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Problem Sllow t hat the quantity speed.
has the units of
Problem 19. W hat is the approximate frequency ran e over which sound with intensity 1 0-l2 W/m can be heard? Consult Fig. 17-3.
f
Solution T he units of pressure (force per unit area) divided by density (mass per unit volome) are ( ~ / m ~ ) / ( k 3) /= ~m ( ~ / k ~ ) ( m ~ =mm/s2)m = ( m / ~ )or ,those of / (~) ~ speed squared.
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Problem 9. A gas w ith density 1.0 kg/m3 and pressure 8.0x104 iV/m2 has sound speed 365 m/s. Are the gas molecules monatomic or diatomic? Solution Solving for 7 in Equation 17-1, we find 7 = p v2/P = (1.0 kg/m3)(365 m /s)*/(8.0~10' ~ / m = 1.67, very ~) close to the value for an ideal monatomic gas. (Actually, 7 - 513 = - 1 . 3 5 ~ 1 0 - ~or this gas.) f
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F IGURE
17-3 For reference.
Solution Inspection of Fig. 17-3 shows that for frequencies approximately between 1 and 6.5 kHz, t he threshold of hearing is at or below 10-l2 w/rn2.
Problem 15. Sound intensity in normal conversation is about 1 p w/m2. W hat is the displacement amplitude of air in a 2.5-kHz sound wave with this intensity? Solution As in Example 17-2, Equation 17-3c, combined with the atmospheric data in Example 17-1, can be used to calculate the displacement amplitude for sound waves of the specified frequency and intensity:
Problem 27. At a distance 2.0 m from a localized sound source you measure the intensity level as 75 dB. How far away must you be for the perceived loudness to drop in half (i.e., to an intensity level of 65 dB)? Solution A change of -10 dB, (i.e., PI - P = -10 dB) corresponds to the intensity decreasing by a factor of one tenth (i.e., I' = 1/10). To see this, note that Equation 17-4 may be written a s /Y - P = (-10 dB)log(I1/I), or 11/1 10(fl'-P)/'~d B + ~a n r = ~ isotropic point source of sound, the intensity falls inversely with the square of the distance (Equation 1 69), so r n = lor? or f = o ( 2 m) = 6.32 m.
2(10-6 w /m2) = 4.44 nm. (1.20 kg/m3) (343 m/s)
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Problem 18. A "tweeter" loudspeaker emitting 5.0 kHz sound has an oscillation amplitude of 1 pm. W hat must be the oscillation amplitude of a "woofer" speaker producing the same sound intensity at 30 Hz?
same air, Equation 17-3c constant. Therefore, a comparison of the woofer and tweeter yields (so), = ( S O ) ~ W * = ( 1 p m)(5 kHz/30 Hz) = 167 pm. /W~
Problem 33. The speed of sound in body tissues is essentially the same as in water. Find the wavelength of 2.0 MHz ultrasound used in medical diagnostics. Solution From Table 17-2, the speed of sound in water is 1497 m/s, s o Equation 16-1 gives t he wavelength of 2.0 MHz ultrasound as X = v / f = (1497 m / s ) s (2.0 MHz) = 0.749 mm.
Solution For the same
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Problem 39. When a stretched string is'clamped at both ends, its fundamental standing-wave frequency is 140 Hz. (a) What is the next higher frequency? (b) If the same string, with the same tension, is now clamped at one end and free at the other, what is the fundamental frequency? (c) What is the next higher frequency in case (b)? Solution-' ( a) The frequencies of the standing-wave of modes a string fixed at both ends are all the (positive) integer multiples of the fundamental frequency, f m = mf for m = 1, 2, . . . . (This follows from Equation 17-8 if we use f m = v/Xm = m (vl2L) = m(vlX1) = m f l.) T hus, f 2 = 21 1 = 2(140 Hz) = 280 Hz. ( b) T he velocity of transverse waves is the same (for the same string under the same tension), but when one end is fixed and the other free, the standing-wave wavelengths are A m = 4L, 4L/3, . . . 4L/(2m - I ) , . . . , where 2m - 1 represents any odd integer for m = 1, 2, . . . . (See Fig. 17-13 and its discussion in the text, or the solution to Problem 41.) Therefore, the fundamental frequency for the string fixed at one end is f l = v/4L = 3(v/2L) = 3(140 Hz) = 70 Hz, i.e., one half the fundamental frequency of the string fixed at both ends. ---
Problem 48. W hat length is necessary for an organ pipe (Fig. 17-27) to produce a 22-Hz tone (a) if the pipe is closed a t one end and (b) if it's open at both ends? Solution T he wavelength of a 22 Hz tone in "normal" air is X = (343 m /s)/(22 Hz) = 15.6 m. (a) In a "closed pipe,"X1 = 4L, so L = 3.90 m. (b) For an "open pipe," X I = 2L, or L = 7.80 m.
Problem 59. You're standing by the roadside as a truck approaches, and you measure the dominant frequency in the truck noise at 1100 Hz. As the truck passes the frequency drops to 950 Hz. What is the truck's speed? Solution T he result of part (a) of the preceding problem gives U/V = (1100 - 950)/(1100 950) = 0.0732. For sound waves' in "normal" air (Example 17-I ), this implies a truck speed of u = 0.0732(343 m/s) = 25.1 m/s = 90.4 km/h. (From Equation 17-10, the frequency, emitted by the truck is f = f i (1 - u/v) = f i(l u/v), where f l and f 2 a re the observed frequencies when the truck is approaching or receding, respectively. The solution of this equation for the source's speed is t"/v = ( fl - f i ) l ( f l + fd.1
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(c) In this case, the standing-wave frequencies are only the odd multiples of the fundamental frequency, f m = (2m - 1 )jl, therefore the second standing-wave mode has frequency f2 = 3f l = 3(70 Hz) = 210 Hz for this string.
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Problem 41. Show that only odd harmonics are allowed on a t aut string with one end tight and the other free. Solution For a string free at one end, the amplitude factor in Equation 17-7 is a maximum for x = L, i.e., 2 Asin kL = f2A. Therefore, kL = (2m - 1)7r/2, where 2m - 1 is an odd integer for m = 1, 2, . . . . In terms of standing-wave wavelengths, kL = (2?r/Xm)L = (2m - l )a/2, or L = (2m - 1)Xm/4, a s stated on page 428. In terms of frequency, f m = v/Xm = (2m - 1 )j l , where j l = v/4L is the frequency of the fundamental. Thus, only odd harmonics occur.
Problem 61. Use t he binomial approximation to show that Equations 17-10 and 17-11 give the same result in the limit u <( v . -.--
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Solution T he binomial expansion for (1 & u/v)-I is 17 u/v . . , s o Equations 17-10 and 11 are the same, for small u lv. (Note the difference in sign convention for u in these two equations.)
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