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2

In our usual model for a pendulum, the frequency is constant down to zero amplitude. The excursions are shorter, but the speed is less and they just compensate. This is a handy feature for clocks. Why do you think the frequency increases? As you get to very small excursions, you probably get more stick/slip, increasing the friction and decreasing the ...

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You could convert h to 4135 e-21 keV/Hz and use that in your calculations, or 4135 keV / (e21 hz), and use that. Your calculation is then 20 keV / 4135 keV * 1e21 Hz or 20 keV / 4.135 *1e18 Hz.

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Let me answer your question with a question. How would you use $E$ as it is to calculate $f$? Suppose you plug $E = 20\text{ keV}$ into the formula and solve it for $f$. \begin{align} f &= \frac{E}{h} \\ &= \frac{20\text{ keV}}{4.135\times 10^{-21}\text{ MeV s}} \\ &= \frac{20}{4.135\times 10^{-21}}\times ... 0 if error in wavelength is \delta \lambda and error in frequency is \delta f then{\delta f \over f} = {\delta \lambda \over \lambda}$$so$${\delta \lambda } = \lambda {\delta f \over f}$$The formula above should give you something more reasonable for the error in the wavelength. More generally if$$ X=kA^n$$where k is a perfectly known ... 0 The speed of the wave = wavelength x frequency, say the frequency is 6.0 x 10^14 Hz calculate the wavelength. NOw use the de Broglie wavelength, Wavelength = h/MV use E=hv to get the energy than, Use E=MC^2 to get the mass Plug that back in to the de Broglie eq you get the same answer 2 As a physics & music major I've thought about this a lot. Our visible light range doesn't even cover one octave (400nm - 700nm), but you can see how 400nm light (violet) almost completes the octave from 700nm light (red). Perhaps if we could see 350nm light we'd perceive it similarly to red? I think there's an evolutionary advantage to our eyes not ... 1 Sadly, there is no relation. While at least some chords have a nice physical basis - octaves are literally harmonics of each other - human perception of color has very little connection to the physics of light. Color is cyclical: we see high-frequency blues as near low-frequency reds with mixed-frequency purples in between, so there is no equivalent of an ... 3 You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction n is given by:$$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}. But Jackson notes that for most optical frequencies (and non-meta-material media), $\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ...

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