16.1: a)  Appmfv then,1000if)b.344.0)Hz100(s)m344(λ 0 0 1000A Therefore, the amplitude is m.102.1 5  maxmax increasing,Sincec) pBkAp  while keeping A constant requires decreasing k, and increasing π , by the same factor. Therefore the new wavelength is Hz.50m,9.6)20(m)688.0( m9.6 sm344 new  f 16.2: m.1021.3or , 12 Hz)1000(Pa) 9 102.2(2 )sm1480(Pa) 2 100.3( 2 max       AA π πBf vp The much higher bulk modulus increases both the needed pressure amplitude and the speed, but the speed is proportional to the square root of the bulk modulus. The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement. 16.3: From Eq. (16.5), .2λ2 max vfπBABAπBkAp  a) Pa.7.78s)m(344Hz)150(m)1000.2(Pa)1042.1(2 55   π b) Pa.778Pa7.78100c) Pa.77.8Pa78.710  The amplitude at Hz1500 exceeds the pain threshold, and at Hz000,15 the sound would be unbearable. 16.4: The values from Example 16.8 are Hz,1000Pa,1016.3 4  fB m.102.1 8 A Using Example 16.5, ,sm295sm344 293K K216 v so the pressure amplitude of this wave is Pa).1016.3( 2 4 max  A v πf BBkAp Pa.108.1m)102.1( sm295 Hz)(10002 38   π This is 0.27Pa)10(3.0Pa)101.8( 23   times smaller than the pressure amplitude at sea level (Example 16-1), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude. 16.5: a) Using Equation (16.7),    2 22 s)400m)(8(so,)λ( BfρvB Pa.1033.1)mkg1300( 103  b) Using Equation (16.8),   2 422 s10(3.9m)5.1()(   ρtLρvY Pa.1047.9)mkg6400( 103  16.6: a) The time for the wave to travel to Caracas was s 579s 39min 9  and the speed was sm10085.1 4  (keeping an extra figure). Similarly, the time for the wave to travel to Kevo was 680 s for a speed of s,m10278.1 4  and the time to travel to Vienna was 767 s for a speed of s.m10258.1 4  The average speed for these three measurements is s.m1021.1 4  Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds. b) From Eq. (16.7), , 2  vB  and using the given value of 33 mkg103.3   and the speeds found in part (a), the values for the bulk modulus are, respectively, Pa.102.5 and Pa105.4 Pa,109.3 111111  These are larger, by a factor of 2 or 3, than the largest values in Table (11-1). 16.7: Use sm1482 water v at C,20 as given in Table   1.16 The sound wave travels in water for the same time as the wave travels a distance m8.20m20.1m0.22  in air, and so the depth of the diver is     m.6.89 sm344 sm1482 m8.20m8.20 air water  v v This is the depth of the diver; the distance from the horn is m.8.90 16.8: a), b), c) Using Eq.   ,10.16       sm1032.1 molkg1002.2 K15.300KmolJ3145.841.1 3 3 2 H      v       sm1002.1 molkg1000.4 K15.300KmolJ3145.867.1 3 3 e H      v       .sm323 molkg109.39 K15.300KmolJ3145.867.1 3 Ar      v d) Repeating the calculation of Example 16.5 at K15.300T gives ,sm348 air v and so airHeairH 94.2,80.3 2 vvvv  and .928.0 airAr vv  16.9: Solving Eq. (16.10) for the temperature,    K,191 KmolJ3145.840.1 hrkm6.3 sm1 0.85 hkm850 mol)kg108.28( 2 3 2                        R Mv T  or C.82 b) See the results of Problem 18.88, the variation of atmospheric pressure with altitude, assuming a non-constant temperature. If we know the altitude we can use the result of Problem 18.88, .1 0 0                  R Mg T y pp Since ,yTT o  m,C106.K,191for 2   T and ).ft.840,44(m667,13K,273 0  yT Although a very high altitude for commercial aircraft, some military aircraft fly this high. This result assumes a uniform decrease in temperature that is solely due to the increasing altitude. Then, if we use this altitude, the pressure can be found: , K273 m)(13,667)mC106(. 