# From Table 17.1, it is apparent that the speed of sound is significantly greater (by roughly a factor of 10) in solids than in gases. This significant.

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From Table 17.1, it is apparent that the speed of sound is significantly greater (by roughly a factor of 10) in solids than in gases. This significant difference can be most easily attributed to a) the difference in density between gases and solids, b) the difference in compressibility between gases and solids, c) both a and b, or d) neither a nor b. (end of section 17.1) QUICK QUIZ 17.1

(b). From Equation 17.1, v = (B/ ), it is apparent that changes in compressibility and density will affect the speed of sound. However, the density of solids is typically a factor of 1000 greater than for gases, and this would contribute to a lower speed of sound. The fact that the speed of sound is higher in solids implies that the incompressibility, or bulk modulus B, must be tremendously greater in solids than in gases to compensate for the density difference. This is, in fact, the case as it is much, much easier to compress a gas than a solid. QUICK QUIZ 17.1 ANSWER

For the periodic sound wave shown in Figure 2, the average velocity of an individual molecule of gas will be a) less than the velocity of the sound wave, b) the same as the velocity of the sound wave, c) greater than the velocity of the sound wave, d) dependent on the region that it is in, rarefaction or compression, or e) zero. (end of section 17.2) QUICK QUIZ 17.2

(e). It is important to remember that, for a sound wave, as was true for the waves discussed in Chapter 15, there is no net transport of matter. Each molecule of the gas oscillates back and forth about its equilibrium position but the equilibrium position does not change. QUICK QUIZ 17.2 ANSWER

Two stations record the sound intensities from a distant explosion. Station A, 2 miles east of the blast, records a sound intensity of 2 x 10 -9 W/m 2. Station B, 6 miles west of the blast, records a sound intensity of 2 x 10 -10 W/m 2. The chief reason for the difference in the two readings is that a) most of the sound is absorbed into the ground before it reaches Station B, b) most of the sound is reflected back towards the source before it reaches Station B, c) the sound does not travel uniformly away from the blast but much more so in an eastward direction, or d) the power from the sound is spread over a much bigger area by the time it reaches Station B. (end of section 17.3) QUICK QUIZ 17.3

(d). Equation 17.7 describes the intensity that is associated with the power from the source, equally distributed over a sphere of radius r. Station B is 3 times as far from the blast as the first station. Since I = 1/r 2, the intensity should be 1/3 2 or 1/9 less at Station B than at Station A. The fact that the decrease in intensity does roughly scale by this amount indicates the decrease is due mainly to the power distributed over a larger sphere. QUICK QUIZ 17.3 ANSWER

Using a sensitive sound meter, you measure the intensity of the sound of a running spider to be –10 dB. The negative sign implies that a) the spider is moving away from you, b) the frequency of the sound is less than is audible to humans, c) the intensity of the sound is less than is audible to humans, or d) this could not happen, the intensity of sound can never be negative, even on the decibel scale. (end of section 17.3) QUICK QUIZ 17.4

(c). Equation 17.8, = 10 log (I/I 0 ), compares the intensity of any sound to the minimum intensity, I 0 = 1.00 x 10 -12 W/m 2, that is audible to humans. A negative value for implies that the numerator, I, is less than the denominator, I 0 and that the sound is not audible to humans. In this case, log (I/I 0 ) = -1, which yields an intensity I = 1.00 x 10 –13 W/m 2. QUICK QUIZ 17.4 ANSWER

You are traveling in a car at a speed of 50 mph toward a stationary ambulance with its siren blaring at a constant frequency. You hear a frequency, f 1. In another situation, you wait on a street corner and listen to the same siren from the same ambulance as it travels toward you at 50 mph. In this case, you hear a frequency f 2 which is a) less than f 1, b) the same as f 1, c) greater than f 1, or d) impossible to determine without knowing the source frequency, f, of the siren. (end of section 17.4) QUICK QUIZ 17.5

(c). Equations 17.9 and 17.11 describe the two situations for a moving observer and moving source. Since the relative velocity, v r = 50 mph, is the same for each situation, we can replace the variables, v O = v S = v r. We have, In all cases, the denominator will be less than the numerator and f 2 will be greater than f 1. QUICK QUIZ 17.5 ANSWER

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