I NSTRUCTOR ’ S S OLUTIONS M ANUAL FOR S ERWAY AND V UILLE ’ S C O L L E G E P H YS I C S N INTH E DITION , V OLUME Charles Teague Emeritus, Eastern Kentucky University Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States 56157_00_fm_pi-viii.indd i 10/13/10 9:54:04 AM www.elsolucionario.org © 2011 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.com ISBN-13: 978-0-8400-6870-5 ISBN-10: 0-8400-6870-0 Brooks/Cole 20 Channel Center Street Boston, MA 02210 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you not accept these conditions, you must return the Supplement unused within 30 days of receipt All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/ or its licensors The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool Printed in the United States of America 14 13 12 11 10 56157_00_fm_pi-viii.indd ii 10/13/10 9:54:05 AM TABLE OF CONTENTS Acknowledgements Preface v vii Part – Mechanics Chapter – Introduction Chapter – Motion in One Dimension Chapter – Vectors and Two-Dimensional Motion Chapter – The Laws of Motion Chapter – Energy Chapter – Momentum and Collisions Chapter – Rotational Motion and the Law of Gravity Chapter – Rotational Equilibrium and Rotational Dynamics Chapter – Solids and Fluids 17 49 89 129 169 209 243 288 Part – Thermodynamics Chapter 10 – Thermal Physics Chapter 11 – Energy in Thermal Processes Chapter 12 – The Laws of Thermodynamics 322 344 371 Part – Vibrations and Waves Chapter 13 – Vibrations and Waves Chapter 14 – Sound 400 426 iii 56157_00_fm_pi-viii.indd iii 10/13/10 9:54:06 AM www.elsolucionario.org 56157_00_fm_pi-viii.indd iv 10/13/10 9:54:06 AM ACKNOWLEDGEMENTS The author would like to thank everyone who has contributed to this work In particular, thanks go to the support staff at Cengage Learning for their excellent guidance and support in all phases of this project Special mention goes to Physics Acquisitions Editor, Charles Hartford; Development Editor, Ed Dodd; Associate Content Project Manager, Holly Schaff; Associate Development Editor, Brandi Kirksey; and Editorial Assistant, Brendan Killion Susan English of Durham Technical Community College served as accuracy reviewer for this manual Her contributions are deeply appreciated Any remaining errors in this work are the responsibility of the author alone I would like to acknowledge the staff of MPS Limited, a Macmillan Company for their excellent work in assembling and typing this manual and preparing diagrams and page layouts Finally, the author would like to thank his wife, Carol, for her patience, understanding, and great support during this effort v 56157_00_fm_pi-viii.indd v 10/13/10 9:54:06 AM www.elsolucionario.org 56157_00_fm_pi-viii.indd vi 10/13/10 9:54:06 AM PREFACE This manual is written to accompany College Physics, Ninth Edition, by Raymond A Serway and Chris Vuille For each chapter in that text, the manual includes solutions to all end-of-chapter problems, more detailed answers to Quick Quizzes and Multiple Choice Questions than available in the main text, and answers to the even-numbered Conceptual Questions Considerable effort has been made to insure that the solutions and answers given in this manual comply with the rules on significant figures and rounding given in the Chapter of the textbook This means that intermediate answers are rounded to the proper number of significant figures when written, and that rounded value is used in all subsequent calculations Users should not be concerned if their answers differ slightly in the last digit from the answers given here Most often, this will be caused by choosing to round intermediate answers at different stages of the solution You are encouraged to keep this manual out of the hands of students as instructors in many colleges throughout the country use this textbook, and many of them use graded problem assignments as part of the final course grade Additionally, even when the problems are not used in such a direct fashion, it is advantageous for students to struggle with some problems in order to improve their problem-solving