On page $519$ of the book Engineering Vibration (which can be downloaded from here), the following wave equation of a variable cross-section rod is derived:

$${ {\partial}\over{\partial}x} \Big( EA(x){ {\partial}w(x,t)\over{\partial}x} \Big)={\rho}A(x) { {\partial}w^2(x,t)\over{\partial}t^2}\tag{1}$$


  • $x$ is the spatial coordinate
  • $t$ is time
  • $E$ is the Young's modulus of the rod
  • $A(x)$ is the variable cross-section function of the rod
  • $\rho$ is the density of the rod
  • $w(x,t)$ is the wave function

Because some of the steps are missing from the derivation process, I tried deriving the same equation myself but failed. Please point out my mistake.

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For the variable cross-section rod in the picture above, I wrote the following dynamics equation: $$\Big(F(x,t)+dF(x,t)\Big) - F(x,t) = \Big(dm(x)\Big){ {\partial}w^2(x,t)\over{\partial}t^2}\tag{2}$$ where $dm(x)$ is the mass of the Infinitesimal element of the rod. Next, I crossed out the forces $F(x,t)$ and defined the mass $dm(x)$ as a product of volume $dV(x)$ and density: $$dF(x,t) = {\rho}\Big(dV(x)\Big){ {\partial}w^2(x,t)\over{\partial}t^2}\tag{3}$$ Since the volume $dV(x)$ is: $$dV(x) = A(x)dx\tag{4}$$ equation $(3)$ can be rewritten as: $$dF(x,t) = {\rho}A(x)dx{ {\partial}w^2(x,t)\over{\partial}t^2}\tag{5}$$ After this, I defined $dF(x,t)$ as: $$dF(x,t) = d\Big(P(x,t)A(x)\Big)=\Big(dP(x,t)\Big)A(x)+ P(x,t)\Big(dA(x)\Big)\tag{6}$$ where $P(x,t)$ is the pressure. By expanding the total derivatives, I obtained: $$dF(x,t) = \Big( { \partial P(x,t) \over \partial x}dx + { \partial P(x,t) \over \partial t}dt \Big)A(x) + P(x,t){ \partial A(x) \over \partial x}dx \tag{7}$$ Hook's law states: $$P(x,t) = E{ {\partial}w(x,t)\over{\partial}x}\tag{8} $$ and because of that, I rewrote equation $(7)$ as: $$dF(x,t) = E\Bigg(\Big( { \partial ^2 w(x,t) \over \partial x^2}dx + { \partial ^2 w(x,t) \over \partial x \partial t}dt \Big)A(x) + w(x,t){ \partial A(x) \over \partial x}dx\Bigg) \tag{9}$$ In the end, I derived the expression: $$E\Bigg(\Big( { \partial ^2 w(x,t) \over \partial x^2}dx + { \partial ^2 w(x,t) \over \partial x \partial t}dt \Big)A(x) + w(x,t){ \partial A(x) \over \partial x}dx\Bigg) = {\rho}A(x)dx{ {\partial}w^2(x,t)\over{\partial}t^2}\tag{10}$$ In equation $(10)$ there is an extra term. Namely, the ${ \partial ^2 w(x,t) \over \partial x \partial t}dt$ term. If that did not exist, I would get equation $(1)$. So, my question is, why does that term not exist in equation $(1)$?

  • $\begingroup$ The only explanation is that the pressure on a line x=cst is constant, $\partial p(x,t)/\partial t |_{x=cst}=0$ $\endgroup$
    – The Tiler
    Commented Sep 26, 2022 at 19:17

1 Answer 1


The problem is in equation 2, it should be the increment in the $x$ direction, the part in parenthesis should be

$$F + \frac{\partial F}{\partial x} dx\, .$$

Also, I would probably take finite values for the size of your element: $\Delta x$. Then you can take the limit, and new derivatives appear.

  • $\begingroup$ You are correct. I found the solution in a nother piece of literature. Thank you Nicoguaro. $\endgroup$ Commented Oct 2, 2022 at 17:14

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