# I Derivation of D'Alembert equation (for pressure waves)

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1. Sep 29, 2018

### fcoulomb

In my textbook there is an explanation of a derivation of D'Alembert equation for pressure waves. ($\frac{\partial^2 y}{\partial x^2}=\frac{\rho}{\beta}\frac{\partial^2 y}{\partial t^2}$)
I put the picture (the only one I found on internet) but I'll call $y_1 ,y_2$ as $\psi_1,\psi_2$ and the second volume will be smaller that the first volume (unlike in the picture) so $V_2<V_1$.

Starting by the fact that there is a pression on the left bigger than pressure on the right (so the volume will be smaller) $P_1 >P_2$, I have $(P_1-P_2)A=ma_x= \rho_0 A dx \frac{\partial^2 y}{\partial t^2}$.

Then I write $P_{1,2}$ as $P_0+dP_{1,2}$, so $dP_1-dP_2= -\frac{\partial(dP)}{\partial x} dx$.

Then I use the fact that $dP=\beta \frac{d\rho}{\rho_0}$ (with $\beta$ the coefficient of compressibility), but I need to write the $d\rho$ as a function of x, so

$\rho_0 V_1=(\rho_0+d\rho)V_2 \rightarrow \rho_0 A dx= (\rho_0+d\rho) A (dx-d\psi)$

Then by book says that I can write $d\psi=\frac{\partial \psi}{\partial x}dx$, but I don't understand how these two variables are related.
Any help?

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2. Sep 29, 2018

### olivermsun

It's just the Chain Rule from calculus. It expresses whatever relationship there is between the variables (or their derivatives, anyway).

3. Sep 29, 2018

### Staff: Mentor

That is the classical wave equation. The solutions of that are waves which vary in time and propagate in space.

Are you asking how to derive the solution to that differential equation? If so, you can find it here:
https://en.wikipedia.org/wiki/Wave_equation#General_solution

edit: wrong link replaced.

4. Sep 29, 2018

### olivermsun

That's why the title of the thread says "D'Alembert equation (for pressure waves)."