# Is the velocity with which molecules of the medium oscillate same as velocity of the sound in that medium?

I wanted to know that under sound field applied molecules of the medium oscillate with some velocity in the direction of sound propagation. Is this velocity same as velocity of the sound in that medium? I am interested in liquid medium.

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This is a good question and while I don't know the answer, I strongly suspect the two velocities are related. It would be interesting to know by what relationship and if it had anything to do with the frequency of the sound. – Dom Mar 16 at 10:31
If a molecule is oscillating there is not a unique velocity vector, by definition. Are you talking about an instantaneous velocity in the direction of sound propagation? that at best will be enlarging and diminishing? an average speed? it is not a well defined quantity – anna v Mar 16 at 11:58
Do you mean average kinetic energy in the bulk flow rest frame (formally the second velocity moment of the distribution function) or something like the quantum vibrational modes of an atom? They are very different phenomena as some answers have eluded to below. – honeste_vivere Mar 16 at 12:33

It can't be, for the celerity of "vibrating" fluid parcels / groups of molecules (for volume or surface waves) also depends on the amplitude of the wave. Moreover $c_\phi = \omega/k$ ( and $c_g = \partial\omega/\partial k$ if dispersive) while the fastest parcels on the fastest moment of the cycle move at $A\omega$, without a dependency on k.

Now at the true molecular scale, even without sounds wave around, the thermal agitation of molecules (not parcels) makes a distribution a velocity around 0 (if no ground motion like wind... or sound waves :-) ) with magnitude peaking around the order of magnitude of sound waves, at least for perfect gas. But it's a wide distribution (responsible for many macroscopic effect, e.g. evaporation/condensation, etc).

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Remarkably enough though, sound speed in a gas is about the same as the mean speed of thermal motion of the particles! This follows from the equipartition theorem. (It's not in fact the same speed; for a monoatomic gas it's about 75%.) – leftaroundabout Mar 16 at 23:21
and it's the (quadratic) mean. I.e. there is a whole large distribution of speeds around. – Fabrice NEYRET Mar 17 at 2:23

I am interested in liquid medium.

The answer is probably not because the speed of sound in most liquids decreases with increasing temperature and yet the molecules in a liquid would have more kinetic energy if the temperature increased. Source: Kaye & Laby Properties of sound in liquids

As ever water is anomalous and shows an increase in speed reach a maximum at about 75 $^\circ$C. Kaye & Laby suggest the following reason "mainly because of the temperature dependence of the adiabatic compressibility of the water molecule itself"

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It is true that as temperature rises molecules of the liquid will have greater kinetic energy and hence higher velocity but I am interested to the velocity with which molecules of the medium oscillate under applied sound field and not its thermal motion which is random and increases with increase of temperature. – Vishal Mar 16 at 11:18
So, does increased kinetic energy of a medium cause increased amplitude of "brownian motion", or increased velocity, or both, or other? – Todd Wilcox Mar 16 at 14:26

I can elaborate on this question with respects to crystalline solids. Usually atoms in solids posses many possible vibrational modes. Information of all of them are collected in the phonon spectrum (see Fig. below).

Those vibrations are not oscillations of individual atoms, but their collective correlated motion. Atoms can vibrate in traverse or longitudinal directions relative to the direction of the wave propagation. These vibrations correspond to optical and acoustic phonons respectively. However, the sound waves are longitudinal waves only. Therefore, from many possible collective oscillations of atoms, only some of them are responsible for the propagation of sound.

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No. For linear acoustics, the speed of sound, $c$, is given by $$c^{2} = \left(\frac{\partial \rho}{\partial p}\right)_{\!s} \, ,$$ and therefore is related to the thermodynamic equation of state. (The subscript $s$ indicates constant entropy.) This is the speed at which the wave propagates through the medium. A typical value for water, for example, is $c \approx 1500 \; \mathrm{ m/s}$.

The amplitude of the velocity oscillation of the molecules is known as the particle velocity, $v$. For a plane wave, the particle velocity is proportional to the acoustic pressure $p$ as $$v = \frac{p}{\rho c}.$$ The denominator, $\rho c$, is called the characteristic impedance. Again for water, $\rho \approx 1000 \; \mathrm{ kg/m^{3} }$, so $\rho c \approx 1.5 \times 10^{6} \; \mathrm{ kg / (m^{2} \, s )}$. A modest acoustic pressure of 1 Pa would then have an associated particle velocity of $v \approx 6.7 \times 10^{-7} \; \mathrm{ m/s }$— Quite different from the sound speed!

You can refer to Chapter 1 of Acoustics by Allan Pierce for derivations and details.

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So, How to now acoustic pressure acting on the fluid? – Vishal Mar 17 at 10:44
Ideally, this is what you measure as a waveform with a hydrophone or transducer. Alternatively, if you know your sound pressure level in dB, you can invert the formula to get the RMS pressure. Make sure to use 1 micropascal as your reference pressure if in water. This all assumes plane waves, though, and there are modifications for wavefront spreading. Do you have a particular geometry or experiment in mind? – Greg Lyons Mar 17 at 14:13
Yes, I wanted to know how to calculate pressure density from the probe sonicator immersed in suspension. From there I will be able to calculate the velocity of the fluid as you suggested earlier. – Vishal Mar 17 at 16:27
I've used sonicators, but I don't know how I'd calculate the pressure field. I'd measure the SPL with an appropriate ultrasonic transducer. The formula I gave is not going to be correct for that geometry: You'll have spreading effects and reflections from the vessel walls. This is becoming significantly different from your original question, though. – Greg Lyons Mar 18 at 1:48

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