# Error of deriving the variable cross section rod wave equation

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:

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

where:

• $$x$$ 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. 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 $$(1)$$?

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

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.