skills Feel free to post answers and solutions to selected questions and problems, but please preserve the manual as a whole You may also encourage students to purchase a copy of the Student Solutions Manual with Study Guide, which provides chapter summaries as well as detailed solutions to selected problems in the main text Attempting to keep the manual of manageable size, and recognizing that the primary users will be instructors well versed in the field, answers and solutions are kept fairly brief Answers to conceptual questions have been shortened by not offering detailed arguments that lead to the answer Problem solutions often omit commentary, intermediate steps, as well as initial steps that could be necessary for clear understanding by students On occasions where selected problem solutions are to be shared with students, you may wish to supply intermediate steps and additional comments as needed An electronic version of this manual can be obtained by requesting the Instructor’s Power Lecture CD from your local Cengage Learning Sales Representative Contact information for your sales representative is available under the “Find Your Rep” tab found at the bottom of the page at www.academic.cengage.com We welcome your comments on the accuracy of the solutions as presented here, as well as suggestions for alternative approaches Charles Teague vii 56157_00_fm_pi-viii.indd vii 10/13/10 9:54:06 AM www.elsolucionario.org 56157_00_fm_pi-viii.indd viii 10/13/10 9:54:06 AM Introduction ANSWERS TO MULTIPLE CHOICE QUESTIONS Using a calculator to multiply the length by the width gives a raw answer of 6783 m , but this answer must be rounded to contain the same number of significant figures as the least accurate factor in the product The least accurate factor is the length, which contains either or significant figures, depending on whether the trailing zero is significant or is being used only to locate the decimal point Assuming the length contains significant figures, answer (c) correctly expresses the area as 6.78 × 10 m However, if the length contains only significant figures, answer (d) gives the correct result as 6.8 × 10 m Both answers (d) and (e) could be physically meaningful Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions According to Newton’s second law, Force = mass × acceleration Thus, the units of Force must be the product of the units of mass (kg) and the units of acceleration ( m s ) This yields kg ⋅ m s 2, which is answer (a) The calculator gives an answer of 57.573 for the sum of the given numbers However, this sum must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places) The required conversion is given by: ⎛ 000 mm ⎞ ⎛ 1.00 cubitus ⎞ h = ( 2.00 m ) ⎜ ⎟⎜ ⎟ = 4.49 cubiti ⎝ 1.00 m ⎠ ⎝ 445 mm ⎠ This result corresponds to answer (c) The given area (1 420 ft ) contains significant figures, assuming that the trailing zero is used only to locate the decimal point The conversion of this value to square meters is given by: 1.00 m ⎞ 2 A = (1.42 × 10 ft ) ⎛⎜ ⎟ = 1.32 × 10 m = 132 m ⎝ 3.281 ft ⎠ Note that the result contains significant figures, the same as the number of significant figures in the least accurate factor used in the calculation This result matches answer (b) You cannot add, subtract, or equate a number apples and a number of days Thus, the answer is yes for (a), (c), and (e) However, you can multiply or divide a number of apples and a number of days For example, you might divide the number of apples by a number of days to find the number of apples you could eat per day In summary, the answers are (a) yes, (b) no, (c) yes, (d) no, and (e) yes 56157_01_ch01_p001-016.indd 10/12/10 1:26:50 PM www.elsolucionario.org The speakers emit sound of wavelength Δy v 343 m s l= = = 0.762 m f 450 Hz l = 0.381 m so r1 1.50 m 14.36 Chapter 14 3.00 m 442 r2 Initially, Δy = 0, and r1 = r2 = O 8.00 m (1.50 m )2 + (8.00 m )2 = 8.14 m To create destructive interference at point O, we move the top speaker upward distance Δy from its original location until we have r1 − r2 = l Since this did