Please consider the following derivation of the wave equation:enter image description here

enter image description here


  • I am just starting to learn about waves so this might be trivial. In the above, why are we interested in the transverse force which seems to be defined as the component of the tension force (which I am assuming is the force in the direction of the curve of the string) in the $f$ axis direction?
  • Why is the wave equation modeled on the vertical displacement of this line segment under tension force?
  • Lastly, is this supposed to model a fixed profile wave moving at a constant velocity? Or is the string supposed to change shape as a function of time? What is the geometric interpretation?

What you looking at is a small segment of the string which you will show will undergo oscillatory motion at right angles to the string in its position of rest.

So in the end you are looking at a transverse wave.

The curvature of the segment changes along its length and so there is a net force downwards acting on the segment trying to restore it to its equilibrium position.

As there is a net force downwards the segment accelerates downwards.

The segment that you are looking at is linked to adjacent segments of the string.

If the wave was travelling to the right then the motion of the segment to the left of the one you are considering would be advanced in phase and the the segment to the right would be retarded in phase relative to the segment under consideration.

So the diagram is indeed a wave profile, a snapshot of the string at an instant of time.

If you took the photograph a little later in time for a right travelling wave you would find the the segment to the right of the segment under consideration would look like the segment under consideration.

So as time goes on the string changes shape with each segment of the string undergoing oscillatory motion but with a progressive phase difference for adjacent segments along the string.

enter image description here

Some of your segments are highlighted in red and you will note that although you get the impression of a wave travelling to the right each of the rod dots is actually oscillating up and down.
Also note that phase difference for dots either side of the red dots.
The diagram also shows you how a wavelength is defined.

  • $\begingroup$ Thanks for your answer. - Please see comment to Nimrod above to confirm my definition of tension. - What do you mean by "it will undergo oscillatory motion at right angles to the string at rest", what exactly is at right angles to the string at rest? - To confirm my understanding, the magnitude of the tension at the end points of the segment is equal for all segments and all endpoints, but the direction changes according to the orientation of the string? $\endgroup$ – user100411 Jan 18 '17 at 12:51
  • $\begingroup$ Imagine the line defined by the string when it is not oscillating. The direction of motion of the moving string will be at right angles to that line. One underlying assumption is that the displacements of the string are small enough so that the magnitude of the tension in the string is constant. The cos formula you quoted in your comment to @NimrodMorag is a small angle (small displacement of string) approximation noting that the second term in the expansion of the cos function will be proportional to theta squared. $\endgroup$ – Farcher Jan 18 '17 at 12:59
  • $\begingroup$ Is my definition of tension fine? All I'm doing is taking the horizontal components of the tension defined in the figure (i.e. $T \cos \theta' - T \cos \theta = 0$). How is this equal to zero? I understand from the animation you provided why it should be, I just don't see how exactly we can conclude this mathematically? $\endgroup$ – user100411 Jan 18 '17 at 13:07
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    $\begingroup$ For small $\theta$ and $\theta '$ the cosine of these angles is approximately one and so the net horizontal force on the segment is zero. $\endgroup$ – Farcher Jan 18 '17 at 13:17
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    $\begingroup$ Have a look at this link. hyperphysics.phy-astr.gsu.edu/hbase/Waves/watwav2.html $\endgroup$ – Farcher Jan 18 '17 at 13:57
  • The component of the tension in the $z$ direction will not cause movement, because it is equal at all points in both directions, so the net force is 0. However, the component perpendicular to $z$ creates a net force and through Newton's Second Law creates movement.

  • The wave propogates in the $z$ direction, where the wave is an up and down movement of all points of the string. Imagine sea waves, the water doesn't travel forward with the wave, only* up and down.

  • The solution for the wave equation is a function $f(z,t)$, so you can set $z_0$ and observe the up and down motion of this point in time. or set $t_0$ and observe the instantaneous profile of the wave at all points at this time. Here is a nice simulation I found online.

*mostly up and down, but for the sake of this example we can ignore tranverse motion.

  • $\begingroup$ Thanks for your answer. A few questions. Would I be right in defining tension of the segment as a pair of vectors at the end points, pointing along the line segment but away from the line segment (as in the figure), hence the pair is of equal magnitude but different direction? Also, how do you know the $z$ components are equal in the direction as you state, i.e. how do you know that $\Delta_{z}F = T \cos \theta' = T \cos \theta = 0$, how do you know that? $\endgroup$ – user100411 Jan 18 '17 at 12:30
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    $\begingroup$ They are not equal but motion of the string along z is neglected for small enough displacement. So we have only transverse waves here. $\endgroup$ – Žarko Tomičić Jan 18 '17 at 13:48
  • $\begingroup$ @ŽarkoTomičić Okay thanks. Would you agree though that it is a bit misleading in the attachment, where they refer to the net vertical force as a 'transverse force'? It does contribute to the motion of the transverse wave but transverse I think would imply a horizontal force. What do you think? $\endgroup$ – user100411 Jan 18 '17 at 14:24

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