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 $u=\frac{dE}{dx}$

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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.

  • 3
    $\begingroup$ This is a good question, there is actually a long-standing argument about this between two senior academics at my University! $\endgroup$
    – Orca
    Commented May 20, 2016 at 19:07
  • $\begingroup$ I think you are forgetting that speed of the oscillation, thus the wavelength and frequency, are dependent upon the local tension in the rope. Even if both ends of the rope are slack/open (i.e., no one is holding on), that it has mass will result in there being a local tension due to the oscillation. I think that may be the source of the non-constant mechanical energy part. A smaller contribution would come from the internal friction produced by the small stretching that occurs within the rope as the tension increases when the wave pulse passes... $\endgroup$ Commented May 20, 2016 at 21:57

3 Answers 3


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 $dE$.

  • $\begingroup$ Your assumption for the value dl is equivalent to assuming that points of the rope only move perpendicular to the x-axis. I see no obvious reason why that would be. Take a standing wave in a string between two nodes: the string is stretched most when the middle is at maximum displacement; according to your derivation, the middle element itself wouldn't be stretched at that point (and have zero potential energy), which seems doubtful to me. $\endgroup$
    – Previous
    Commented May 26, 2016 at 7:56
  • 1
    $\begingroup$ What is the logic behind writing dU = Tdl? $\endgroup$
    – User
    Commented Oct 22, 2017 at 16:28
  • $\begingroup$ @InternetGuy The term $Tdl$ is an infinitesimal work, it is the force $T$ acting on the element $dm$ times the displacement $dl$ of this element. This work done increases the energy of the system and this gain is stored in the form of potential energy. $\endgroup$
    – Diracology
    Commented Oct 24, 2017 at 10:24
  • $\begingroup$ @Diracology In your initial answer you said that dl is the extension in, and not the displacement of, the element being considered. In any case, according to me, the work done in displacing the element converts into kinetic energy and the work done in extending the element converts into its potential energy. What I am failing to understand is why Tdl equal to the work done in extending the element. Shouldn't the element provide some sort of resistance to extension which should be considered? $\endgroup$
    – User
    Commented Oct 25, 2017 at 1:38
  • $\begingroup$ @InternetGuy Sorry, in my previous comment I meant the extension and not the displacement. In fact the element provide resistance to extension and this resistance is given by the tension. Thus this resistance times the extension $dl$ gives work and the associated energy is stored as potential energy. $\endgroup$
    – Diracology
    Commented Oct 25, 2017 at 16:00

The second derivation is correct, as explained by Diracology.

However, the first derivation is 'sort of' correct, in the sense that the location of potential energy can be ambiguous. For example, consider the three following systems.

  • A mass on a stretched spring.
  • A mass sitting on a table.
  • A charged mass next to another charge.

These three systems have elastic potential energy, gravitational potential energy, and electrical potential energy. But in an intro physics class, you'll get three different answers if you ask "where" the energy is. In the first case, it's the spring; in the second, it's "the system of Earth and mass"; in the third, it's "in the electric field between them".

In all three of these cases, as long as we're not doing GR, it makes no difference where we say the potential energy is stored, because it can only be extracted one way: by moving the mass. You could say the potential energy is stored behind Jupiter if you want.

This is why you'll see various conventions for 'where' the energy is stored in a rope. However, in this case, there is an unambiguous correct answer, because a rope has many degrees of freedom, unlike a mass. You can extract the potential energy in the rope by taking any individual section of it and un-stretching it, implying that the potential energy $dU$ of a small piece of rope $dx$ is well-defined.

Worse, for non-sinusoidal waves, the first definition gives the incorrect answer. Imagine a wave $y(x)$ that has $y = 1$ for $0 < x < L$ and $y = 0$ elsewhere. By the correct definition, there's only potential energy at $x = 0$ and $x = L$. By the incorrect definition, the total potential energy is proportional to $L$, which is wrong.

An even worse example: if I just hold the rope at $y = 1$ forever, the incorrect definition says the amount of potential energy is infinite! The correct answer is zero.


There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and small mass elements of it. From the Abstract:

We consider the energy density and energy transfer in small amplitude, one-dimensional waves on a string, and find that the common expressions used in textbooks for the introductory physics with calculus course give wrong results for some cases, including standing waves. We discuss the origin of the problem, and how it can be corrected in a way appropriate for the introductory calculus based physics course.

To extract the result from this work, instead of

$$dE=dK+dU=\frac 12\mu \left(\frac{\partial \xi}{\partial t}\right)^2dx+\frac{T}{2}\left(\frac{\partial \xi}{\partial x}\right)^2dx$$

it should read

$$dE=dK+dU=\frac 12\mu \left(\frac{\partial \xi}{\partial t}\right)^2dx+\frac{T}{2} \xi \left(\frac{\partial^2 \xi}{\partial x^2}\right)dx$$

According to the paper the two expressions give the same result for the global energy content of the string, but the energy densities are not the same. In Section IV it is demonstrated how this fixes some of the issues with an element not having constant energy in particular for standing waves.

  • $\begingroup$ +1, I also recently asked a question about the same kind of thing. Waves on a string actually are extremely complicated! $\endgroup$
    – knzhou
    Commented May 24, 2016 at 21:44
  • 1
    $\begingroup$ This is definitely a subtle subject. I shall mention that E. Butikov says exactly the opposite (he indeed cites LM Burko). He claims that $-\frac{T}{2}\xi\left(\frac{\partial^2\xi}{\partial x^2}\right) $ cannot be interpreted as potential energy density stored in the string, the true expression being $-\frac{T}{2}\left(\frac{\partial\xi}{\partial x}\right)^2 $. Globally both expressions give the same total potential energy. Source: butikov.faculty.ifmo.ru/WaveEnergyPS.pdf $\endgroup$
    – Diracology
    Commented May 24, 2016 at 22:17
  • $\begingroup$ You might also want to look at section 2 in the paper by Bretherton and Garrett (1969): Wavetrains in Inhomogeneous Moving Media $\endgroup$
    – Nick P
    Commented May 25, 2016 at 4:55

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