# Derivation of longitudinal wave in a solid rod

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:

• Commented Aug 9, 2021 at 19:03

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).

• For the computation in x you can use better \Bigr| or \Biggr| that work in mathjax Commented Aug 8, 2021 at 12:25
• @Sebastiano thank you, edited Commented Aug 8, 2021 at 12:32
• You are welcome 🙂 Commented Aug 8, 2021 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). Commented Aug 9, 2021 at 16:41
• @prslv04 I added a clarification on the sing of the forces Commented Aug 9, 2021 at 17:29