The total energy of an harmonic oscillator is constant along the cycle, only changing from potential (elastic for example) to kinetics, and from kinetics to potential.

In an elastic travelling wave in a string, the points of maximum and minimum height of the cycle have zero kinetic energy as in an oscillator. The point of zero transverse displacement has maximum kinetic energy.

My first question is: has that zero displacement point also zero tension? It seems logical, because that way each point of the string exchanges potential and kinetic energy as an oscillator, but this reference says the opposite on fig 19, page 19.

Edit May,5th The author really adresses my objections in a remark on the same page!

Basically, his (previous) conclusion that the maximum strech is located where the velocity is maximum relies on the hypothesis that the displacement is 100% transverse.

It is enough to allow a small amount of longitudinal displacement, and that strech can be distributed along the string, or even be located in the point of zero speed.

Another related point is about EM waves. The energy density is:

$T = \frac{1}{2}(E^2 + B^2)$

As $E$ and $B$ oscillates in phase, each point in space changes energy along a cycle. Particularly when $E = B = 0$, $T = 0$. Unless what matters is only the average energy in the cycle as in this post. Or what matters is that each point of spacetime has always the same energy.

My second question: what is the correct interpretation of the energy density variation?

Sorry for asking 2 question in post. But they are related by the notion of energy conservation on waves.

  • $\begingroup$ Note: "each point in spacetime has always the same energy" only applies for an ideal electromagnetic plane wave, which already, by definition, fills all of space. For any other wave shape, this is not the case. $\endgroup$ May 4, 2020 at 4:08

3 Answers 3


In a harmonic oscillator, at the minimum displacement (zero) from equilibrium, $KE$ is max. At this point, force $F = 0$, where $k$ is spring constant and $A$ is amplitude. Continuing this analogy over to a plane wave on a string, at every $x$ (direction along length of string), the $y$ value (direction perpendicular to length of string) can be thought of as performing SHM (simple harmonic motion). At the max. speed point, $y = 0$ and consequently, net force on the string element is zero, similar to a harmonic oscillator. This can be seen from the equation of motion for a plane wave:

$$\frac{\partial^2y}{\partial t^2} = {\frac{T}{\mu}}\frac{\partial^2y}{\partial x^2}$$

At maxima of velocity in y-direction, $\frac{\partial y}{\partial t}$,

$$\frac{\partial^2y}{\partial t^2} = 0 \Rightarrow \frac{\partial^2y}{\partial x^2} = 0$$

This implies that the net force on the string element at $y=0$ is zero. This does not however mean that "tension" $T$ in the string is zero, as that is assumed to be constant everywhere in the string. When they refer to the point $y=0$ in your reference as max. "stretch" though, they are referring to the stretching in the length of a string element $dl$ calculated using Eq. (47) of the reference:

$$dl = \sqrt{dx^2+dy^2} = dx\sqrt{1+\left(\frac{\partial y}{\partial x}\right)^2} \approx dx\left[1+\frac{1}{2}\left(\frac{\partial y}{\partial x}\right)^2\right] $$

The "stretch" is then $(dl - dx)$, as $dx$ is the original unstretched length of the string element if it were at rest:

$$\frac{dx}{2}\left(\frac{\partial y}{\partial x}\right)^2$$

At every point in the plane wave $y = f(x\pm vt)$, it holds:

$$\frac{\partial y}{\partial t} = \pm v \frac{\partial y}{\partial x}$$

This implies that at max. y-velocity, $\frac{\partial y}{\partial t} $ is max. and so is $\frac{\partial y}{\partial x}$. Thus, the "stretch", $\frac{dx}{2}\left(\frac{\partial y}{\partial x}\right)^2$ is also max. Note that this "stretch" is not the same as the displacement in the y-direction harmonic oscillator motion, which is zero $(y=0)$. This "stretch" is for the string element $dl$, not for the wave on the string as a whole. Similarly, one can see why the reference fig. calls the max displacement (max $y$) point as zero "stretch".

Now, coming to the EM wave, there is a nice answer which goes into the energy density of EM waves. Classical interpretation is limited to average values of the energy density and hence, an instantaneous value for $E=0, B=0$ is not well-defined.

Edit: I realized later that I misinterpreted your first question and I have altered my answer accordingly.

  • $\begingroup$ In an elastic string: $\sigma = k \epsilon$. So, If tension is uniform, the streching $\epsilon$ must be uniform. If $dL$ is the streched elementary length, when we look at the max. displacement point, the difference $dL - dL_0$ also must be the same. We can not take a $dx$ there and assume it is unstrechted only because it remains horizontal. $\endgroup$ May 4, 2020 at 16:33
  • $\begingroup$ As explained on page 2 of reference posted by OP, the longitudinal displacement of string element $dx$ is second order in the slope, i.e., $dl-dx \propto \left(\frac{\partial y}{\partial x}\right)^2$. The slope is assumed to be small for plane waves on a string and so, the longitudinal displacement can be ignored, which gives equal $T$ at every point. Hence, upto first order in the slope, the longitudinal displacement of each point, "stretch", is zero. $\endgroup$ May 4, 2020 at 20:47

The reference you linked says that they are maximally stretched, but doesn't say that they are under maximum tension. Suppose you slowly pull on a rubber band, and at some time let go on both ends simultaneously. That point would be the most stretched point, but there is instantaneously no tension.

Similarly, from the maximum displacement leading up to the point of zero displacement, there is an increasing amount of "stretch". This can be shown by the fact that the magnitude of the slope increases and the so the length is longer. At zero displacement, the amount of stretch starts to decrease, and so the zero point is an inflection. As you correctly guessed, that means there must be zero tension there, as it is the point where stretching changes to compressing.


This is outdated, but might not be too late for further discussion since the same puzzle tortured me for years.

Note that the following discussion is not intended to be a full answer but comments which are too long to edit into comments. So please feel free to correct or make remarks.

It is probably better to use different variables for instance $\psi$ for the transversely perturbed quantity and longitudinal $\varphi$ from the coordinate variables $y$ and $x$. By the way, for wave equation, one cannot neglect d'Alembert's work, and maybe his operator, i.e., \begin{align} \Box\psi&=\left(\frac{1}{c^{2}}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}\right)\psi=0. \end{align}

Follow the eq.(3) from above reference, where $\cos\theta=1-\theta^2/2\approx1-(\text{d}\psi/\text{d}x)^2/2$ and apply the Newton's law in the longitudinal direction, \begin{align} \mu\text{d}x\frac{\partial^2\varphi}{\partial t^2}&=T\cos\theta|_{x+\text{d}x}-T\cos\theta|_{x} \end{align} After dividing $\text{d}x$ and the limiting process.

\begin{align} \mu\frac{\partial^2\varphi}{\partial t^2}&=\frac{\partial(T\cos\theta)}{\partial x} &\approx 2T\frac{\partial\psi}{\partial x}\frac{\partial^2\psi}{\partial x^2}=2T\frac{\partial\psi}{\partial x}\frac{\partial^2\psi}{c^2\partial t^2}. \end{align} where the transverse wave equation has been used in the last step. Therefore, one can see the longitudinal displacement $\varphi$ has one order higher than the transverse $\psi$, by integrating both sides w.r.t. time $t$.


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