I'm confused about energy driven by a wave. Consider a sinousoidal wave moving in a rope.
In my view each element $dm$ of the rope follows a simple harmonic motion in time. That means that the mechanical energy $dE=dK+dU$ of the single element $dm$ is constant.
Nevertheless on Halliday-Resnik-Krane I found this explanation.
Despite the analogies with simple harmonic motion the mechanical energy of the element $dm$ is not constant. [...] That's not surprising since the element $dm$ is not an isolated system and its motion is the result of the action of the rest of the rope on it.
I really do not understand how this can be possible. A similar doubt is for the energy density per unit length.
To sum up I found two cotrasting description of the energy and the energy density in a mechanical wave on a rope.
$1.$ (This is the one I'm ok with) The mechanical energy of the single element $dm$ of the rope is constant and equal to $$dE=\frac{1}{2} dm v_{max}^2$$
From here the linear energy density, defined as
$$u=\frac{dE}{dx}=\frac{1}{2} \mu \omega^2 A^2$$
is constant.
$2.$ (Hallyday-Resnik-Krane) The mechanical energy of the single element of the rope is $$dE=\frac{1}{2} dm (\frac{\partial \xi}{\partial t})^2+\frac{1}{2} T(\frac{\partial \xi}{\partial x})^2 dx $$
($T$ is the tension of the rope)
The mechanical energy of the element of mass $dm$ is not constant since the element is not isolated from the rest of the rope.
From here the linear energy density is not constant either and its expression is $u=\frac{dE}{dx}$
Which of these two is the correct one and why?
In the description $2.$ I'm ok with the expression of the mechanical energy but I do not agree with the fact that $dE$ and $u$ are not constant.
Is the mechanical energy of $dm$ really not constant? If so, what can be an explanation for that?
Is this somehow related to the fact that the energy of a wave is not concentrated in a single point but somehow spread in all the rope continuously?
Any suggestion on this topic is really appreciated.
Answer
The energy of an element of a traveling wave is not constant. Halliday-Resnick-Krane is right. For a string of density $\mu$ and tension $T$ the kinetic energy of an element $dx$ is $$dK=\frac 12\mu dx\left(\frac{\partial \xi}{\partial t}\right)^2.$$ For the potential energy we have $$dU=Tdl,$$ where $dl$ is the stretched amount of the string. A small section $dh$ of the string is the hypotenuse of a right angle triangle with basis $dx$ and height $\frac{\partial \xi}{\partial x}dx$. Hence the amount stretched is $$dl=\sqrt{dx^2+\left(\frac{\partial \xi}{\partial x}\right)^2dx^2}-dx=\frac{dx}{2}\left(\frac{\partial \xi}{\partial x}\right)^2.$$ In the last equation we neglect higher order terms in $\left(\frac{\partial \xi}{\partial x}\right)$ since we assume small displacements. Then $$dU=\frac{Tdx}{2}\left(\frac{\partial \xi}{\partial x}\right)^2.$$ Therefore $$dE=dK+dU=\frac 12\mu \left(\frac{\partial \xi}{\partial t}\right)^2dx+\frac{T}{2}\left(\frac{\partial \xi}{\partial x}\right)^2dx.$$
For a progressive harmonic wave $\xi(x,t)=A\cos(kx-\omega t)$ we get $$dE=\frac 12\mu\omega^2 A^2\sin^2(kx-\omega t)dx+\frac{1}{2}Tk^2A^2\sin^2(kx-\omega t)dx.$$ Using $v^2=T/\mu$ and $\omega=vk$ we get $$dE=\mu\omega^2A^2\sin^2(kx-\omega t)dx,$$ which is not constant.
Remember that energy is being transmitted by the wave along the rope. The source of energy being the harmonic oscillator that generates the wave at one end of the string. So it is not a problem that the energy at each point is not constant. Another important point: The reason the result is quite different to what we expect when we think about simple harmonic motion (which gives constant energy for the particle) is that the element of the rope is not only moving transversely. A wave in a rope always have a longitudinal component. This was implicit when we computed the potential energy and assumed the stretched amount was the $dl=dh-dx$. Notice that when the element $dx$ has displacement $A$, it is at rest having vanishing kinetic energy. Moreover this element is not stretched, $\frac{\partial \xi}{\partial x}=0$, giving vanishing elastic potential energy. This does agree with the equation obtained for $dE$.
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