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### ch1106

Course: CHEM 101, Spring 2009
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11 Vibrations Chapter and Waves 1 4 T 1 4 2 mk 2.62 kg 2 81.5 N m 0.282 s 13. (a) The total energy of an object in SHM is constant. When the position is at the amplitude, the speed is zero. Use that relationship to find the amplitude. Etot 1 mv 2 1 kx 2 1 kA2 2 2 2 A m k v2 x2 3.0 kg 280 N m 0.55 m s 2 0.020 m 2 6.034 10 2 m 6.0 10 2 m (b) Again use conservation of energy. The energy is all kinetic...

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11 Vibrations Chapter and Waves 1 4 T 1 4 2 mk 2.62 kg 2 81.5 N m 0.282 s 13. (a) The total energy of an object in SHM is constant. When the position is at the amplitude, the speed is zero. Use that relationship to find the amplitude. Etot 1 mv 2 1 kx 2 1 kA2 2 2 2 A m k v2 x2 3.0 kg 280 N m 0.55 m s 2 0.020 m 2 6.034 10 2 m 6.0 10 2 m (b) Again use conservation of energy. The energy is all kinetic energy when the object has its maximum velocity. 2 Etot 1 mv 2 1 kx 2 1 kA2 1 mvmax 2 2 2 2 vmax A k m 6.034 10 2 m 280 N m 3.0 kg 0.5829 m s 0.58 m s 14. The spring constant is found from the ratio of applied force to displacement. F 80.0 N k 4.00 102 N m x 0.200 m Assuming that there are no dissipative forces acting on the ball, the elastic potential energy in the loaded position will become kinetic energy of the ball. Ei Ef 1 2 2 kxmax 1 2 2 mvmax vmax xmax k m 0.200 m 4.00 102 N m 0.180 kg 9.43 m s 15. (a) The work done to compress a spring is stored as potential energy. 3.0 2 J 2W W 1 kx 2 k 416.7 N m 4.2 102 N m 2 2 2 x 0.12 m (b) The distance that the spring was compressed becomes the amplitude of its motion. The k A . Solve this for the mass. maximum acceleration is given by amax m k k 4.167 10 2 N m amax A m A 0.12 m 3.333 kg 3.3 kg m amax 15 m s 2 16. The general form of the motion is x (a) The amplitude is A A cos t 0.45 cos 6.40t . 6.40 s 2 1 xmax 0.45 m . 1 (b) The frequency is found by (c) The total energy is given by Etotal 1 2 2 mvmax 1 2 2f 2 6.40 s 1 f 1.019 Hz 2 1.02 Hz m A 2 1 2 0.60 kg 6.40 s 0.45 m 2.488 J 2.5 J (d) The potential energy is given by Epotential 1 2 kx 2 1 2 m x2 1 2 0.60 kg 6.40 s 12 0.30 m 2 1.111 J 1.1 J 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be repr oduced, in any form or by any means, without permission in writing from the publisher. 274
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Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThe kinetic energy is given by Ekinetic Etotal Epotential 2.488 J 1.111 J 1.377 J1 .4 J17. If the energy of the SHO is half potential and half kinetic, then the potential energy is half the tot
Arcadia University - CHEM - 101
Chapter 11Vibrations and Waves21. The equation of motion is x (a) The amplitude is A0.38 sin 6.50 tA sin t .1xmax0.38 m .2f 6.50 s1(b) The frequency is found byf1f6.50 s(c) The period is the reciprocal of the frequency. T (d) The total energ
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(d) Because the mass started at the equilibrium position of x = 0, the position function will be proportional to the sine function.x 0.231 m sin 21 2 2 mvmax 1 22.04 Hz t2x0.231 m sin 4.08
Arcadia University - CHEM - 101
Chapter 11Vibrations and WavesThe stretch of the bungee cord needs to provide a force equal to the weight of the jumper when he is at the equilibrium point. 65.0 kg 9.80 m s2 mg k x mg x 5.60 m k 113.73 N m Thus the unstretched bungee cord must be 25.0
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionvbottom2 gL 1 cos1 2 2 mvbottom 1 2o2 9.80 m s 220.760m 1 cos12.0o0.571 m s(c) The total energy can be found from the kinetic energy at the bottom of the motion.Etotal0.365 kg 0.571 m s
Arcadia University - CHEM - 101
Chapter 11Vibrations and Waves37. The distance between wave crests is the wavelength of the wave. v f 343 m s 262 Hz 1.31 m 38. To find the wavelength, use AM: FM:v f.545 m3.41 mv13.00 10 m s 550 103 Hz3.00 108 m s 88.0 106 Hz8v23.00 108 m s
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionm Vv FT mLm d2 LFT v2 m L2m Ld 277.5 m s27.8 103 kg m321.5 10 2 m 221.378 kg m1.378 kg m8.3 103 NB , where B is the bulk modulus of water, from 43. The speed of the water wave is
Arcadia University - CHEM - 101
Chapter 11Vibrations and Waves47. (a) Assuming spherically symmetric waves, the intensity will be inversely proportional to the square of the distance from the source. Thus Ir 2 will be constant. 2 I near rn2ear I far rfarI near I far2 rfarr2 near2
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionffingeredv 22 33v L 2 2L3 2f unfingered3 2294 Hz441 Hz54. Four loops is the standing wave pattern for the 4th harmonic, with a frequency given byf44 f1280 Hz . Thus f170 Hz , f2 140
Arcadia University - CHEM - 101