not change r2, we must now have r1 = r2 + l = 8.14 m + 0.381 m = 8.52 m But, after moving the speaker, this gives r1 = 14.37 (1.50 m + Δy ) + (8.00 m ) 2 or (1.50 m + Δy ) Thus, Δy = 8.59 m − 1.50 m = 1.43 m = 8.52 m = (8.52 m ) − (8.00 m ) = 8.59 m 2 d The wavelength of the sound is l= (a) v 343 m s = = 0.500 m f 686 Hz At the first relative maximum (constructive interference), r x r = x + l = x + 0.500 m Using the Pythagorean theorem, r = x + d 2, or FIGURE P14.37 (modified) ( x + 0.500 m )2 = x + ( 0.700 m )2 giving x = 0.240 m (b) At the first relative minimum (destructive interference), r = x + l = x + 0.250 m Therefore, the Pythagorean theorem yields ( x + 0.250 m )2 = x + ( 0.700 m )2 or 56157_14_ch14_p426-456.indd 442 x = 0.855 m 10/12/10 2:51:57 PM Sound 14.38 443 In the fundamental mode of vibration, the wavelength of waves in the wire is l = L = ( 0.700 m ) = 1.400 m If the wire is to vibrate at f = 261.6 Hz, the speed of the waves must be v = l f = (1.400 m )( 261.6 Hz ) = 366.2 m s The mass per unit length of the wire is m= m 4.300 × 10 −3 kg = = 6.143 × 10 −3 kg m L 0.700 m and the required tension is given by v = F m as F = v m = ( 366.2 m s ) ( 6.143 × 10 −3 kg m ) = 823.8 N 14.39 In the third harmonic, the string of length L forms a standing wave of three loops, each of length l = L The wavelength of the wave is then l/2 l/4 2L 16.0 m l= = ≈ 5.33 m 3 (a) 3l/2 l 3l /4 5l/4 The nodes in this string, fixed at each end, will occur at distances of 0, l = 2.67 m, l = 5.33 m, and 3l = 8.00 m from the end Antinodes occur halfway between each pair of adjacent nodes, or at distances of l = 1.33 m, 3l = 4.00 m, and 5l = 6.67 m from the end (b) The linear density is m= m 40.0 × 10 −3 kg = = 5.00 × 10 −3 kg m L 8.00 m and the wave speed is v= F = m 49.0 N = 99.0 m s 5.00 × 10 −3 kg m Thus, the frequency is 14.40 f= v 99.0 m s = = 18.6 Hz l 5.33 m In the fundamental mode, the distance from the finger of the cellist to the far end of the string is one-half of the wavelength for the transverse waves in the string Thus, when the string resonates at 449 Hz, l = ( 68.0 cm − 20.0 cm ) = 96.0 cm The speed of transverse waves in the string is therefore v = l f = ( 0.960 m )( 449 Hz ) = 431 m s continued on next page 56157_14_ch14_p426-456.indd 443 10/12/10 2:52:00 PM www.elsolucionario.org 444 Chapter 14 For a resonance frequency of 440 Hz, the wavelength would be v 431 m s = = 0.980 m = 98.0 cm f ′ 440 Hz l′ = To produce this tone, the cellist should position her finger at a distance of x= L− l 98.0 cm = 68.0 cm − = 19.0 cm 2 from the nut Thus, she should move her finger 1.00 cm toward the nut 14.41 When the string vibrates in the fifth harmonic (i.e., in five equal segments) at a frequency of f5 = 630 Hz, we have L = 5(l5 2), or the wavelength is l5 = 2L The speed of transverse waves in the string is then v = l5 f5 = (2L 5) f5 For the string to vibrate in three segments (i.e., third harmonic), the wavelength must be such that L = 3(l3 2) or l3 = 2L The new frequency would then be f3 = 14.42 v (2L 5) f5 3 = = f5 = ( 630 Hz ) = 378 Hz l3 2L If a wire of length is fixed at both ends, the wavelength of the fundamental mode of vibration is l1 = The speed of transverse waves in the wire is v = F m , where F is the tension in the wire and m is the mass per unit length of the wire The fundamental frequency for the wire is then f1 = v = l1 F ⎛ 1⎞ = ⎜ ⎟ m 2⎝ ⎠ F m If we have two wires with the same mass per unit length, one of length L and under tension F while the second has length 2L and tension 4F, the ratio of the fundamental frequencies of the two wires is f1, long f1, short = (1 2L ) (1 L ) 4F m = =1 F m or the two wires have the same fundamental frequency of vibration If this frequency is f1 = 60 Hz, then the frequency of the second harmonic for both wires is f2 = f1 = ( 60 Hz ) = 120 Hz 14.43 m= m 25.0 × 10 −3 kg = = 1.85 × 10 −2 kg m L 1.35 m (a) The linear density is (b) In a string fixed at both ends, the fundamental mode has a node at each end and a single antinode in the center, so that L = l 2, or