16.6.1 Predicting the Speed of Sound

Let’s return to our small, cylindrical piece of fluid and this time, in FIGURE 16.24 , apply Newton’s second law to it. The fluid pressure at position x is $p (x, t)$ , and this pressure pushes the cylinder to the right with force $a p (x, t)$ . At the same time, the fluid at position $x + Δ x$ has pressure $p (x + Δ x, t)$ , and it pushes the cylinder to the left with force $a p (x + Δ x, t)$ . The net force is

F_{net x} = a p (x, t) - a p (x + Δ x, t) = - a (p (x + Δ x, t) - p (x, t))

(16.47)

The minus sign arises because the pressure forces push in opposite directions.

The figure illustrates the fluid pressure that exerts a net force on a cylinder.

FIGURE 16.24 Full Alternative Text

Newton’s second law is $F_{net x} = m a_{x}$ . The cylinder’s mass is $m = ρ V = ρ a Δ x$ , where $ρ$ is the fluid density. (Don’t confuse area a with acceleration $a_{x}!$ ) The acceleration, just as in our analysis of a string, is the second partial derivative of displacement with respect to time. Thus the second law for our little cylinder of fluid is

F_{net x} = - a (p (x + Δ x, t) - p (x, t)) = m a_{x} = ρ a Δ x \frac{\partial^{2} D}{\partial t^{2}}

(16.48)

The area a cancels, and a slight rearrangement gives

\frac{\partial^{2} D}{\partial t^{2}} = - \frac{1}{ρ} \frac{p (x + Δ x, t) - p (x, t)}{Δ x} \to - \frac{1}{ρ} \frac{\partial p}{\partial x}

(16.49)

where, once again, we have a derivative in the limit $Δ x \to 0$ .

Fortunately, we already found, in Equation 16.43, that $p = - B \partial D / \partial x$ . Substituting this for p in Equation 16.49 gives

\frac{\partial^{2} D}{\partial t^{2}} = \frac{B}{ρ} \frac{\partial^{2} D}{\partial x^{2}}

(16.50)

Equation 16.50 is a wave equation! Just as in our analysis of a string, applying Newton’s second law to a small piece of the medium has led to a wave equation. We’ve already shown that a sinusoidal traveling wave, Equation 16.44, is a solution, so we don’t need to prove it again. Further, by comparing Equation 16.50 to the general wave equation, Equation 16.35, we can predict the speed of sound in a fluid:

v_{sound} = \sqrt{\frac{B}{ρ}}

(16.51)

TABLE 16.3 gives values of the bulk modulus for several common fluids.

TABLE 16.3 Bulk moduli of common fluids

Medium	B (Pa)
Mercury	$2.9 \times 10^{10}$
Water	$2.2 \times 10^{9}$
Ethyl alcohol	$1.1 \times 10^{9}$
Helium (1 atm)	$1.69 \times 10^{5}$
Air (1 atm)	$1.42 \times 10^{5}$

EXAMPLE 16.7 The speed of sound in water

Predict the speed of sound in water.

SOLVE From Table 16.3, the bulk modulus of water is $2.2 \times 10^{9} Pa$ . The density of water is $1000 {kg/m}^{3}$ . Thus we predict

v_{sound} = \sqrt{\frac{2.2 \times 10^{9} Pa}{1000 {kg/m}^{3}}} = 1500 m/s

This matches the value given earlier in Table 16.1 to two significant figures, which is all we can compute with the data provided.

For gases, both B and $ρ$ are proportional to the pressure, so their ratio is independent of pressure. At $0 ° C$ and 1 atm, the density of air is $ρ_{0} = 1.29 {kg/m}^{3}$ . Thus the speed of sound in air at $0 ° C$ is

v_{sound in air} = \sqrt{\frac{B_{0}}{ρ_{0}}} = \sqrt{\frac{1.42 \times 10^{5} Pa}{1.29 {kg/m}^{3}}} = 331 m/s (at 0 ° C)

(16.52)

exactly as shown in Table 16.1.

You can use the ideal-gas law to show that the density (at constant pressure) of a gas is inversely proportional to its absolute temperature T in kelvins. If the density at $0 ° C$ and 1 atm is $ρ_{0}$ , then the density at temperature T is

ρ_{T} = ρ_{0} \frac{273}{T (K)} = ρ_{0} \frac{273}{T (° C) + 273}

(16.53)

where we used $0 ° C = 273 K$ to convert kelvin to $° C$ . Thus a general expression for the speed of sound in air is

v_{sound in air} = \sqrt{\frac{B}{ρ}} = \sqrt{\frac{B_{0}}{ρ_{0}} \frac{T (°C) + 273}{273}} = 331 m/s \times \sqrt{\frac{T (°C) + 273}{273}}

(16.54)

This was the expression given without proof in Section 16.5. Now we see that it comes from the wave equation for sound with a little help from the ideal-gas law.