16. Sound and Light Waves 16.18 Since sound wave amplitude increases with piston displacement, low-frequency speakers need bigger displacements for comparable power output. Mechanical design requires a bigger system to produce the greater displacements. b) Wavelength: closed at one end 16.19 The wave corresponding to the fundamental frequency is sketched hi each tube. open organ pipe Frequency: AI =4L close one end In the open pipe case: closed pipe case: Ai =2L! = 41 So, since the wavelength is inversely proportional to frequency, it follows that the frequency is cut in half when the pipe is closed. /i = -2- = ~.\ L = ~ AI 4L 4/1 This is half the answer of part a): L 2.6 m 16.24 /i = 3.1 khz 16.25 a) Pressure amplitude P* is given by: p* o f ^S Q = P*[pov s 2irfr l 10.2 Pa_ (1.3 kg / m 3 ) (330 m / s) 27r (10 3 Hz) = 3.8 x 10~ 6 m b) so = (3.8 x 10-6 m) (1 "" OHz) = 76 x 10~ 6 50 Hz m 0 p* = 6.3 mg/m 3 ; v 0 = 1.7 mm/s; s 0 = 53 nm \ 16.21/At 0 C the speed of sound is 330 m / s. Speed of the jet craft, «jet - (0.82) (330 m/s) = 970^h f i km ' 1000 m 1 16.22 L = 1.3 m; 196 Hz; 327 Hz; 458 Hz 16.23 a) Wavelength is AI = 2L. open organ pipe 16.27 The chamber must have a displacement node at each end. Vacuum chamber cylinder F mdamental " L = 3.5 m " In general the wavelength of a standing wave is: An = 2L \ X _ 2L AS - -3-3 r A4 -- \ L f n Frequency: Thus L is: VIA 1L 5,2 in 2/i 2(32.7 Hz) Frequency is related to wavelength and the speed of sound v s by: \ f e v s _nv s An/n - V. -> / - - if v s = 340 m/s, then the fundamental frequency is 340 m/s 2(3.5 m) -
The next harmonic has A = f and / 3 = 5/ 0 = 5 (65.4 Hz) = 327 Hz The third harmonic has frequency:. / 4 = 7/ 0 = 7(65.4 Hz) =45 8 Hz 16.24 The ear canal acts like a closed pipe. Thus the fundamental frequency is related to the length by v ' u» 340 m/s damental frequency is about 3 khz in this model. *f Q_16Jj6^The pipe is "open at one end", so it is presumably closed at the other end. Thus in the fundamental mode, A = 4 and / = v s /X, so v, = \f = 4Lf = 4 (25 x 10~ 2 m) (340 Hz) = 340 m/s Using the given formula: So: Thus T = (546 C) (1.0303-1) = 16.5 C But we only have 2 sig fig, so T = 17 C. 16.28 The refractive index is n c/v, so c 3.0 x 10 8 m/s = 1.25x 10 s m/s 2.4 The speed of light in diamond is 1.3 x 10 s m/s. 16.30 In 1 y light travels a distance: d = ct = (3.00 x 10 a m/s) (365.25 d) * So 1 light year equals 9.47 xlo 15 m. 3 = 9.47 x 10 15 m Tf- Cjl6.32/rhe radio communications between astronauts and mission control travelled at the speectof light In each communcation, signals had to travel twice the Earth/Moon distance. Thus d 4xlO«m _8^_ 2-4 - 2 C- 2 3TIo5^-3 S ~ 2 -' S The time delay experienced by the astronauts was 3 s. 1634 Let's assume that gaps and teeth have equal width. If the wheel rotates so that the
16. Sound and Light Waves 315 L6.40 /visible = 0.066 W/m 2 16.41 Spherical loudspeaker where air density po and the speed of sound v s are given typical values: po = 1.3kg m v s = 330 m/s SIL D = 10 db Iog 10 SIL D = +20dBlog 10 (^ radius. R= 10.0 cm =.100 m frequency, / = 3.0x 10 2 Hz amplitude, s 0 =.050 mm = 5.0 x 10~ 5 m Now So us 0 R = (27r)(3.0 x 10 2 Hz) x(5.0 x 10~ 5 m)(.10 m) = 9.43 xlo 3 s SIL is given by: SIL D = 10dBlog 10 Where the reference intensity is / ref = 10-12 W/m 2 I. The difference in SIL at the speaker and at distance D(SILo) is: SIL D - SIL R = I = 53 db Yes, this is audible. 2(iO- ia -2 16.42 5.6 m "^^x 16.43/The explosion releases 1.0 x 10 7 J in 1 second, so the power of the explosion is: -10dBlog 10 10dBlog 10 According to the inverse square law, intensity is inversely proportional to distance squared. Thus, / oc -s 20dBlog 10 (- At the speaker, intensity is related to frequency and displacement amplitude, so by Eqn. 16.16: J = -p 0 v s (us 0 ) 2 \ of the energy is energy is converted to lund waves (the other half presumably is converted to light and mechanical deformation). The sound waves have power: Psound = (.5)(1.0X10 7 W) ' = 5.0 x 10 6 W Assuming spherical wave fronts, the intensity is,- Jsound 4-Trr 2 where r is the distance at which we wish to fine me intensity, r = no iu I = 5.0 x 106 W 47r(110 m) 2 = 33 W/m 2
