I am currently studying out of Vibrations and Waves by AP French. I have been trying to solve this problem but seem to be having trouble deriving mathematical equations for it:

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I have a qualitative understand about certain parts of the question that I will explain shortly, but is there any law that I can use to relate pressure of the gas in the tube and the displacement of a particle in the tube during an oscillation?

Here is my qualitative understanding of the question:

For part a, if particles were not at the same pressure as the pressure of the surrounding medium, particles would naturally flow with the concentration gradient until the concentration of air particles at the open end of the tube would be about the same as the pressure of the surroundings. This won't be exactly the same (I think due to oscillations still occurring, so the gas at the end of the tube won't always be exactly the same pressure of the outside medium at any given time), but it will be relatively close. I'm not exactly sure why this would result in a maximum movement of air, though.

For part b I have a better understanding I believe. Particles would collect at the end of the tube due to there being a physical barrier over which particles cannot cross, and thus we would have a node where particles must stop their motion going towards the barrier and turn around in the opposite direction. Because particles must stop at the barrier, the concentration of particles must change from very high (in comparison to other non-mode spots in the tube) to very low (in comparison, again, to other non-node spots in the tube) over time. Thus we would have a maximum pressure variation at this nodes.

Is this intuition correct, and if not, how can I remedy this (hopefully with some mathematics)? Thank you for any help you can offer!


1 Answer 1


The pressure and density changes that take place in a gas when it is a medium for stationary or progressive sound waves are more or less adiabatic. Hence treating the gas as ideal, $$pV^\gamma = \text{constant},\ \ \ \text{leading to} \ \ \ \ \frac {dp}{dV}=-\gamma \frac pV\ \ \ \ \text{so for small changes}\ \ \ \frac {\Delta p}{\Delta V}=-\gamma \frac pV. $$ In a tube of cross-sectional area $A$ we now take $V$ as the small volume of air between $x$ and $x+\Delta x$ along the tube, so $V = A \Delta x$.

Suppose that, because of the wave, the gas at $x$ moves forward by $\xi$, and the gas at $x+\Delta x$ moves forward by $\xi + \frac{d\xi}{dx}\Delta x$. So the increase in volume of the gas that had volume $V$ will be $$\Delta V =A\frac{d\xi}{dx}\Delta x=V\frac{d\xi}{dx}$$ So substituting in our equation for adiabatic changes, $$\Delta p=-\gamma p \frac{d\xi}{dx}$$ So the pressure change at a point due to the sound wave is proportional to the gradient of the displacement due to the sound wave. This applies at any instant, but since pressure differences will cause accelerations, $\Delta p$ and $\xi$ will also depend on time, so we ought to write $$\Delta p=-\gamma p\frac{\partial\xi}{\partial x}$$

You can, for example, substitute a sinusoidal variation of displacement with distance and derive the pressure change as a function of distance.

  • $\begingroup$ This answer’s logic makes a lot of sense so thank you very much! My main question is, though: if p is about constant for a point at the tube’s open end, the the left side of your final equation would be 0 and thus theres 0 change in displacement with respect to position in the tube. This could make sense if we’re thinking gas near the open end would move together. But how could this relate to the maximum movement of air as the question states? Would gas moving together mean more gas is able to move at the open end than gas at other points in the tube? $\endgroup$
    – Lightbulb
    Oct 11, 2020 at 20:24
  • $\begingroup$ Similarly, if the gas must stop at the barrier, then the right side is now zero. So there would be high pressure consistently at the end of the tube because the change in pressure there is consistently 0? That seems slightly unphysical to me but I could definitely me wrong $\endgroup$
    – Lightbulb
    Oct 11, 2020 at 20:25
  • $\begingroup$ At the open end $\Delta p$ is zero so there is no change of displacement wrt distance – which is consistent with a maximum or minimum (that is maximum in the other direction!) of displacement. This accords with the slogan, "A pressure node is a displacement antinode". At the closed end of the pipe we have a displacement node and a pressure antinode. $\endgroup$ Oct 11, 2020 at 20:34
  • $\begingroup$ Awesome, this makes complete sense. Thank you very much! $\endgroup$
    – Lightbulb
    Oct 12, 2020 at 13:57

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