(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that [tex]v = \sqrt{B/\rho}[/tex]

2. Relevant equations

3. The attempt at a solution

Let the density of the medium be ρ.

Suppose you have a single pulse of a longitudinal wave inside a pipe traveling with speed v.

At any time, consider the location of the pulse. Originally, the mass at the location was m. Suppose the pulse increases the mass of its current location by Δm. With the pulse the total mass there becomes m + Δm.

The mass at the location of the pulse must be compressed; otherwise, its volume would expand by ΔV to keep the density constant. We have

[tex] \frac{m+\Delta m}{V + \Delta V} = \frac{m}{V} \implies \rho = \frac{\Delta m}{\Delta V} [/tex]

So the mass at that location must be compressed by ΔV. From the definition of the bulk modulus, we have

[tex] B = \frac{\Delta P}{-\Delta V / V} = \frac{V}{\Delta m} \cdot \Delta P \cdot \frac{\Delta m}{-\Delta V} [/tex]

Then we have

[tex] B = \frac{V}{\Delta m} \cdot \Delta P \cdot \rho \implies \frac{B}{\rho} = \frac{V}{\Delta m} \cdot \Delta P [/tex]

ΔP must be the pressure difference between a stationary pulse and a moving one, so we have by Bernoulli that

[tex] P_{stationary} = P_{pulse} + \frac{1}{2} \left(\frac{m+\Delta m}{V}\right)v^2 [/tex]

so

[tex] \Delta P = \frac{1}{2} \left(\frac{m+\Delta m}{V}\right)v^2 [/tex]

We end up with:

[tex] \frac{B}{\rho} = \frac{V}{\Delta m} \cdot\frac{1}{2} \left(\frac{m+\Delta m}{V}\right)v^2 = \frac{m+\Delta m}{2\Delta m}v^2 [/tex]

Something must have gone wrong somewhere.

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