Speed of Sound Waves

Sound waves are longitudinal waves which need a medium to travel. Here we find the speed of sound wave in a fluid such as a gas. We use the impulse-momentum theorem to find the speed of sound in a fluid. If $\vec J$ denotes the impulse for time $t$, the impulse momentum theorem is the change in momentum $\Delta \vec p$ in that time:

\[\vec J = \Delta \vec p \]

We consider a cylinder of cross-sectional area $A$ containing gas as shown in Figure 3 and the sound wave travels in positive x-direction of our coordinate system. The cylinder has a movable piston attached at its left end. Initially the cylinder contains a gas of uniform density $\rho$, but when the piston is pushed inward, the gas near the piston is compressed which is separated by a dashed line.

Initially when the gas was in undisturbed condition, the force on the piston by the gas was $ - PA\hat i$ and by Newton's third law the piston also exerts the same magnitude of force on the gas in opposite direction.

The dashed line separates the compressed and undisturbed gas. As the piston is pushed inward with the total force $(P + \Delta P)A\widehat i$ in such a way that all the particles of the moving gas move with the same constant velocity $v_\text{x}$, the piston moves a distance of $v_\text{x} t$ in time $t$.

When the piston is moved, a disturbance is created in the gas which travels with the wave speed $v$ and the dashed line also moves with this same speed. Therefore, in time $t$ the dashed line moves a distance of $vt$.

Figure 3 Speed of longitudinal wave (sound wave) in a gas.

The net or resultant force on the piston which compresses the part of the gas in the cylinder is $\Delta PA\hat i$, and in time $t$ the impulse due to this force is $\vec J = \Delta PAt\widehat i$. Now note the change in momentum of the compressed gas.

The volume of the gas on the left of the dashed line which is $V = vtA$ is decreased when the piston compresses the part of the gas and the volume of the compressed gas in time $t$ is $V + dV= vtA - v_\text{x}tA$ where $dV = - v_\text{x}tA$ but the mass of the compressed gas $m = vtA\rho$ remains the same.

The initial momentum of the compressed gas is zero as the gas was undisturbed and no gas particles were compressed. As the piston is moved inward, a part of the gas near the piston is compressed and moved with velocity $v_\text{x}$ and hence the final momentum of the compressed gas in time $t$ is $mv_\text{x} = vtA\rho v_\text{x}$. Using above equation of impulse-momentum theorem you will find,

\[\begin{align*} \Delta PAt &= vtA\rho {v_x}\\ {\rm{or,}}\quad \Delta P &= v\rho {v_x} \tag{1} \label{1} \end{align*}\]

If $B$ is the bulk modulus of the gas which is the ratio of bulk stress to strain, you have:

\[\begin{align*} B = - \frac{{\Delta PV}}{{\Delta V}} &= - \frac{{\Delta PvtA}}{{ - {v_x}tA}}\\ {\rm{or,}}\quad \Delta P &= \frac{{B{v_x}}}{v} \tag{2} \label{2} \end{align*}\]

Using Equations \eqref{1} and \eqref{2}, you will get,

\[\begin{align*} v\rho {v_x} = \frac{{B{v_x}}}{v}\\ {\rm{or,}}\quad v = \sqrt {\frac{B}{\rho }} \tag{3} \label{3} \end{align*}\]

The above Equation \eqref{3} gives the speed of sound wave in a fluid. If we had considered a liquid instead of gas in the cylinder, we would have got the same expression again. The Equation \eqref{3} shows that the wave speed of sound increases as the bulk modulus of the gas increases.

Bulk modulus is an elastic property of the material and as it increases the wave speed increases. On the other hand the wave speed decreases as the density of the fluid increases which seems reasonable because the density is the inertial property of the medium which resists the change in motion.

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