l = 2L = (1.10 m ) = 2.20 m Then, the desired wave speed in the wire is v = l f = ( 2.20 m )( 41.2 Hz ) = 90.6 m s (c) The speed of transverse waves in a string is v = F m , so the required tension is F = m v = (1.85 × 10 −2 kg m ) ( 90.6 m s ) = 152 N continued on next page 56157_14_ch14_p426-456.indd 444 10/12/10 2:52:02 PM Sound (d) l = 2L = (1.10 m ) = 2.20 m (e) The wavelength of the longitudinal sound waves produced in air by the vibrating string is lair = 14.44 445 (a) [See part (b) above.] vair 343 m s = = 8.33 m 41.2 Hz f A string fixed at each end forms standing wave patterns with a node at each end and an integer number of loops, each loop of length l 2, with an antinode at its center Thus, L = n ( l ) or l = 2L n If the string has tension T and mass per unit length m, the speed of transverse waves is v = l f = T m Thus, when the string forms a standing wave of n loops (and hence n antinodes), the frequency of vibration is f= (b) T m v n = = l 2L n 2L ⇒ fA = nA 2L A TA mA Assume the length is doubled, LB = 2L A, and a new standing wave is formed having nB = n A and TB = TA Then nB 2LB fB = (c) T m nA TB = m A ( 2L A ) 1⎛ n TA = ⎜ A m A ⎝ 2L A TA ⎞ f = A ⎟ mA ⎠ Solving the general result obtained in part (a) for the tension in the string gives T = 4m f L2 n Thus, if fB = f A, LB = L A, and nB = n A + 1, we find ⎛ 4m A f A2 L2A ⎞ ⎛ n ⎞ 4m A fB2 L2B 4m A f A2 L2A n A2 n A2 TB = = = TA = ⎜ A ⎟ TA = 2 ⎜ 2 ⎟ nB n ⎝ nA + 1⎠ ⎠ ( n A + 1) ( n A + 1) ( n A + 1) ⎝ A (d) If now we have fB = f A, LB = L A 2, and nB = 2n A, then TB = 14.45 (a) 2 4m A fB2 L2B 4m A ( f A ) ( L A ) ⎛ 4m A f A2 L2A ⎞ = = ⎜ ⎟⎠ = 16 TA 16 ⎝ nB2 n A2 ( 4nA2 ) or TB = TA 16 From the sketch at the right, notice that when d = 2.00 m, L= 5.00 m − d = 1.50 m, d L and ⎛ d / 2⎞ q = sin −1 ⎜ = 41.8° ⎝ L ⎟⎠ Then evaluating the net vertical force on the lowest bit of string, gives the tension in the string as F= (b) → F q → F → mg ΣFy = 2F cosq − mg = (12.0 kg) (9.80 m s2 ) mg = = 78.9 N cosq cos ( 41.8° ) The speed of transverse waves in the string is v= F = m 78.9 N = 2.81× 10 m s 0.001 00 kg m continued on next page 56157_14_ch14_p426-456.indd 445 10/12/10 2:52:06 PM www.elsolucionario.org 446 Chapter 14 For the pattern shown, ( l ) = d, so l = 2d 4.00 m = 3 Thus, the frequency is f= 14.46 (a) v ( 2.81× 10 m s ) = = 2.11× 10 Hz l 4.00 m For a standing wave of loops, ( l ) = L, or l = L = (2.0 m) The speed of the waves in the string is then ⎛ 2.0 m ⎞ v=l f =⎜ (150 Hz ) = 1.0 × 10 m s ⎝ ⎟⎠ Since the tension in the string is F = mg = ( 5.0 kg ) ( 9.80 m s ) = 49 N, v = F m gives m= (b) F 49 N = = 4.9 × 10 −3 kg m v (1.0 × 10 m ) If m = 45 kg, then F = ( 45 kg ) ( 9.80 m s ) = 4.4 × 10 N, and 4.4 × 10 N = 3.0 × 10 m s 4.9 × 10 −3 kg m v= Thus, the wavelength will be l= v 3.0 × 10 m s = = 2.0 m f 150 Hz and the number of loops is (c) n= 2.0 m L = = l 1.0 m If m = 10 kg, the tension is F = (10 kg ) ( 9.80 m s ) = 98 N, and v= Then, l = and n = 98 N = 1.4 × 10 m s 4.9 × 10 −3 kg m v 1.4 × 10 m s = = 0.93 m f 150 Hz 2.0 m L = is not an integer, l 0.47 m so no standing wave will form 14.47 The speed of transverse waves in the string is v= F 50.000 N = = 70.711 m s m 1.000 × 10 −2 kg m continued on next page 56157_14_ch14_p426-456.indd 446 10/12/10 2:52:11 PM Sound 447 The fundamental wavelength is l = L = 1.200 m and its frequency is f1 = v 70.711 m s = = 58.926 Hz l1 1.200 m The harmonic frequencies are then fn = nf1 = n ( 58.926 Hz ), with n being an integer The largest one under 20 000 Hz is f339 = 19 976 Hz = 19.976 kHz 14.48 The distance between adjacent nodes (and between adjacent antinodes) is one-quarter of the circumference d NN = d AA = so and l 20.0 cm = = 5.00 cm l = 10.0 cm = 0.100 m, f= v 900 m s = = 9.00 × 10 Hz = 9.00 kHz l 0.100 m The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it 14.49 Assuming an air temperature of T = 37°C = 310 K, the speed of sound inside the pipe is v = ( 331 m s ) TK 310 = ( 331 m s ) = 353 m s 273 K 273 In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is l = 4L Thus, for the whooping