116 Lea & Burke Physics: The Nature of Things Sound intensity in decibels is the SIL: 16. 46 = 1.5 x SIL - 10dBl 0glo ( 10 _ 12 7 w/m2 ) = 140 db 32.88 L6.44 130 db; 130 Pa; 2.5 x 10~ 4 m; 500 W 16.45 Signal withfrequency / = 250 Hz. Amplifier 5.0 W of power area A Tube of diameter 10 cm. = d filled with air Intensity of the sound wave generated: The power incident on any cross sectional area of the tube, A, is P = 5.0 W. Cross-sectional area is -(I)' Thus intensity is 5.0 W A f (.1 m) 2 W = 636.6 W I = 640-^ Displacement amplitude and pressure amplitude of wave: Intensity is related to pressure amplitude P, by: P 2, 1 = P.= ^2Ip 0 v s Using air density po = 1.2 kg / m 3 and speed of sound v a = 340 m/s P. = N/2(637W/m 2 )(1.2 kg/m 3 )(340 m/s) = 720 Pa Intensity is related to displacement amplitude so by: 1 '^Wvr 2 / 2^ 16.47 SLL is given by: = 10dBlo glo, -'ref where reference intensity Intensity is proportional to inverse square of distance: I oc -= r 2 The difference in two sound intensity levels 16.48 20% A5IL - - 10dBlog 10 (A") _ 10dBlo glo ( - V-'ref/ Vrrf In terms of distance: If r 2 = 2r 1; then AS7L = 20 dblog 10 ( i ) = -6 db 16.49 The speed of light is the same in all frames c = 3 x 10 8 m / s 16.50 If light is blue-shifted, then the object is approaching. If light is redshifted, then the object is receding. Jupiter is spinning with respect tc Earth. The blueshifted side is spinning towards Earth. 16.51 T = period ' v w T = 1.1 mm Above the threshold of pain. 637 W/m 2 2(1.2 kg/m 3 )(340 m/s)
If the man with whistle bicycles toward the friend at 10.0 m/s, then we have a source with velocity 3 m/s (in the air frame) toward the observer, and the observer has a velocity -7 m/s in the air frame. Thus the Doppler shifted frequency is (equation 16.22a): 1 + 7/330 / = (2.00 khz) = (2.00 khz) 1.0306 = 2.06 khz 1-3/330 If we use the approximate formula 16.22b, we get / = (2.00 khz) l - = (2.00 khz) = 2. 06 khz which is the same result to 3 sig fig. f 16.50 Following the method of Example 16.9: 1 + Thus: (3.8T7) 2 = Thus: 3.8T7 2-1 = 0.87524 c 3.87T 2 +1 v, = 0.87524 (2.998 x 10 s m/s) = 2.624 x 10 s m/s 16.58 U&ve leaving A at time t = 0reaches0 after a time Ai = r/u w = t t leaving B at time t = T travels a shorter distance rj and arrives at time ta where: Since s = v e T -C r, we may expand the square root: Thus: = ri/l - 2- cos 5 + f-) 2 ~ rfl - - V r XT-/ \ r cos d The time interval between the arrival times of waves from A and B is: r f a \ r s v e T tg t^ = TH 11 cost)} = T cos9 = T cos9 v w \ r I v w v w v w The observed frequency is (time between passage of wavefronts)" 1, so A S B v. v u / = - : - /0 equation 16.6p^Mfe divide the problem into two parts: outgoing sound and returning sound.
/.- " / _^ JZ &-?/ Outgoing: ^fe have bom a moving source (the torpedo) and a moving observer (the submarine), so we use equation 16.22. The speed of sound in water is 1531 m/s (Table 16.1), much greater than the speed of either source or observer, so let's use 16.22b. Thus: (16-32) m/sn / 16 \ 1531 m/s )~ \ ' 1531 / Returning: The reflected pulse is Doppier shifted again. Both source and observer are moving opposite the direction that the wave is moving. Now the submarine is the source and the torpedo is the observer. Thus: Thus the frequency received by me torpedo is: 2-12 - 3kHz 16.62, 64, 66 See solutions manual 16.68 When light is incident at the critical angle, the angle of refraction is 90. Thus SnelFs law becomes: m this case, HI 1.31 (ice) and n 2 = 1.00 (air), so: am*.. IS.ig.. 0.78338 HI l.ol Thus 0 e = sin- 1 0.76336 = 0. 8685 rad = 49. 8 From Example 16.13, 9 a = 41 for glass. So glass is a better material for light pipes since more of the light is totally internally reflected. 16.70 At the first side, ni = 1 (air) and n-i = n (glass), so the angle of refraction <p is given by: sin# = nsin< (1) At the far side of the slab, n\ = n and n 2 = 1. The angle of incidence is <b (see diagram) so n3in(p = sin0 / (2) Comparing equations 1 and 2, we see that 9 f = 9. Thus the light emerges at the same angle that it enters. However it has been displaced, and emerges from the slab at point 3. The undisplaced ray intersects the side of the slab at point C. We want to find the distance T cl ft Jt