From multiple online sources I read that $$E \propto A^2$$ but when I mentioned this in class, my teacher told me I was wrong and that it was directly proportional to amplitude instead.

As far as I know, every website I stumbled upon concerning this said that is the case. My teacher has a Ph.D and seems pretty experienced, so I don't see why he would make a mistake, are there cases where $E \propto A$?

I also saw this derivation:

$$\int_0^A {F(x)dx} = \int_0^A {kx dx} = \frac{1}{2} kA^2$$

located here, does anyone mind explaining it in a bit more detail? I have a basic understanding of what an integral is but I'm not sure what the poster in the link was saying. I know there is a pretty good explanation here, but it seems way too advanced for me (gave up once I saw partial derivatives, but I see that they're basically the same later on). The first one I linked seems like something I could understand.

  • 4
    $\begingroup$ You are asking the right questions and thinking the right way. Forget about the PhD and instead ask your teacher to explain in detail why he thinks $E\propto A$. Galileo had something apposite to say here: " ... the authority of a thousand is not worth the humble reasoning of a single individual". Energies in linear systems are quadratic functions of generalised co-ordinates, as in Kyle's answer. $\endgroup$ Feb 22 '14 at 0:32

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 wrong. The total energy in a wave comes from the sum of the the changes in potential energy, $$\Delta U=\frac12\left(\Delta m\right)\omega^2y^2,\tag{PE}$$ and in kinetic energy, $$\Delta K=\frac12\left(\Delta m\right)v^2\tag{KE}$$ where $\Delta m$ is the change in mass. If we assume that the density of the wave is uniform, then $\Delta m=\mu\Delta x$ where $\mu$ is the linear density. Thus the total energy is $$E=\Delta U+\Delta K=\frac12\omega^2y^2\,\mu\Delta x+\frac12v^2\,\mu \Delta x$$ As $y=A\sin\left(kx-\omega t\right)$ and $v=A\omega\cos(kx-\omega t)$, then the energy is proportional to the square of the amplitude: $$E\propto\omega^2 A^2$$

  • $\begingroup$ This probably easily available somewhere on wikipedia or something, but can I ask where you go the PE equation you listed? $\endgroup$
    – D. W.
    Aug 24 '14 at 6:28
  • $\begingroup$ @D.W.: Sorry for the very late reply, you can see it on this Hyperphysics site. You can use the fact that $U\sim kx^2\sim m\omega^2x^2$ and the change in $U$ would be associated with a mass change in the wave, $\Delta m\sim \mu\Delta x$ (with $\mu$ the linear density). $\endgroup$
    – Kyle Kanos
    Feb 7 '15 at 20:08

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