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I am trying to understand the derivation of the wave equation for a longitudinal wave in a solid rod (with a given Young's modulus).

One of the derivations gives the two forces acting on a solid segment as such:

The magnitude of F1 is taken to be responsible for the deformation (stretching) of the element. But then F2 is equated to F1 plus what appears to be the same term. Why is that?

Other derivations of this wave equation have been equally confusing, which is why I am asking here.

Edit: Here is the rest of the derivation:

pll04
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1 Answers1

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It is a Taylor expansion and first order approximation. $$F_1 = AY \frac{\partial \eta}{\partial x}\Bigr|_{x}$$

$$F_2 = AY \frac{\partial \eta}{\partial x}\Bigr| _{x + \Delta x} = AY \left(\frac{\partial \eta}{\partial x}\Bigr| _{x} + \Delta x \frac{\partial^2 \eta}{\partial x^2}\Bigr| _{x} + O(\Delta x^2)\right) \approx F_1 + AY\Delta x \frac{\partial^2 \eta}{\partial x^2}\Bigr| _{x}$$

The picture is misleading imo. Since the two forces are calculated in the same way, just in different points of the rod, it think they should be drawn pointing in the same direction. The actual sign shall be given only by the sign of ${\partial \eta \over \partial x}$, that may be different if calculated in different points.

That said, I understand what the picture is trying to convey. The authors want to show that the segment is being stretched. From the third law, we know that for every force there is an opposite one. This means that in every point the authors can choose to draw the force with which the segment is being pulled (which will be directed to the right) or the force with which the segment is resisting the pull (which will be directed to the left).

They chose to do the former for $F_2$ and the latter for $F_1$. The reason is that the stretch of the segment is obviously given by the difference of the magnitude of the forces $|F_2| - |F_1|$, and their drawing choice might facilitate this idea.

I know, this was a convoluted argument, but the bottom line is simple: when working with forces it is often easier to understand what is physically going on, work with the magnitude of the forces, add and subtract them as physically required and only at the end of the calculation put a sign to the result (if needed).

Prallax
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  • For the computation in x you can use better \Bigr| or \Biggr| that work in mathjax – Sebastiano Aug 08 '21 at 12:25
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    @Sebastiano thank you, edited – Prallax Aug 08 '21 at 12:32
  • You are welcome – Sebastiano Aug 08 '21 at 12:37
  • But shouldn't the term equated with F1 have a minus sign? If we define positive displacement to be to the right, then a positive F1 force should cause negative displacement (-d nu). – pll04 Aug 09 '21 at 16:41
  • @prslv04 I added a clarification on the sing of the forces – Prallax Aug 09 '21 at 17:29
  • Comment on the update: As a full wave passes, each point will experience a changing (sinusoidal) force that goes from positive to negative (pointing to the right and then to the left) as the point oscillates about its undisturbed position. So I disagree: I think F1 and F2 are both pulling forces, and can change sign. After all we must only consider forces acting on the segment, and not forces the segment exerts on its surroundings. – pll04 Aug 09 '21 at 18:27
  • @prslv04 I don't know, it might well be as you say. There might be more than one way to look at the problem and tell what is pulling and what is pushing. The important thing is that to obtain the strain you subtract the magnitudes of the forces, as I guess the authors will proceed to do afterwards, don't you agree? – Prallax Aug 09 '21 at 19:20
  • I just updated the question to show the rest. The strain is actually already calculated by considering the infinitesimal stretching (d nu) each force does (one d nu at x and a different one at x+dx). The next part is to subtract the forces for Newton's second law. But this is unique to this specific case. This is why I think it is important that the forces acting on the segment can be both positive and negative, and so can be the extensions/compressions they cause. – pll04 Aug 09 '21 at 19:32