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Chapter 20 - PHYS 132 WEEK 2 20.2 Model This is a wave...

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Unformatted text preview: PHYS 132, WEEK 2 20.2. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second. Visualize: Please refer to Figure Ex20.2. This snapshot graph shows the wave at all points on the x-axis at t = 2 s. You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for all later times. 20.9. Model: The wave is a traveling wave on a stretched string. Solve: The wave speed on a string whose radius is R, length is L, and mass density is r is 22mVRLRLLLrrpmrp==== stringS vTm= with If the string radius doubles, then rp==== ()stringSstring2280 m/s140 m/s222vTvR 20.12. Model: The wave is a traveling wave. Solve: = (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, w = 124 rad/s, and 00 radf . The frequency is 124 rad/s19.7 Hz22fwpp=== (b) The wavelength is 222.33 m2.7 rad/mkppl=== (c) The wave speed 45.9 m/svfl== . 20.16. Visualize: Please refer to Figure Ex20.16. Solve: The amplitude of the wave is the maximum displacement which is 6.0 cm. The period of the wave is 0.60 s, so 110.60 s1.67 HzfT=== the frequency . The wavelength is 2 m/s1.2 m1.667 Hzvfl=== 20.19. Visualize: Solve: For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28: ()()()212122240 m30 m10 mrrpppffflllfD=-=-=-fi=D 2fpD= 2fpD= 4fpD= for two points on adjacent wavefronts and for two points separated by 2 l . Thus, l = 10 m when 4fpD= , and l = 5 m when . The crests corresponding to these two wavelengths are shown in the figure. One can see that a crest of the wave passes the 40 m-listener and the 30 m-listener simultaneously. The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus, the lowest frequency is 1340 m/s34 Hz10 mvfl=== 2 68 Hzf= . The next highest frequency is 20.20. Visualize: Solve: (a) Because the same wavefront simultaneously reaches listeners at x = -7.0 m and x = +3.0 m, ()212120 radrrrrpflD==-fi= Thus, the source is at x = -2 m, so that it is equidistant from the two listeners. ()()225 m2 m4.58 m-= . (b) The third person is also 5 m away from the source. Her y-coordinate is thus y = 20.23. Visualize: Solve: The explosive's sound travels down the lake and into the granite, and then it is reflected by the oil surface. The echo time is thus equal to echowater downgranite downgranite upwater upgranitegranitegranite500 m500 m0.94 s793 m1480 m/s6000 m/s6000 m/s1480 m/stttttddd=+++= 20.25. Solve: (a) The frequency is air 343 m/s1715 Hz0.20 mvfl=== (b) The frequency is 89 3.010 m/s1.5010 Hz1.50 GHz0.20 mcfl==== (c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be 7water91480 m/s9.8710 m987 nm1.5010 Hzvfl-==== 20.26. Model: Light is an electromagnetic wave that travels with a speed of 3 108 m/s. Solve: (a) The frequency of the blue light is 814blue9 3.010 m/s6.6710 Hz45010 mcfl-=== (b) The frequency of the red light is 814red9 3.010 m/s4.6210 Hz65010 mf-== (c) Using Equation 20.30 to calculate the index of refraction, vacuummaterialnll= vacuummaterial650 nm1.44450 nmnllfi=== 20.29. Model: Light is an electromagnetic wave. Solve: (a) The time light takes is () 33118glass 3.0 mm3.010 m3.010 m1.5010 s3.010 m/s1.50tvcn---==== (b) The thickness of water is () 811waterwater 3.010 m/s1.5010 s3.38 mm1.33cdvttn-==== 20.34. Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities at distances r1 and r2 is 212221 IrIr=2350 m2 m2 m1.61050 mII-^fi==~˯ ()()() 3233250 m2 m 1.6102.0 W/m1.6103.210 W/mII---fi=== Assess: The power generated by the sound source is P = I 2m [4p(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a significant amount of power. 20.38. Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observer approaching the source and decreases for an observer receding from a source. Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The minimum speed you need to travel is calculated as follows: ()000120 kHz121 kHz343 m/svvffv-^^=-fi=-fi~~˯˯ 016.3 m/sv= Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle. 20.52. Solve: The difference in the arrival times for the P and S waves is SPSP ddtttvvD=-=- 11120 s4500 m/s8000 m/sd^fi=-~˯ 61.2310 m1230 kmdfi== Assess: d is approximately one-fifth of the radius of the earth and is reasonable. 