crane l = ( 5.0 ft ) = 2.0 × 101 ft and 14.50 (a) f= v ( 353 m s ) ⎛ 3.281 ft ⎞ = ⎜ ⎟ = 58 Hz l 2.0 × 101 ft ⎝ m ⎠ In the fundamental resonant mode of a pipe open at both ends, the distance between antinodes is d AA = l = L Thus, l = L = ( 0.320 m ) = 0.640 m (b) 14.51 f= d AA = l ⎛ v ⎞ ⎛ 343 m s ⎞ = 4.29 × 10 −2 m = 4.29 cm = = 2 ⎜⎝ f ⎟⎠ ⎜⎝ 000 Hz ⎟⎠ Hearing would be best at the fundamental resonance, so l = 4L = ( 2.8 cm ) and 56157_14_ch14_p426-456.indd 447 v 343 m s = = 536 Hz l 0.640 m and f= v 343 m s ⎛ 100 cm ⎞ = ⎜ ⎟ = 3.1× 10 Hz = 3.1 kHz l ( 2.8 cm ) ⎝ m ⎠ 10/12/10 2:52:14 PM www.elsolucionario.org 448 14.52 14.53 Chapter 14 (a) To form a standing wave in the tunnel, open at both ends, one must have an antinode at each end, a node at the middle of the tunnel, and the length of the tunnel must be equal to an integral number of half-wavelengths [L = n ( l ) or l = 2L n] The resonance frequencies of the tunnel are then vsound ⎛ ⎞ 343 m s 343 m s in air fn = = = n⎜ ⎟ = n ( 0.085 Hz ) n = 1, 2, 3, … 2L n ln ⎝ ( 2.00 × 10 m ) ⎠ (b) It would be good to make such a rule Any car horn would produce several closely spaced resonance frequencies of the air in the tunnel, so the sound would be greatly amplified Other drivers might be frightened directly into dangerous behavior or might blow their horns also (a) The fundamental wavelength of the pipe open at both ends is l1 = L = v f1 Since the speed of sound is 331 m s at 0°C, the length of the pipe is L= (b) v 331 m s = = 0.552 m f1 ( 300 Hz ) At T = 30°C = 303 K, v = ( 331 m s ) TK 303 = ( 331 m s ) = 349 m s 273 273 and f1 = 14.54 (a) Observe from Equations 14.18 and 14.19 in the textbook that the difference between successive resonance frequencies is constant, regardless of whether the pipe is open at both ends or is closed at one end Thus, the resonance frequencies of 650 Hz or less for this pipe must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, and 50.0 Hz, with the lowest or fundamental frequency being f1 = 50.0 Hz (b) Note, from the list given above, the resonance frequencies are only the odd multiples of the fundamental frequency This is a characteristic of a pipe that is open at only one end and closed at the other (c) The length of a pipe with an antinode at the open end and a node at the closed end is onequarter of the wavelength of the fundamental frequency, so the length of this pipe must be L= 14.55 v v 349 m s = = = 316 Hz l1 L ( 0.552 m ) l1 vsound 343 m s = = = 1.72 m ( 50.0 Hz ) 4 f1 In a string fixed at both ends, the length of the string is equal to a half-wavelength of the fundamental resonance frequency, so l1 = 2L The fundamental frequency may then be written as f1 = v = l1 2L F = m F 4L2 m If a second identical string with tension F ′ < F is struck, the fundamental frequency of vibration would be f1′ = ⎛ F ⎞ F′ F′ F′ = ⎜ ⎟ = f1 4L m F 4L2 m F ⎝ ⎠ continued on next page 56157_14_ch14_p426-456.indd 448 10/12/10 2:52:17 PM Sound 449 When the two strings are sounded together, the beat frequency heard will be ⎛ ⎛ 5.40 × 10 N ⎞ F′ ⎞ = 1.10 × 10 fbeat = f1 − f1′ = f1 ⎜ 1− Hz 1− ( ) ⎜ ⎟ = 5.64 beats s 6.00 × 10 N ⎠ F ⎟⎠ ⎝ ⎝ 14.56 By shortening her string, the second violinist increases its fundamental frequency Thus, f1′ = f1 + fbeat = (196 + 2.00 ) Hz = 198 Hz Since the tension and the linear density are both identical for the two strings, the speed of transverse waves, v = F m , has the same value for both strings Therefore, l′1 f1′ = l1 f1, or l′1 = l1 ( f1 f1′) The fundamental wavelength of a string fixed at both ends is l = 2L, and this yields ⎛ f ⎞ ⎛ 196 ⎞ = 29.7 cm L ′ = L ⎜ ⎟ = ( 30.0 cm ) ⎜ ⎝ 198 ⎟⎠ ⎝ f1′⎠ 14.57 The commuter, stationary relative to the station and the first train, hears the actual source frequency fO,1 = fS = 180 Hz from the first train The frequency the commuter hears from the second train, moving relative to the station and commuter, is given by ( ) fO,2 = fS ± fbeat = 180 Hz ± Hz = 178 Hz or 182 Hz ( ) This stationary observer ( vO = ) hears the lower frequency fO,2 = 178 Hz if the second train