Chapter 11Vibrations and WavesvFT mL. The tension in the string will be the same as the weight of the masses hung from the endof the string, FT mg . Combining these relationships gives an expression for the masses hung from the end of the string. 4L2
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionsin sin22 1v2 v1331 0.60T2 331 0.60T1 24osin2sin 25o331 0.6010331 0.60 10sin 25o325 3370.4076sin 1 0.407665. The angle of refraction can be found from the law of refraction, Equati
Arcadia University - CHEM - 101
Chapter 11Vibrations and Wavesk 2.011 10 N m (b) To find the amount of stretch given a starting height of 35 m, again use conservation of energy. x , and there is no kinetic energy at the top or bottom positions. Note that ybottomFkxmgxmg65 kg 9.8
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition73. Relative to the fixed needle position, the ripples are moving with a linear velocity given by rev 1 min 2 0.108 m v 33 0.373 m s min 60 s 1 rev This speed is the speed of the ripple waves mov
Arcadia University - CHEM - 101
Chapter 11Vibrations and Waves77. (a) The overtones are given by f nnf1 , n2, 3, 4f3 3 392 Hz 1180 HzG : f22 392 Hz784 HzA : f 2 2 440 Hz 880 Hz f3 3 440 Hz 1320 Hz (b) If the two strings have the same length, they have the same wavelength. The f
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition80. For the pebble to lose contact with the board means that there is no normal force of the board on the pebble. If there is no normal force on the pebble, then the only force on the pebble is t
Arcadia University - CHEM - 101
Chapter 11Vibrations and WavesTpen Tver Tpen2l0 2mg k mkgl0mg k mg k1l0 k mg1Tver , by a factor of 1l0 k mg84. Block m stays on top of block M (executing SHM relative to the ground) without slipping due to static friction. The maximum static
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition87. The car on the end of the cable produces tension in the cable, and stretches the cable according to 1F Lo , where E is Youngs modulus. Rearrange this equation to see that Equation (9-4), L EA
Arcadia University - CHEM - 101
CHAPTER 12: Sound Answers to Questions1. Sound exhibits several phenomena that give evidence that it is a wave. The phenomenon of interference is a wave phenomenon, and sound produces interference (such as beats). The phenomenon of diffraction is a wave
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionparticular pipe is fixed, the frequency is also a function of temperature. Thus when the temperature changes, the resonant frequencies of the organ pipes change as well. Since the speed of sound
Arcadia University - CHEM - 101
Chapter 12Soundamplitude (anti-node). Thus the interference can be described as interference in time. To experience the full range from constructive interference to destructive interference, the time of observation must change, but all observations coul
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition2.The round trip time for sound is 2.5 seconds, so the time for sound to travel the length of the lake is 1.25 seconds. Use the time and the speed of sound in water to determine the depth of the
Arcadia University - CHEM - 101
Chapter 12SoundThe speed of sound in concrete is obtained from Equation (11-14a), Table (9-1), and Table (10-1).vconcreteE20 109 N m2 2.3 103 kg m32949 m sdvair tair343 m s2949 m s 2949 m s 343 m s1.1 s427m4.3 102 m7.The 5 second rule says
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition13. (a) According to Table 12-2, the intensity in normal conversation, when about 50 cm from the speaker, is about 3 10 6 W m2 . The intensity is the power output per unit area, and so the power
Arcadia University - CHEM - 101
Chapter 12Sound17. The intensity is proportional to the square of the amplitude. I A2. A A2.0 2.0 dB 10 log 2.0 10 log 220 20 log 2.0 100.1 1.259 I0 A0 A0 A01.318. (a) The intensity is proportional to the square of the amplitude, so if the amplitude i
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition24. For a vibrating string, the frequency of the fundamental mode is given by fv 2L1 2LFT mL.f1 2LFT mLFT =4Lf 2 m4 0.32 m 440 Hz23.5 10 4 kg87 N25. (a) If the pipe is closed at one
Arcadia University - CHEM - 101
Chapter 12Sound29. (a) We assume that the speed of waves on the guitar string does not change when the string is v fretted. The fundamental frequency is given by f , and so the frequency is inversely 2L proportional to the length. 1 f fL constant L f 33
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionfv 2LLv 2f343 m s 2 294 Hz0.583 m33. (a) At T20o C , the speed of sound is 343 m s . For an open pipe, the fundamental frequency is v given by f . 2L v v 343 m s f L 0.583 m 2L 2 f 2 294 H
Arcadia University - CHEM - 101
Chapter 12Sound(b) The harmonics for the closed pipe are f n, n odd. Again, they must be below 20 kHz. 4L 4 2.14 m 2 10 4 Hz nv 4 2 10 Hz n 499.1 4L 343 m s The values of n must be odd, so n = 1, 3, 5, , 499. There are 250 harmonics, and so there are2
