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I'm learning physics in the online textbook "The Physics classroom". The chapter on energy of waves ("Energy Transport and the Amplitude of a Wave") makes the energy proportional to the square of the amplitude. And goes on: "Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only affect the amplitude of that pulse". Every other place talks about energy depending on amplitud AND frequency. This textbook has been very reliable for me (until today :)) Unfortunately it has not a questions and answers section. Am I misunderstanding something?

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A single pulse of a given shape has a particular spectrum of frequencies. Changing its amplitude does not change that spectrum. Or, put another way, if the frequency spectrum were to change then it would be a different shaped pulse. Thus, for a given pulse shape, the only way to give it more energy is to increase its amplitude.

Continuous or cyclic waves are different. The energy in these does also depend on frequency, for the simple reason that if you send n times as many cycles, they will carry n times as much energy between them. So you can double the energy either by doubling the amplitude or by doubling the frequency.

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Energy of a classical wave is proportional to the square of the amplitude and is independent of frequency. However it’s more helpful to talk about power transmission of a wave (per unit area) because the total energy in a wave is spread out.

The power transmitted will in general depend on the frequency as well as the amplitude of the wave.

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  • $\begingroup$ The power transmitted will depend on frequency? What? A red laser and a blue laser with the same intensity will have the same electric field amplitude, meaning it depends only on amplitude, not freq. $\endgroup$ May 10, 2020 at 15:12
  • $\begingroup$ @OfekGillon with optical waves the time period (fm) is so low that in timescales we are interested in (at best ns), the frequency dependence averages out (to a factor of 1/2). This is why in electrodynamics we almost always deal with the time averaged Poynting vector than the Poynting vector itself. $\endgroup$ May 10, 2020 at 16:05
  • $\begingroup$ This is also true for 10 cycles of 532nm vs 5 cycles of 1064nm. Meaning the frequency doesn't play a role, only the time period of the pulse and the amplitude. There is no explicit dependence on $\omega$ $\endgroup$ May 10, 2020 at 16:12
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In a string with the same tension, the only way to increase energy is by increasing amplitude. A guitar string should be pulled strongly for example.

But if the tension in the string is increased, and it is pulled with the same amplitude, the energy is greater. The same guitar string, but more stressed, has a higher pitch, higher sound frequency. So, we can say that energy depends also on the frequency.

Energy in a string is: $$E = \frac{1}{2}\mathbf {|T|}\left(\frac {\partial y}{\partial x}\right)^2$$ Small amplitudes have also small slopes $\left(\frac {\partial y}{\partial x}\right)$ and less energy. But tension also plays a role.

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