is moving away from the station vS = − vS , so fO = fS [(v + v O ) (v − vS )] gives the speed of the receding second train as ( ) ⎛ 343 m s + 178 Hz = (180 Hz ) ⎜ ⎜⎝ 343 m s − − vS ( ) ⎞ ⎛ 343 m s + ⎞ ⎟ = (180 Hz ) ⎜ ⎟ ⎟⎠ ⎝ 343 m s + vS ⎠ or ⎛ 180 Hz ⎞ 343 m s + vS = ( 343 m s ) ⎜ ⎝ 178 Hz ⎟⎠ and ⎡⎛ 180 Hz ⎞ ⎤ vS = ( 343 m s ) ⎢⎜ ⎟ − 1⎥ = 3.85 m s ⎣⎝ 178 Hz ⎠ ⎦ so one possibility for the second train is vS = 3.85 m s away from the station ( ) ) The other possibility is that the second train is moving toward the station vS = + vS and the commuter is detecting the higher of the possible frequencies fO,2 = 182 Hz In this case, fO = fS [(v + v O ) (v − vS )] yields ( ⎛ 343 m s + ⎞ 182 Hz = (180 Hz ) ⎜ ⎟ ⎝ 343 m s − vS ⎠ or and ⎛ 180 Hz ⎞ 343 m s − vS = ( 343 m s ) ⎜ ⎝ 182 Hz ⎟⎠ ⎡ ⎛ 180 Hz ⎞ ⎤ vS = ( 343 m s ) ⎢1− ⎜ ⎟ ⎥ = 3.77 m s ⎣ ⎝ 182 Hz ⎠ ⎦ In this case, the velocity of the second train is vS = 3.77 m s toward the station 14.58 The temperatures of the air in the two pipes are T1 = 27°C = 300 K and T2 = 32°C = 305 K The speed of sound in the two pipes is v1 = ( 331 m s ) T1 273 K and v2 = ( 331 m s ) T2 273 K Since the pipes have the same length, the fundamental wavelength, l = 4L, is the same for them Thus, from f = v l , the ratio of their fundamental frequencies is seen to be f2 f1 = v2 v1, which gives f2 = f1 ( v2 v1 ) continued on next page 56157_14_ch14_p426-456.indd 449 10/12/10 2:52:20 PM www.elsolucionario.org 450 Chapter 14 The beat frequency produced is then ⎛ T ⎞ ⎛v ⎞ fbeat = f2 − f1 = f1 ⎜ − 1⎟ = f1 ⎜ − 1⎟ ⎝ v1 ⎠ ⎝ T1 ⎠ or 14.59 (a) ⎛ 305 K ⎞ fbeat = ( 480 Hz ) ⎜ − 1⎟ = 3.98 Hz ⎝ 300 K ⎠ First consider the wall a stationary observer receiving sound from an approaching source having velocity va The frequency received and reflected by the wall is freflect = fS [v (v − va )] Now consider the wall as a stationary source emitting sound of frequency freflect to an observer approaching at velocity va The frequency of the echo heard by the observer is ⎛ v ⎞ ⎛ v + va ⎞ ⎛ v + va ⎞ ⎛ v + va ⎞ = fS ⎜ fecho = freflect ⎜ ⎟⎠ = fS ⎜ ⎜⎝ ⎟ ⎝ v ⎟⎠ v ⎝ v − va ⎠ ⎝ v − va ⎟⎠ Thus, the beat frequency between the tuning fork and its echo is ⎛ 2va ⎞ ⎛ v + va ⎞ ⎛ (1.33) ⎞ = ( 256 Hz ) ⎜ fbeat = fecho − fS = fS ⎜ − 1⎟ = fS ⎜ = 1.99 Hz ⎟ ⎝ 343 − 1.33 ⎟⎠ ⎝ v − va ⎠ ⎝ v − va ⎠ (b) When the student moves away from the wall, va changes sign so the frequency of the echo heard is fecho = fS [(v − va ) (v + va )] The beat frequency is then ⎛ ⎛ va ⎞ v − va ⎞ = fS ⎜ fbeat = fS − fecho = fS ⎜ 1− ⎟ ⎟ v + va ⎠ ⎝ ⎝ v + va ⎠ giving va = v fbeat fS − fbeat The receding speed needed to observe a beat frequency of 5.00 Hz is va = 14.60 (343 m s )(5.00 Hz ) = 3.38 m s ( 256 Hz ) − 5.00 Hz The extra sensitivity of the ear at 000 Hz appears as downward dimples on the curves in Figure 14.29 of the textbook At T = 37°C = 310 K, the speed of sound in air is v = ( 331 m s ) TK 310 = 353 m s = ( 331 m s ) 273 273 Thus, the wavelength of 000 Hz sound is l= v 353 m s = = 0.118 m f 000 Hz For the fundamental resonant mode in a pipe closed at one end, the length required is L= 56157_14_ch14_p426-456.indd 450 l 0.118 m = = 0.0295 m = 2.95 cm 4 10/12/10 2:52:25 PM Sound 14.61 451 At normal body temperature of T = 37.0°C, the speed of sound in air is v = ( 331 m s ) 1+ TC 37.0 = ( 331 m s ) 1+ 273 273 and the wavelength of sound having a frequency of f = 20 000 Hz is v (331 m s ) 1+ 37.0 = 1.76 × 10 −2 m = 1.76 cm = f ( 20 000 Hz ) 273 l= Thus, the diameter of the eardrum is 1.76 cm 14.62 From the defining equation for the decibel level, b = 10 log(I I ), the intensity of sound having a decibel level b is ( I = 10 b 10 )I Thus, the intensity of a 40 dB sound is I 40 = (10 4.0 )I 0, while that of a 70 dB sound is I 70 = (10 7.0 )I Since the combined intensity of sound from a swarm of n mosquitoes is I swarm = nI 40, we must require that I swarm = nI 40 = I 70 or n= 7.0 I 70 (10 ) I = = 10 3.0 = 000 I 40 (10 4.0 ) I We conclude that the swarm should contain ∼ 000 mosquitoes to yield a 70 dB sound 14.63 (a) With a decibel level of 103 dB, the