27
$\begingroup$

I understand that the speed of sound is inversely proportional to the density of the medium as shown here and as answered for this question.

The problem now is that the speed of sound in air actually decreases with altitude although the density of the air decreases. This is shown here and here.

I understand that the speed of sound also depends on the elasticity, but I'm not sure how this can change for air.

So what is actually happening? How can the speed of sound decrease although the density has also decreased?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Good question! A small comment: When linking to graphs as sources, it would help to link the article where the image is from - that helps the reader to understand what the graph means and how the values were determined. $\endgroup$
    – JiK
    Sep 13, 2019 at 19:40

4 Answers 4

Reset to default
39
$\begingroup$

Wikipedia gives a pretty much straightforward answer. In an ideal gas, the speed of sound depends only on the temperature:

$$ v = \sqrt{\frac{\gamma \cdot k \cdot T}{m}} $$

So it neither decreases, nor increases with altitude, but just follows air temperature as can be seen in this graph:

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob ♦
    Sep 13, 2019 at 14:18
  • $\begingroup$ "The speed of sound in an ideal gas depends only on its temperature and composition. The speed has a weak dependence on frequency and pressure in ordinary air, deviating slightly from ideal behavior." - Wikipedia $\endgroup$
    – Vinicius ACP
    Mar 26, 2021 at 15:10
  • 1
    $\begingroup$ @ViniciusACP According to ideal gas law, temperature maps to pressure directly. Agree, that there's a sound wave dispersion, but for sake of reason these can be ignored, as long as graph shows strong correlation only between $T$ and $v$. $\endgroup$
    – Agnius Vasiliauskas
    Mar 26, 2021 at 20:59
14
$\begingroup$

The speed of sound in a gas is given by $\sqrt{ \dfrac {\gamma \,P}{\rho}}= \sqrt{\gamma \, R \, T}$ where the temperature, $T$, is in kelvin, $\gamma$ is the ratio of the specific heat capacities of a gas at constant pressure and constant volume and $R$ is the specific gas constant.

With increasing altitude there is a decrease in the density but also a decrease in the pressure, but not at the same rate because there is also a change in the temperature.

As the altitude increases, the temperature decreases and so does the speed of sound and then when the temperature increases so does the speed.

enter image description here

Source - Engineering Toolbox

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Have you built this table? Very good and my best complimens. $\endgroup$
    – Sebastiano
    Sep 13, 2019 at 21:27
  • $\begingroup$ On the left-hand side of your equation, $\rho$ is mass density, not molar density. This means that the right-hand side is incorrect as well — it's dimensionally inconsistent. I believe that the correct expression would be $\sqrt{ \gamma R T/m_\text{mol}}$, where $m_\text{mol}$ is the molar mass. $\endgroup$
    – Michael Seifert
    Sep 13, 2019 at 21:36
  • 1
    $\begingroup$ @MichaelSeifert I understand what you have written and the apparent error is due to the fact that I did not define my gas constant $R$ as the specific gas constant with units of $\rm J\, kg^{-1} \, K^{-1}$. $\endgroup$
    – Farcher
    Sep 13, 2019 at 21:48
  • 1
    $\begingroup$ @Sebastiano The table comes from the Engineering Toolbox website. I have also put the source of the table in my answer. $\endgroup$
    – Farcher
    Sep 16, 2019 at 10:10
5
$\begingroup$

Sound is simply a compression wave. The velocity of the wave is inversely related to the square root of fluid pressure, and directly proportional to the materiel’s Young modulus. A lower density means a lower pressure, which increases the wave velocity as you noticed.

Traveling through a compressible medium such as air, the simple equation for the velocity (v) of a compression wave is the square root of the Young’s modulus (B) divided by pressure (p)

$$v=\sqrt{\frac{B}{p}}.$$

Please see the following link for the derivation of this formula.

https://youtu.be/qVusackhzBs

But as the materiel becomes less dense, it reduces the surface area where the force is applied, the Young modulus decreases.

Increasing altitude thins the air out (reduces the surface area over which the force is applied) faster than it reduces pressure, resulting in a net decrease in the ratio, and therefore wave speed.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Welcome to Stack Exchange! You can enter equations using MathJax, I've converted for you. $\endgroup$
    – uhoh
    Sep 13, 2019 at 1:12
  • 1
    $\begingroup$ $B$ is the bulk modulus not Young’s modulus. $\endgroup$
    – Farcher
    Sep 13, 2019 at 8:13
0
$\begingroup$

In laymen terms, sound propagation depends on air particles colliding. The closer the particles to each other the quicker they can collide. In a less dense environment, the particles must travel farther to collide. Particle size has an effect as well as can be heard in glove boxes containing Helium vs Argon. Temperature also has an effect in the amount of energy available to the particles and therefore excitability. This is very a simplistic and generalized answer without any mathematical or scientific proofs.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.