# Tag Info

## New answers tagged frequency

0

If the source is moving faster than the speed of sound then ahead of the source it produces a shock wave instead of a sound wave. This is the infamous sonic boom associated with faster than sound flight. Your equation is giving you a silly result because no propagating sound wave exists, so there is no frequency to be calculated. If you're interested in ...

3

Classically (since rob has done a thorough hop on the quantum picture), the amplitude of a light wave is not related to any physical extent. It is not the size of the wave in space, it is the strength of the fields (electric and magnetic). We often draw wavy lines, but if you look closely the transverse axes will be label differently for, say, waves on a ...

2

If you twisted my arm and forced me to assign an amplitude to a single photon, I'd do it this way: The energy density of a classical electromagnetic field is \begin{align} U &= \frac12 \left( \epsilon_0 E^2 + \frac1{\mu_0} B^2 \right) \\ &= \epsilon_0 E^2 &\text{(only for light in a vacuum)} \end{align} where $E,B$ are the amplitudes of the ...

0

The maximum of $$\left|\cfrac{V_2}{V_1}\right| = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}$$ is $1$ at $\omega=\pm \infty$, and you find the half power frequency by solving: $$\frac{1+(\omega RC)^2}{4+(\omega RC)^2}=\frac{1}{2}$$ which gives $\omega=\pm \sqrt{2}/RC$

1

As a planetary science and aviation enthusiast I can offer these tidbits, although a bit late for the 2013 posted question.... http://www.wired.com/2010/05/gallery-clouds/ This shows mountain-induced Van Karman vortex street (Strouhal instability) in a cloud layer as viewed from space. and so does this: ...

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