intensity of the sound at 1.60 m from the speaker is found from b = 10 ⋅ log ( I I ) as I = I ⋅10 b 10 = (1.00 × 10 −12 W m ) ⋅1010.3 = 1.00 × 10 −1.7 W m If the speaker broadcasts equally well in all directions, the intensity (power per unit area) at 1.60 m from the speaker is uniformly distributed over a spherical wave front of radius r = 1.60 m centered on the speaker Thus, the power radiated is P = IA = I ( 4p r ) = (1.00 × 10 −1.7 W m ) 4p (1.60 m ) = 0.642 W 14.64 Poutput efficiency = (a) At point C, the distance from speaker A is rA = Pinput = 0.642 W = 0.004 or 0.43% 150 W (b) C ( 3.00 m )2 + ( 4.00 m )2 = 5.00 m and the intensity of the sound from this speaker is IA = PA 1.00 × 10 −3 W = 4p rA2 4p ( 5.00 m ) rA 4.00 m rB A = 3.18 × 10 −6 W m B 3.00 m 2.00 m The sound level at C due to speaker A alone is then ⎛I ⎞ ⎛ 3.18 × 10 −6 W m ⎞ = 65.0 dB b A = 10 ⋅ log ⎜ A ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m ⎟⎠ ⎝ I0 ⎠ continued on next page 56157_14_ch14_p426-456.indd 451 10/12/10 2:52:28 PM www.elsolucionario.org 452 Chapter 14 (b) The distance from point C to speaker B is rB = ( 2.00 m ) + ( 4.00 m ) = 4.47 m and the intensity of the sound from this speaker alone is IB = PB 1.50 × 10 −3 W = = 5.97 × 10 −6 W m 2 4p rB2 4p ( 4.47 m ) The sound level at C due to speaker B alone is therefore ⎛I ⎞ ⎛ 5.97 × 10 −6 W m ⎞ = 67.8 dB b B = 10 ⋅ log ⎜ B ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m ⎟⎠ ⎝ I0 ⎠ (c) If both speakers are sounded together, the total sound intensity at point C is I AB = I A + I B = 3.18 × 10 −6 W m + 5.97 × 10 −6 W m = 9.15 × 10 −6 W m and the total sound level in decibels is ⎛I ⎞ ⎛ 9.15 × 10 −6 W m ⎞ = 69.6 dB b AB = 10 ⋅ log ⎜ AB ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m ⎟⎠ ⎝ I0 ⎠ 14.65 We assume that the average intensity of the sound is directly proportional to the number of cars passing each minute If the sound level in decibels is b = 10 ⋅ log ( I I ), the intensity of the sound is I = I ⋅10 b 10, so the average intensity in the afternoon, when 100 cars per minute are passing, is I100 = I ⋅1080.0 10 = (1.00 × 10 −12 W m ) ⋅108.00 = 1.00 × 10 −4 W m The expected average intensity at night, when only cars pass per minute, is given by the ratio I I100 = 100 =1 20, or I5 = I100 1.00 × 10 −4 W m = = 5.00 × 10 −6 W m 20 20 and the expected sound level in decibels is ⎛I ⎞ ⎛ 5.00 × 10 −6 W m ⎞ = 67.0 dB b = 10 ⋅ log ⎜ ⎟ = 10 ⋅ log ⎜ ⎝ 1.00 × 10 −12 W m ⎟⎠ ⎝ I0 ⎠ 14.66 The well will act as a pipe closed at one end (the bottom) and open at the other (the top) The resonant frequencies are the odd integer multiples of the fundamental frequency, or fn = ( 2n − 1) f1 , where n = 1, 2, 3, … Thus, if fn and fn +1 are two successive resonant frequencies, their difference is ( ) fn +1 − fn = [ ( n + 1) − 1] f1 − ( 2n − 1) f1 = 2n + − − 2n + f1 = f1 In this case, we have 60.0 Hz − 52.0 Hz = f1, giving the fundamental frequency for the well as f1 = 4.00 Hz In the fundamental mode, the well (pipe closed at one end) forms a standing wave pattern with a node at the bottom and the first antinode at the top, making the depth of the well d= 56157_14_ch14_p426-456.indd 452 l1 ⎛ vsound ⎞ ⎛ 343 m s ⎞ = = ⎜ ⎟ = 21.4 m 4 ⎜⎝ f1 ⎟⎠ ⎝ 4.00 Hz ⎠ 10/12/10 2:52:31 PM Sound 14.67 453 If r1 and r2 are the distances of the two observers from the speaker, the intensities of the sound at their locations are I1 = P 4p r12 I2 = and P 4p r22 where P is the power output of the speaker The difference in the decibel levels for the two observers is ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ r2 ⎞ ⎛r ⎞ ⎛r ⎞ b1 − b = 10 log ⎜ ⎟ − 10 log ⎜ ⎟ = 10 log ⎜ ⎟ = 10 log ⎜ 22 ⎟ = 10 log ⎜ ⎟ = 20 log ⎜ ⎟ ⎝ I1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ Since b1 = 80 dB and b = 60 dB, we find that 80 − 60 = 20 log(r2 r1 ) This yields log(r2 r1 ) = 1.0 We also know that and r2 r1 = 10 r2 = 10r1 [1] r1 + r2 = 36.0 m Substituting Equation [1] into [2] gives: Then, Equation [2] yields 14.68 or [2] 11r1 = 36.0 m or r1 = 36.0 m 11 ≈ 3.3 m r2 = 36.0 m − r1 = 36.0 m − 3.3 m = 32.7 m We take