1p m)C10K)(.6molJ315.8( )smmol)(9.8kg108.28( 2 o 2 23                        p and ,p13.)70(.p o 66.5 o p or about .13 atm. Using an altitude of 13,667 m in the equation derived in Example 18.4 gives ,p18. o p which overestimates the pressure due to the assumption of an isothermal atmosphere. 16.10: As in Example K,294.15C21 with 5,-16 T s.m80.344 molkg108.28 K)K)(294.15molJ3145.8)(04.1( 3        M RT v The same calculation with s,m344.22gives283.15T so the increase is s.m58.0 16.11: Table 16.1 suggests that the speed of longitudinal waves in brass is much higher than in air, and so the sound that travels through the metal arrives first. The time difference is s.208.0 )mkg(8600Pa)1090.0( m0.80 sm344 m0.80 311 Brassair    v L v L t 16.12: mol)kg108.28( K)K)(260.15molJ3145.8)(40.1( mol)kg108.28( K)K)(300.15molJ3145.8)(40.1( 33       s.m24 (The result is known to only two figures, being the difference of quantities known to three figures.) 16.13: The mass per unit length  is related to the density (assumed uniform) and the cross-section area ,by AρμA  so combining Eq. (15.13) and Eq. (16.8) with the given relations between the speeds,  900 so 900 Υ AF Αρ F ρ Υ 16.14: a) m.0.16 Hz220 )mkg10(8.9Pa)100.11( 3310    f ρΥ f v b) Solving for the amplitude A (as opposed to the area ) 2 πra  in terms of the average power , av ΙaP  2 av )2( ρΥω aP A  m.1029.3 ))Hz220(2(Pa)100.11)(mkg109.8( )m)10(0.800(W))1050.6(2 8 210 3 3 2-26       π π c) s.m104.55m)10289.3)(Hz220(22 58   πΑfπω 16.15: a) See Exercise 16.14. The amplitude is 2 2 ρΒω Ι  m.1044.9 Hz))3400(2(Pa)1018.2)(mkg1000( )mW1000.3(2 11 293 26       π The wavelength is m.434.0 Hz3400 )mkg(1000Pa)1018.2( 39    f ρB f v b) Repeating the above with Pa1040.1γ 5  pB and the density of air gives m.0.100andm1066.5 9   A c) The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense air molecules must have a larger amplitude to transfer the same amount of energy. 16.16: From Eq. (16.13), ,2 2 max BvpI  and from Eq. (19.21), . 2  Bv  Using Eq. (16.7) to eliminate   .22, 2 max 2 max ρBpBpρBIv  Using Eq. (16.7) to eliminate B, .2)(2 2 max 22 max ρvpρvvpI  16.17: a) Pa.95.1 s)m(344 m)10(5.00Hz)(150Pa)1042.1(22 max 65   π v πBfA BkAp b) From Eq. (16.14), .mW104.58s))m344)(mkg2.1(2(Pa)95.1(2 23322 max   ρvpI c)   dB.6.96log10 12 3 10 104.58     16.18: (a) The sound level is dB.57or ,logdB)(10so,logdB)10( 212 2 0 mW10 mW0.500 I I   βββ μ b) First find v, the speed of sound at C,0.20  from Table 16.1, s.m344v The density of air at that temperature is .mkg20.1 3 Using Equation (16.14), .mW1073.2or , s)m344)(mkg20.1(2 )mN150.0( 2 25 3 222 max   I ρv p I Using this in Equation (16.15), dB.74.4or , mW10 mW102.73 logdB)10( 212 25      ββ 16.19: a) As in Example 16.6, dB.40.6.mW104.4 212 s)m344)(mkg20.1(2 Pa)100.6( 3 25     βI 16.20: a)   dB.0.6log10 4  I I b) The number must be multiplied by four, for an increase of 12 kids. 16.21: Mom is five times further away than Dad, and so the intensity she hears is 2 25 1 5   of the intensity that he hears, and the difference in sound intensity levels is dB.14log(25)10  16.22: dB25dB90dB75level)(Sound  i f I I I I I I f 0 i 0 log10log10log10level)(Sound  Therefore i I I f log10dB25  35.2 102.310 i f   I I 16.23: ,0.20Thus,.dB)log(10dB13or ,dB)log10( 0 00  IIβ I I I I or the intensity has increased a factor of 20.0. 