20.53. Solve: The time for the wave to travel from California to the South Pacific is 6 8.0010 m5405.4 s1480 m/sdtv=== 1480.28 m/s.dvt== A time decrease to 5404.4 s implies the speed has changed to Since the 4.0 m/s increase in velocity is due to an increase of 1C, an increase of 0.28 m/s occurs due to a temperature increase of ()1C0.28 m/s0.07C4.0 m/s^=~˯ Thus, a temperature increase of approximately 0.07C can be detected by the researchers. 20.54. Model: This is a sinusoidal wave.0, sinDytAkytwf=++ ()() Solve: (a) The equation is of the form , so the wave is traveling along the y-axis. Because it is +wt rather than -wt the wave is traveling in the negative y-direction. (b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the air molecules are oscillating18.96 mk-=forth along the y-axis. back and (c) The wave number is , so the wavelength is 122=0.70 m8.96 mkppl-== 1 3140 sw-= , so the wave's frequency is 13140 s=500 Hz22fwpp-== The angular frequency is Thus, the wave speed v = l f = (0.70 m)(500 Hz) = 350 m/s. The period T = 1/f = 0.0020 s = 2.0 ms. (d) The interval t = 0 s to t = 4 ms is exactly 2 cycles of the wave. The initial value at y = 1 m is ()()()141 m, 0 s0.02 mmsin8.96Dytp===+ = -0.0063 mm Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of 343 m/s, so the air temperature must be greater than 20. 20.56. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction. Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and w = 638 rad/s. Using the equation for the wave speed in a stretched string, 22233SstringSstring638 rad/s5.010 kg/m12.6 N12.57 rad/mTvTvkwmmm-^^=fi====~~˯˯ () ()max, 2.0 cmDxt= (b) The maximum displacement is the amplitude (c) From Equation 20.17, . ()()2 max638 rad/s2.010 m12.8 m/syvAw-=== 20.62. Model: We have a sinusoidal traveling wave on a stretched string. Solve: (a) The wave speed on a string and the wavelength are calculated as follows: S20 N100.0 m/s100.0 m/s1.0 m0.002 kg/m100 HzTvvflm===fi=== (b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant can be obtained from Equation 20.15 as follows: ()00 m, 0 ssinDAf= ()01.0 mm1.0 mmsinffi-= 0 rad2pffi=- ()()0, sinDxtAkxtwf=-+ (c) The wave (as distinct from the oscillator) is described by number and angular frequency are 222 rad/m1.0 mkpppl=== . In this equation the wave ()()100.0 m/s2 rad/m200 rad/svkwpp=== Thus, the wave's displacement equation is ()()()()12, 1.0 mmsin2 rad/m200 rad/s radDxtxtppp=--ΰ (d) The displacement is ()()()()()()120.50 m, 0.015 s1.0 mmsin2 rad/s0.50 m200 rad/s0.015 sDppp=--ΰ = -1.00 mm 20.66. Model: The wave is traveling on a stretched string. Solve: The wave speed on the string is S 50 N100 m/s0.005 kg/mTvm=== The velocity of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated as follows: ()100 m/s20.030 m9.42 m/s2.0 mp^==~˯ max22yvvAfAAwpplfi=== ()0cosyvAkxtwwf=--+ 20.68. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun's rays when they impinge upon the earth is () 262sunsunearth22211earth 410 W1420 W/m4441.49610 mPPIIrrppp=fi=== With r sun-Venus = 1.082 1011 m 610 W/mII=== 2.279 1011 m, the intensities of electromagnetic waves at these and rsun-Mars 22VenusMars2720 W/m and planets are . 20.74. Model: The sound generator's frequency is altered by the Doppler effect. The frequency increases as the generator approaches the student, and it decreases as the generator recedes from the student. Solve: The generator's speed is ()()S10021.0 m2 rev/s10.47 m/s60vrrfwpp^====~˯ The frequency of the approaching generator is Hz619 Hz10.47 m/s11343 m/sffvv+===-0S600 Doppler effect for the receding generator, on the other hand, is 0S 600 Hz582 Hz10.47 m/s11343 m/sffvv-===++ Thus, the highest and the lowest frequencies heard by the student are 619 Hz and 582 Hz. 20.82. Model: The wave pulse is a traveling wave on a stretched wire. Visualize: Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upward tension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point ()TwMgygm=== where m = m/L is the linear density of the entire rope. Using Equation 20.2, we get Tygvgymmm=== ,vgy= (b) The time to travel a distance dy at y, where the wave speed is dydydtvgy== is Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other: 000122TLLdytdtyLgygg^D====~˯ 2LtgfiD= ...
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