epsilon and Mu of the medium in which ...

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I don't know where you got this formula but I think it's wrong. See https://en.wikipedia.org/wiki/Sound_intensity Sound Intensity is given by $I = p \cdot v$ where p is the sound pressure; v is the particle velocity. Sufficiently far away from any source or diffraction object the relation ship between particle velocity and pressure is $v=\rho \cdot c ... 0 The frequency is a function of the dimensions of the bar and its Young's modulus. You need to know what mode of oscillation you are exciting in your bar - there is a hug difference between the flexural and longitudinal modes. If the rod is bending, you can find the formulas here. The derivation goes on and on... but you should be able to use the formula on ... 0 Yes. Materials that absorb electromagnetic radiation and emit it in a different frequency are known as fluorescent. You probably see them as the coating on the inside of fluorescent tubes, where they absorb ultra-violet light and emit a lower frequency visible light. 11 Because the frequency of a sound wave is defined as "the number of waves per second." If you had a sound source emitting, say, 200 waves per second, and your ear (inside a different medium) received only 150 waves per second, the remaining waves 50 waves per second would have to pile up somewhere — presumably, at the interface between the two media. ... 0 There is a system in which the frequency will change when the medium changes: a string fixed at both ends, such as a guitar string. If you pluck a guitar string, then change the medium status by changing the tension, the pitch you hear will change. This is because the wavelengths are fixed (2L, L, L/2, L/3, etc) but the speed of the wave is changing. ... 8 This has to do with continuity of the wave motion. Imagine you had a change in frequency going from medium A to medium B - say 10 Hz become 20 Hz. How do you make something move at 20 Hz? You need to apply a driving force at 20 Hz of course. But the incoming wave is going at 10 Hz. To add energy to the wave we must be pushing when it it moving away from us ... 2 Frequency, in physics, is the number of crests that pass a fixed point in the medium in unit time. So it should depend on the source not on the medium. If I take a source who vibrates faster than yours then number of crests that my source can create per second (for example) will be more than yours. But speed of the wave depend on the properties of the ... 0 After some investigation it turns out that my question is a bit of a convoluted way to ask for the momentum, which contains the information as follows. The momentum-energy relation $$E(t)^2/c^2 - \vec{p}(t)^2 = m_0^2$$ with$\vec{p}$the momentum,$E$the engery and$m_0$the invariant rest mass, simplifies to $$\vec{p}(t)^2 = E(t)^2/c^2$$ for ... 1 It's always possible to expand the potential in Taylor series around any local minima (in this example$U(x) $has local minima at$x_0$, thus$U'(x_0)=0 $) $$U(x) \approx U(x_0)+\frac{1}{2}U''(x_0)(x-x_0)^2$$ Setting$ U(x_0)=0 $and$ x_0=0\$ (for simplicity, the result don't depend on this) and equating to familiar simple harmonic oscillator ...

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