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionf f1 2L f f0.98 FT mL f10.98 0.981 2LFT mL0.98 f 0.98 3.0 Hz294 Hz 144. Beats will be heard because the difference in the speed of sound for the two flutes will result in two different fr
Arcadia University - CHEM - 101
Chapter 12Sound48. To find the beat frequency, calculate the frequency of each sound, and then subtract the two frequencies. vv 1 1 f beat f1 f 2 343 m s 5.649 5.6 Hz 2.64 m 2.76 m 1 2 49. (a) Observer moving towards stationary source. v 30.0 m s f 1 ob
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(c) For the 300 m/s relative velocity: 1 fsource f 2000 Hz vsrc moving 1 vsnd1 1 300 m s 343 m s 300 m s 343 m s16.0 103 Hzf observermovingf1vsrc vsnd2000 Hz13.75 103 HzThe difference i
Arcadia University - CHEM - 101
Chapter 12SoundThen the wall can be treated as a stationary source emitting the frequency f wall , and the bat as a moving observer, flying toward the wall. v v v v 1 f bat f wall 1 bat f bat 1 bat f bat snd bat vsnd vsnd vsnd vbat v 1 bat vsnd 3.00 10
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionvbloodvsndf 2 f original f1.54 103 m s500 Hz 2 2.25 106 Hz 500 Hz0.171 m sIf instead we had assumed that the heart was moving towards the original source of sound, we would f get vblood vsn
Arcadia University - CHEM - 101
Chapter 12Sound61. (a) The Mach number is the ratio of the objects speed to the speed of sound. 1m s 1.5 104 km hr 3.6 km hr vobs M 119.05 120 vsound 35 m s (b) Use Eq. 12.5 to find the angle. v 1 1 sin 1 snd sin 1 sin 1 0.48o vobj M 119.05 62. From Eq.
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition66. Each octave is a doubling of frequency. The number of octaves, n, can be found from the following. 20, 000 Hz 2 n 20 Hz 1000 2 n log1000 nlog2n log 1000 log 2 9.97 10 octaves67. Assume that
Arcadia University - CHEM - 101
Chapter 12Sound71. Relative to the 1000 Hz output, the 15 kHz output is 10 dB. P P P 5 kHz 1 10 dB 10 log 15 kHz 1 log 15 kHz 0.1 150 W 150 W 150 WP kHz 1515 W72. The 140 dB level is used to find the intensity, and the intensity is used to find the p
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionL f1 22L 343 m s 0.540 m2 0.395 m 0.125 m0.540 mv635 Hz77. The frequency of the guitar string is to be the same as the third harmonic (n = 3) of the closed tube. nv The resonance frequenci
Arcadia University - CHEM - 101
Chapter 12Sound82. The sound is Doppler shifted up as the car approaches, and Doppler shifted down as it recedes. The observer is stationary in both cases. The octave shift down means that f approach 2 f recede .f approach f enginef engine 1 vcar vsnd
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionff originalf detectorf originalf originalvsnd vsndvblood vbloodf original2vblood vsnd vblood5.50 106 Hz2 0.32 m s 1.54 10 m s 0.32 m s32.29 103 Hz86. Use Eq. 12-4, which applies when
Arcadia University - CHEM - 101
Chapter 12Sound(b) The wavelength of sound in the rod is twice the length of the rod, 1.80 m . (c) The wavelength of the sound in air is determined by the frequency and the speed of sound in air. v 34 3 m s 0 .1 2 m f 2833 Hz 91. Eq. 11-18 gives the rel
Arcadia University - CHEM - 101
CHAPTER 13: Temperature and Kinetic Theory Answers to Questions1. Because the atomic mass of aluminum is smaller than that of iron, an atom of aluminum has less mass than an atom of iron. Thus 1 kg of aluminum will have more atoms than 1 kg of iron. Prop
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory10. The lead floats in the mercury becauseHgPb. As the substances are heated, the density ofboth substances will decrease due to volume expansion (see problem 17 for the derivation of this result). The density
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition18. Charless law states that the volume of a fixed mass of gas increases proportionately to the absolute temperature, when the pressure is held constant. As the temperature increases, the molecul
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryAlso, the violent bursting forth of steam propels some of the overheated water out of the radiator as well, which can spray onto the person opening the cap and again cause serious burns. 27. Exhaled air contains w
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition6.Assume that the temperature and the length are linearly related. The change in temperature per unit length change is as follows. T 100.0o C 0.0o C 9.066 Co cm L 22.85 cm 11.82 cm Then the temp
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory13. The amount of water that can be added to the container is the final volume of the container minus the final volume of the water. Also note that the original volumes of the water and the container are the same.