16.24: Open Pipe: Hz594 2 1 1 v f v L  Closed at one end: f v L  4 1 Taking ratios: Hz297 2 Hz594 Hz594 4 2   f fv v L L 16.25: a) Refer to Fig. (16.18). i) The fundamental has a displacement node at m600.0 2  L , the first overtone mode has displacement nodes at m300.0 4  L m900.0and 4 3  L and the second overtone mode has displacement nodes at m000.1andm600.0m,200.0 6 5 26  L LL . ii) Fundamental: 0,:First m.200.1,0 L m.200.1m,600.0 2  L L m.200.1m,800.0m,400.00,:Second 3 2 3  L LL b) Refer to Fig. (16.19); distances are measured from the right end of the pipe in the figure. Pressure nodes at: Fundamental: m200.1L . First overtone: m.200.1m,400.03  LL Second overtone: m, 240.05 L m.200.1,m720.053  LL Displacement nodes at Fundamental: .0 First overtone: m.800.032,0 L Second overtone: m, 480.052 ,0 L m960.054 L 16.26: a) Hz,382 m)450.0(2 )344( 2 1  sm L v f Hz,11473Hz,7642 131  fff Hz.15294 14  ff b) Hz.13387Hz,9565Hz,5733Hz,191 171513 4 1  fffffff L v Note that the symbol “ 1 f ” denotes different frequencies in the two parts. The frequencies are not always exact multiples of the fundamental, due to rounding. c) Open: ,3.52 1 000,20  f so the 52 nd harmonic is heard. Stopped; ,7.104 1 000,20  f so 103 rd highest harmonic heard. 16.27: Hz.25295 Hz,15173Hz,506 1312 m)4(0.17 m/s)344( 1  fffff 16.28: a) The fundamental frequency is proportional to the square root of the ratio M  (see Eq. (16.10)), so Hz,767 00.4 8.28 )57( 3)5( Hz)262( He air air He airHe     M M ff b) No; for a fixed wavelength , the frequency is proportional to the speed of sound in the gas. 16.29: a) For a stopped pipe, the wavelength of the fundamental standing wave is ,m56.04 L and so the frequency is   kHz.0.614m)56.0(sm344 1 f b) The length of the column is half of the original length, and so the frequency of the fundamental mode is twice the result of part (a), or 1.23 kHz. 16.30: For a string fixed at both ends, Equation   ,,33.15 2L nv n f  is useful. It is important to remember the second overtone is the third harmonic. Solving for v, , 2 n Lf n v  and inserting the data,     3 /s588m635.2 v , and .sm249v 16.31: a) For constructive interference, the path difference m00.2d must be equal to an integer multiple of the wavelength, so ,λ n nd   .Hz172 m2.00 sm344 λ n nn d v n d vnv f n         Therefore, the lowest frequency is 172 Hz. b) Repeating the above with the path difference an odd multiple of half a wavelength,     .Hz172 2 1  nf n Therefore, the lowest frequency is   .0nHz86  16.32: The difference in path length is     .2or ,2 xLxxLxxLx  For destructive interference, ))λ21((  nx ,and for constructive interference, λ.nx  The wavelength is   m670.1)Hz206(sm344λ  fv (keeping an extra figure), and so to have 34,0  nLx for destructive interference and 44  n for constructive interference. Note that neither speaker is at a point of constructive or destructive interference. a) The points of destructive interference would be at m.1.42m,58.0x b) Constructive interference would be at the points m.1.83 m,1.00 m,17.0x c) The positions are very sensitive to frequency, the amplitudes of the waves will not be the same (except possibly at the middle), and exact cancellation at any frequency is not likely. Also, treating the speakers as point sources is a poor approximation for these dimensions, and sound reaches these points after reflecting from the walls , ceiling, and floor. 