Physics · Ch 11 — Waves

Velocity of Longitudinal Waves in an Elastic Medium

11.3.2

Velocity of Longitudinal Waves in an Elastic Medium

For a longitudinal wave travelling through an elastic medium -- taking air held inside a long cylindrical tube of cross-sectional area $A$ as the concrete case -- the wave speed can again be derived from Newtonian momentum-impulse reasoning. A piston at one end of the tube is set moving at speed $u$ into the initially undisturbed fluid; in a short time interval $\Delta t$ the piston itself moves a distance $\Delta d = u\Delta t$ , while the disturbance it creates (the leading edge of the compressed region) moves a much greater distance $\Delta x = v\Delta t$ , since the wave speed $v$ is generally much larger than the piston speed $u$ . The mass of fluid that has been set into motion in this time is $\Delta m = \rho A(v\Delta t)$ , so the momentum imparted to it by the piston's motion is $\Delta p = [\rho A(v\Delta t)]u$ ; since this momentum change equals the impulse delivered by the small excess pressure $\Delta P$ acting over area $A$ for time $\Delta t$ , equating the two gives $\Delta P = \rho v u$ . Relating this pressure change to the fractional volume change via the medium's bulk modulus $K$ (defined by $\Delta P = K\,\Delta V/V$ ) and simplifying leads to the general elastic-wave speed formula $v = \sqrt{K/\rho}$ , or more generally $v=\sqrt{E/\rho}$ using whichever elastic modulus $E$ is appropriate for the medium in question. For a thin one-dimensional solid rod, only Young's modulus $Y$ matters, giving $v=\sqrt{Y/\rho}$ (steel, with $Y\approx2\times10^{11}\ \text{N/m}^2$ , carries sound at roughly 5000 m/s); for a genuine three-dimensional solid, both the bulk modulus $K$ and the rigidity (shear) modulus $\eta$ contribute, $v=\sqrt{(K+\tfrac{4}{3}\eta)/\r …

Figure 11.16Longitudinal waves in the fluid produced by displacing the fluid using a piston

What this figure shows. A long cylindrical tube of cross-sectional area A is shown filled with a fluid of density rho, initially at rest and at pressure P, with a piston at the left end. The figure shows the piston having been set in motion toward the right at speed u for a short time interval, so that a compressed region of fluid -- shown with force $F=(P+\Delta P)A$ acting on its leading face, ahead of the undisturbed fluid still at force $F=PA$ -- has moved a distance $v\Delta t$ (the disturbance's own speed) while the piston itself has moved a shorter distance $u\Delta t$ . This figure sets up the momentum-impulse argument used to derive the general elastic-wave speed formula: by equating the impulse delivered by the small excess pressure $\Delta P$ over the time $\Delta t$ to the momentum gained by the mass of fluid set into motion, the derivation arrives first at $\Delta P = \rho v u$ and then, combining with th …

Table 11.2Speed of sound in various media

S.No.	Medium	Speed (m/s)
Solids
1.	Rubber	1600
2.	Gold	3240
3.	Brass	4700
4.	Copper	5010
5.	Iron	5950
6.	Aluminum	6420
Liquids at 25°C
1.	Kerosene	1324
2.	Mercury	1450
3.	Water	1493
4.	Sea Water	1533
Gas (at 0°C)
1.	Oxygen	317
2.	Air	331

Misc Example 11.7Speed of sound in a steel rod

Worked out. A steel rod has Young's modulus $Y=2\times10^{11}\ \text{N/m}^2$ and density $\rho=7800\ \text{kg/m}^3$ , and the task is to find the speed of sound travelling along it. Using the one-dimensional solid-rod formula $v=\sqrt{Y/\rho}$ gives $v=\sqrt{(2\times10^{11})/7800}\approx\sqrt{2.564\times10^7}\approx5.06\times10^3\ \text{m/s}$ , i.e. roughly 5064 m/s. This confirms numerically what Table 11.2 shows in general: sound travels dramatically faster through a stiff solid like steel than through a liquid or gas, which is exactly why a shepherd can detect an approaching train earlier by pressing …

Misc Example 11.8Bulk modulus and speed of sound in water from a compressibility measurement

Worked out. A 100 kPa increase in pressure is found to shrink a certain volume of water by 0.005% of its original volume, and the task is to compute both the bulk modulus of water and the resulting speed of sound (compressional waves) in it. The bulk modulus is defined as $B=-\Delta P/(\Delta V/V)$ , so substituting the given fractional volume change of $0.005\times10^{-2}=5\times10^{-5}$ and pressure change $\Delta P=100\times10^3\ \text{Pa}$ gives $B=(100\times10^3)/(5\times10^{-5})=2000\times10^6\ \text{Pa}=2000\ \text{MPa}$ . The speed of sound then follows from the liquid formula $v=\sqrt{B/\rho}$ with water's density $\rho=1000\ \text{kg/m}^3$ : $v=\sqrt{(2000\times10^6)/1000}=\sqrt{2\times10^6}\approx1414\ \text{m/s}$ , a value close to the 1493 m/s listed for water in Table 11.2, illustrating how a simple compressibility experiment lets the speed of sound in a …

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