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) The fractional change in density is T 0 T 87 100 06oC40o C 25o C5.7 103This is a 0.57% increase . 18. Assume that each dimension of the plate changes according to Equation 13-1a. A A
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryOriginal volume for glass bulb and Hg in bulb: V0bulb Change in glass bulb volume: Change in Hg volume in glass bulb:VglassV0bulbglassTVHgVbulb 0HgTNow find the additional volume of Hg, and use that to f
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(c) For concrete, repeat the calculation with the expansion coefficient and elastic modulus for concrete.Stress FA ET 12 106Co20 109 N m260 C o1.4 107 N m26 2 The ultimate tensile strength
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory29. Assume the gas is ideal. Since the amount of gas is constant, the value ofPV Tis constant.PV1 1 T1P2V2 T2V2V1P T2 1 P2 T13.00m31.00 atm 3.20 atm273 38 K 273 K1.07 m330. Assume the air is an ideal g
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition34. (a) Assume that the helium is an ideal gas, and then use the ideal gas law to calculate the volume. Absolute pressure must be used, even though gauge pressure is given. 18.75 mol 8.315 J mol
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryOne factor limiting the maximum altitude would be that as the balloon rises, the density of the air decreases, and thus the temperature required gets higher. Eventually the air would be too hot and the balloon fab
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) 6.55 1022 moles6.02 1023 molecules 1 mol4 1046 molecules44. The net force on each side of the box will be the pressure difference between the inside and outside of the box, times the area
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory48. The rms speed is given by Equation 13-9, vrms3kT m .vrms vrms2 13kT2 m 3kT1 mT2 T1373 K 273 K1.1749. The rms speed is given by Equation 13-9, vrms 3kT m . Since the rms speed is proportional to the squ
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition55. The temperature of the nitrogen gas is found from the ideal gas law, and then the rms speed is found from the temperature. 2.1 atm 1.013 105 Pa atm 8.5 m3 PV PV nRT T 167.3 K nR 1300 mol 8.31
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory59. The pressure can be stated in terms of the ideal gas law, PNkT V . Substitute for the temperaturefrom the expression for the rms speed, vrms 3kT m T mvr2ms 3k . The mass of the gas is the mass of a molecule
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition66. The relative humidity of 40% with a partial pressure of 530 Pa of water gives a saturated vapor pressure of 530 Pa 0.40 Psaturated 530 Pa Psaturated 1325 Pa 0.40 From Table 13-3, the temperat
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory71. From Example 13-19, we have an expression for the time to diffuse a given distance. Divide the distance by the time to get the average speed.t C C x Dx2 1 21.00 0.40 mol m3 1.00 0.40 mol m315 10 6 m 95 10
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThus 6 290.20721% of the original gas remains in the cylinder.76. Assume the air is an ideal gas, and that the pressure is 1.0 atm. PV NkTN PV kT 1.013 105 Pa 6.5 3.1 2.5 m3 1.38 1023J K 27
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory80. The temperature can be found from the rms speed by Equation 13-9, vrmsvrms 3kT m23kT m .T2 mvrms28 1.66 10 kg274 10 km h2341m s 3.6 km h3k3 1.38 10JK1.4 105 K81. From the ideal gas law, if the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionM N2 4 REarth P46.38 106 m21.01 105 Pag 5.27 1018 kg 1 mole9.80 m s 2 29 10 3 kg 1 mole5.27 1018 kg 1.1 1044 molecules6.02 1023 molecules84. The temperature of the nitrogen gas is found