Derivation of D'Alembert equation (for pressure waves)

  • #1
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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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Answers and Replies

  • #2
olivermsun
Science Advisor
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...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?
It's just the Chain Rule from calculus. It expresses whatever relationship there is between the variables (or their derivatives, anyway).
 
  • #4
olivermsun
Science Advisor
1,244
118
That is the classical wave equation. The solutions of that are waves which vary in time and propagate in spac.

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

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