16.33:     m500.0Hz688sm344λ  fv To move from constructive interference to destructive interference, the path difference must change by 2.λ If you move a distance x toward speaker B, the distance to B gets shorter by x and the difference to A gets longer by x so the path difference changes by 2x. 2λ2 x and m0.1254λ x 16.34: We are to assume     .m00.2Hz172m/s344λso,sm344  fvv If mr A 00.8 and B r are the distances of the person from each speaker, the condition for destructive interference is   λ, 2 1  nrr AB where n is any integer. Requiring   0λ 2 1  nrr AB gives     ,4m00.2m00.8 2 1  A rn so the smallest value of B r occurs when ,4n and the closest distance to B is     m.00.1m00.24-m00.8 2 1  B r 16.35:     m400.0Hz860sm344  fv 5.3 differencepath m.4.1m0.12m13.4isdifferencepath The    The path difference is a half-integer number of wavelengths, so the interference is destructive. 16.36: ,Sincea) beat ba fff  the possible frequencies are 440.0  Hz5.1Hz Hz441.5or Hz5.438 b) The tension is proportional to the square of the frequency. Therefore   Hz440 Hz5.12 2 2 i).So .2and     T T f f T T ffTfT .1082.6 3    .1082.6ii) 3 Hz440 Hz5.12     T T 16.37: a) A frequency of   Hz110Hz112Hz108 2 1  will be heard, with a beat frequency of 112 Hz–108 Hz = 4 beats per second. b) The maximum amplitude is the sum of the amplitudes of the individual waves,   .m100.3m105.12 88   The minimum amplitude is the difference, zero. 16.38: Solving Eq. (16.17) for v, with L v = 0, gives   ,sm775sm0.25 Hz1240Hz1200 Hz1240 S LS L            v ff f v or 780 sm to two figures (the difference in frequency is known to only two figures). Note that ,0 S v since the source is moving toward the listener. 16.39: Redoing the calculation with +20.0 sm for LS for m/s0.20and vv  gives 267 Hz. 16.40: a) From Eq.   Hz.375,sm0.15,0 with ,17.16 LS    A fvv b) Hz.371,sm0.15 ,sm35.0 With LS    B fvv c) )in figureextraan (keepingHz4 ABA fff      . The difference between the frequencies is known to only one figure. [...]...  f0  f beat  f 0 1   vv   v  v   w  w    16. 75: Refer to Equation (16. 31) and Figure (16. 38) The sound travels a distance vT and the plane travels a distance vsT before the boom is found So, h 2  (vT ) 2  (vsT ) 2 , or hv v vsT  h 2  v 2T 2 From Equation (16. 31), sin   Then, vs  2 vs h  v 2T 2 16. 76: a) b) From Eq (16. 4), the function that has the given p( x, 0) at t  0... other 16. 44: For a stationary source, vS  0, so f L  which gives vL  v  fL fS  v vL v  vS fS  1  vL v  fS , 490 Hz  1  344 m s  520 Hz  1  19.8 m/s This is negative because the listener is moving away from the source 16. 45: a) vL  18.0 m/s, vS  30.0 m s , and Eq 16. 29 gives f L   362 262 Hz  314  302 Hz b) vL  18.0 m s , vS  30.0 m/s and f L  228 Hz 16. 46: a) In Eq (16. 31),... 36.0 b) As in Example16.20, 950 m  t  2.23 s (1.70) (344 m s) ( tan(36.0)) 16. 47: a) Mathematically, the waves given by Eq (16. 