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I have trouble understanding some derivations about longitudal waves. I need to derive $\Delta p=B \frac{\partial s}{\partial x}$ from $\Delta p=B \frac{\Delta V}{V}$ knowing that $\Delta V=A\Delta s $ and $ V= A\Delta x$.

Where $B= \frac{\Delta p}{\Delta V/V }$, $\Delta p$ is pressure difference, $\Delta V$ volume difference, $\Delta s$ longitudial displacement caused by wave, $\Delta x$ is interval of of length of tube filled with liquid.

I don't understand why $\Delta V=A\Delta s $.

This derivation is needed to derive $\Delta p=Bks_m \sin(kx-wt)$ from $s=s_m cos(kx-wt)$.

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You need to find the change in volume of air in that small tube of $\Delta x$ length. So, if at $x$ the displacement of air particles is $s$ and at $x+\Delta x$ it is $s+\Delta s$, how much change in volume has occurred to the column of air that was initially in the $\Delta x$ interval?

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  • $\begingroup$ You just paraphrased part of my question. $\endgroup$
    – Alex Alex
    Commented May 20, 2021 at 8:28
  • $\begingroup$ No...I'm answering. The new volume is definitely $(A\Delta x-As+A(s+\Delta s))$. So change is $A\Delta s$. Instead of doing it myself, I was giving you a hint. $\endgroup$
    – Ritam_Dasgupta
    Commented May 20, 2021 at 9:03
  • $\begingroup$ Still, I don't understand. $\endgroup$
    – Alex Alex
    Commented May 20, 2021 at 9:27
  • $\begingroup$ As I said, $V_{new}=A(\Delta x + \Delta s)$ and $V_{old}=A\Delta x$. So $\Delta V= A \Delta s$. $\endgroup$
    – Ritam_Dasgupta
    Commented May 20, 2021 at 9:33

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