The speed of any mechanical wave, transverse or longitudinal, depends on both an inertial property of the medium (to store kinetic energy) and an elastic property of the medium (to store potential energy). Thus, we can generalize Eq. 16-26, which gives the speed of a transverse wave along a stretched string, by writing
where (for transverse waves) τ is the tension in the string and μ is the string’s linear density. If the medium is air and the wave is longitudinal, we can guess that the inertial property, corresponding to μ, is the volume density ρ of air. What shall we put for the elastic property?
In a stretched string, potential energy is associated with the periodic stretching of the string elements as the wave passes through them. As a sound wave passes through air, potential energy is associated with periodic compressions and expansions of small volume elements of the air. The property that determines the extent to which an element of a medium changes in volume when the pressure (force per unit area) on it changes is the bulk modulus B, defined (from Eq. 12-25) as
Here ΔV/V is the fractional change in volume produced by a change in pressure Δp. As explained in Section 14-3, the SI unit for pressure is the newton per square meter, which is given a special name, the pascal (Pa). From Eq. 17-2 we see that the unit for B is also the pascal. The signs of Δp and ΔV are always opposite: When we increase the pressure on an element (Δp is positive), its volume decreases (ΔV is negative). We include a minus sign in Eq. 17-2 so that B is always a positive quantity. Now substituting B for τ and ρ for μ in Eq. 17-1 yields
as the speed of sound in a medium with bulk modulus B and density ρ. Table 17-1 lists the speed of sound in various media.
The density of water is almost 1000 times greater than the density of air. If this were the only relevant factor, we would expect from Eq. 17-3 that the speed of sound in water would be considerably less than the speed of sound in air. However, Table 17-1 shows us that the reverse is true. We conclude (again from Eq. 17-3) that the bulk modulus of water must be more than 1000 times greater than that of air. This is indeed the case. Water is much more incompressible than air, which (see Eq. 17-2) is another way of saying that its bulk modulus is much greater.
Formal Derivation of Eq. 17-3
We now derive Eq. 17-3 by direct application of Newton’s laws. Let a single pulse in which air is compressed travel (from right to left) with speed ν through the air in a long tube, like that in Fig. 16-2. Let us run along with the pulse at that speed, so that the pulse appears to stand still in our reference frame. Figure 17-3a shows the situation as it is viewed from that frame. The pulse is standing still, and air is moving at speed ν through it from left to right.
Let the pressure of the undisturbed air be p and the pressure inside the pulse be p + Δp, where Δp is positive due to the compression. Consider an element of air of thickness Δx and face area A, moving toward the pulse at speed ν. As this element enters the pulse, the leading face of the element encounters a region of higher pressure, which slows the element to speed ν + Δν, in which Δν is negative. This slowing is complete when the rear face of the element reaches the pulse, which requires time interval
Fig. 17-3 A compression pulse is sent from right to left down a long air-filled tube. The reference frame of the figure is chosen so that the pulse is at rest and the air moves from left to right. (a) An element of air of width Δx moves toward the pulse with speed ν. (b) The leading face of the element enters the pulse. The forces acting on the leading and trailing faces (due to air pressure) are shown.
Let us apply Newton’s second law to the element. During Δt, the average force on the element’s trailing face is pA toward the right, and the average force on the leading face is (p + Δp)A toward the left (Fig. 17-3b). Therefore, the average net force on the element during Δt is
The minus sign indicates that the net force on the air element is directed to the left in Fig. 17-3b. The volume of the element is A Δx, so with the aid of Eq. 17-4, we can write its mass as
The average acceleration of the element during Δt is
Thus, from Newton’s second law (F = ma), we have, from Eqs. 17-5, 17-6, and 17-7,
which we can write as
The air that occupies a volume V (= Aν Δt) outside the pulse is compressed by an amount ΔV (= A Δν Δt) as it enters the pulse. Thus,
Substituting Eq. 17-9 and then Eq. 17-2 into Eq. 17-8 leads to
Solving for ν yields Eq. 17-3 for the speed of the air toward the right in Fig. 17-3, and thus for the actual speed of the pulse toward the left.
One clue used by your brain to determine the direction of a source of sound is the time delay Δt between the arrival of the sound at the ear closer to the source and the arrival at the farther ear. Assume that the source is distant so that a wavefront from it is approximately planar when it reaches you, and let D represent the separation between your ears.
(a) Find an expression that gives Δt in terms of D and the angle θ between the direction of the source and the forward direction.
Solution: The situation is shown (from an overhead view) in Fig. 17-4, where wavefronts approach you from a source that is located in front of you and to your right. The Key Idea here is that the time delay Δt is due to the distance d that each wavefront must travel to reach your left ear (L) after it reaches your right ear (R). From Fig. 17-4, we find
where ν is the speed of sound in air. Based on a lifetime of experience, your brain correlates each detected value of Δt (from zero to the maximum value) with a value of θ (from zero to 90°) for the direction of the sound source.
(b) Suppose that you are submerged in water at 20°C when a wavefront arrives from directly to your right. Based on the time-delay clue, at what angle θ from the forward direction does the source seem to be?
Solution: The Key Idea here is that the speed of the sound is now the speed νw in water, so in Eq. 17-10 we substitute νw for ν and 90° for θ, finding that
Fig. 17-4 A wavefront travels a distance d (= D sin θ) farther to reach the left ear (L) than to reach the right ear (R).
νw for ν and 90° for θ, finding that
Since νw is about four times ν, delay Δtw is about one-fourth the maximum time delay in air. Based on experience, your brain will process the water time delay as if it occurred in air. Thus, the sound source appears to be at an angle θ smaller than 90°. To find that apparent angle, we substitute the time delay D/νw from Eq. 17-11 for Δt in Eq. 17-10, obtaining
Then, to solve for θ we substitute ν = 343 m/s and νw = 1482 m/s (from Table 17-1) into Eq. 17-12, finding
and thus
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