1) and Eq (16. 4) are out of phase Physically, at a displacement node, the air is most compressed or rarefied on either side of the node, and the pressure gradient is zero Thus, displacement nodes are pressure antinodes b) (This is the same as Fig (16. 3).) The solid curve... 100.0 dB  10  log (1.45) 2  96.8 dB, so the airliner is not in violation of the ordinance 16. 49: a) Combining Eq (16. 14) and Eq (16. 15), 2 pmax  2 ρvI 010( β 10)  2(1.20 kg m 3 )(344 m s)(1012 W m )105.20  1.144  10 2 Pa, or  1.14  102 Pa, to three figures b) From Eq (16. 5), and as in Example 16. 1, pmax pmax v (1.144  102 Pa) (344 m s)   A  7.51  10 9 m 5 2π (1.42  10 Pa)587... source, Eq (16. 27) becomes vS  v  0.12 m   0.32 m s   0.25 m s 1.6 s TS b) Using the result of part (a) in Eq (16. 18), or solving Eq (16. 27) for v S and substituting into Eq (16. 28) (making sure to distinguish the symbols for the different wavelengths) gives   0.91 m  v  vL  fL    v  v  fS  S   a) The direction from the listener to source is positive, so vS   v 2 and vL  0 16. 43:... especially Chapters 4 and 17.) 16. 56: a) The cross-section area of the string would be a  (900 N) (7.0  108 Pa)  1.29  106 m 2 , corresponding to a radius of 0.640 mm (keeping extra figures) The length is the volume divided by the area, L (4.00  103 kg) V m ρ    0.40 m (7.8  103 kg m3 )(1.29  10 6 m 2 ) a a b) Using the above result in Eq (16. 35) gives f 1  377 Hz, or 380 Hz to two figures 16. 57:... (1) and (2) together: v  0.858 m s 16. 65: a) A  R  pmax  BkA  pmax  2π ρB fR, I  2πBA 2πBAf  In air v   v B Therefore ρ p 2 max  2π 2 ρB f 2 (R ) 2 2 ρB b) PTot  4πR 2 I  83 ρB f 2 R 2 (R ) 2 c) I  PTot 4 πd 2  2π 2 ρB f 2 R 2 ( R ) 2 d2 p max  (2 ρB I )1 2  , 2π ρB fR (R) p max R(R ) ,A  d d 2π ρB f 16. 66: (See also Problems 16. 70 and 16. 74) Let f 0  2.00 MHz be the frequency... the final calculation, f beat is negligible compared to f 0 16. 67: a)   v f  (1482 m s) (22.0  103 Hz)  6.74  102 m or Problem 16. 70; the difference in frequencies is b) See Problem 16. 66  2vW  24.95 m s  3 f  fS   v  v   22.0  10 Hz 1482 m s   4.95 m s   147 Hz  W     The reflected waves have higher frequency 16. 68: a) The maximum velocity of the siren is ωP AP  2πf P... v  vb    1   f L  v  vb       Letting f L  f refl and fS  f bat gives the result b) If f bat  80.7 kHz, f refl  83.5 kHz, and vbat  3.9 m s , vinsect  2.0 m s 16. 70: (See Problems 16. 66, 16. 74, 16. 67) a) In a time t, the wall has moved a distance v1t and the wavefront that hits the wall at time t has traveled a distance vt, where v  f 0 λ0 , and the number of wavecrests in.. .16. 41: In terms of wavelength, Eq (16. 29) is v  vs L  S  v  vL a) vL  0, vS  25.0 m and  L   319  344 m s  400 Hz   0.798 m This is, of 344 course, the same result as obtained directly from Eq (16. 27) vS  25.0 m s and vL  369 m s  400 Hz   0.922 m The frequencies corresponding to these wavelengths are c) 431 Hz and d) 373 Hz 16. 42: a) In terms of the . would be unbearable. 16. 4: The values from Example 16. 8 are Hz,1000Pa,1 016. 3 4  fB m.102.1 8 A Using Example 16. 5, ,sm295sm344 293K K 216 v so the pressure. energy. 16. 16: From Eq. (16. 13), ,2 2 max BvpI  and from Eq. (19.21), . 2  Bv  Using Eq. (16. 7) to eliminate   .22, 2 max 2 max ρBpBpρBIv  Using Eq. (16. 7)