toward the east as the positive direction, so the velocity of the sea water relative to Earth is v W E = −10.0 km h The velocity of the trailing ship, which is the sound source (S), relative the propagation medium (sea water) is then ⎛ 0.278 m s ⎞ = 20.6 m s v SW = v SE − v W E = + 64.0 km h − ( −10.0 km h ) = +74.0 km h ⎜ ⎝ km h ⎟⎠ The velocity of the leading ship, the observer (O) in this case, relative to the water is ⎛ 0.278 m s ⎞ = 15.3 m s v OW = v OE − v W E = + 45.0 km h − ( −10.0 km h ) = +55.0 km h ⎜ ⎝ km h ⎟⎠ With the source moving toward the observer, but the observer moving away from the source, the frequency detected by the observer is given by Equation 14.12 as ⎛ v + (− v OW ) ⎞ ⎛ v + vO ⎞ = fS ⎜ f0 = fS ⎜ ⎟ ⎟ ⎝ v − vS ⎠ ⎝ v − (+ v SW ) ⎠ The speed of sound in sea water is v = 533 m s (Table 14.1) and the frequency emitted by the source is fS = 200.0 Hz, so the observed frequency should be ⎛ 533 m s − 15.3 m s ⎞ = 204 Hz f0 = (1 200.0 Hz ) ⎜ ⎝ 533 m s − 20.6 m s ⎟⎠ 14.69 This situation is very similar to the fundamental resonance of an organ pipe that is open at both ends The wavelength of the standing waves in the crystal is l = t, where t is the thickness of the crystal, and the frequency is f= 56157_14_ch14_p426-456.indd 453 v 3.70 × 10 m s = = 2.62 × 10 Hz = 262 kHz l ( 7.05 × 10 −3 m ) 10/12/10 2:52:34 PM www.elsolucionario.org 454 14.70 Chapter 14 The distance from the window ledge to the man’s head is Δy = d − h = 20.0 m − 1.75 m = 18.3 m The time for a warning to travel this distance is t1 = (18.3 m) (343 m s) = 0.0534 s, so the total time needed to receive the warning and react is t1′ = t1 + 0.300 s = 0.353 s The elapsed time when the pot, starting from rest, reaches the level of the man’s head is t2 = ( Δy ) = ay ( −18.3 m ) = 1.93 s −9.80 m s Thus, the latest the warning should be sent is at t = t2 − t1′ = 1.93 s − 0.353 s = 1.58 s into the fall At this time, the pot has fallen Δy = 2 g t = ( 9.80 m s )(1.58 s ) = 12.2 m 2 and is 20.0 m − 12.2 m = 7.8 m above the sidewalk 14.71 On the weekend, there are one-fourth as many cars passing per minute as on a weekday Thus, the intensity, I , of the sound on the weekend is one-fourth of that, I1, on a weekday The difference in the decibel levels is therefore ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ b1 − b = 10 log ⎜ ⎟ − 10 log ⎜ ⎟ = 10 log ⎜ ⎟ = 10 log(4) = dB ⎝ I2 ⎠ ⎝ Io ⎠ ⎝ Io ⎠ so, b = b1 − dB = 70 dB − dB = 64 dB 14.72 (a) At T = 20°C = 293 K, the speed of sound in air is v = ( 331 m s ) 1+ TC 20.0 = 343 m s = ( 331 m s ) 1+ 273 273 The first harmonic or fundamental of the flute (a pipe open at both ends) is given by l1 = 2L = v 343 m s = = 1.31 m f1 261.6 Hz Therefore, the length of the flute is L= (b) λ1 1.31 m = = 0.655 m 2 In the colder room, the length of the flute, and hence its fundamental wavelength, is essentially unchanged (that is, l1′ = l1 = 1.31 m) However, the speed of sound, and thus the frequency of the fundamental, will be lowered At this lower temperature, the frequency must be f1′ = f1 − fbeat = 261.6 Hz − 3.00 Hz = 258.6 Hz The speed of sound in this room is v ′ = l1′ f1′ = (1.31 m )( 258.6 Hz ) = 339 m s From v = ( 331 m s ) 1+ TC 273, the temperature in the colder room is given by ⎡⎛ ⎤ ⎡⎛ 339 m s ⎞ ⎤ ⎞ v T = ( 273°C ) ⎢⎜ − = 273°C ( ) ⎥ ⎢⎜ ⎟ − 1⎥ = 13.4°C ⎟ ⎢⎣⎝ 331 m s ⎠ ⎢⎣⎝ 331 m s ⎠ ⎦⎥ ⎦⎥ 56157_14_ch14_p426-456.indd 454 10/12/10 2:52:38 PM Sound 14.73 The maximum speed of the oscillating block and speaker is k 20.0 N m = ( 0.500 m ) = 1.00 m s m 5.00 kg vm ax = Aw = A (a) When the speaker moves away from the stationary observer, the source velocity is vS = −vm ax and the minimum frequency heard is (f ) O (b) m in ⎛ ⎞ ⎛ v ⎞ 343 m s = 439 Hz = ( 440 Hz ) ⎜ = fS ⎜ ⎝ 343 m s + 1.00 m s ⎟⎠ ⎝ v + vm ax ⎟⎠ When the speaker (sound source) moves toward the stationary observer, then vS = +vm ax and the maximum frequency heard is (f ) O 14.74 455 m ax ⎛ ⎞ ⎛ v ⎞ 343 m s = 441 Hz = ( 440 Hz ) ⎜ = fS ⎜ ⎟ ⎝ 343 m s − 1.00 m s ⎟⎠ ⎝ v − vm ax ⎠ The speed of transverse waves in the wire is F = m vT = F⋅L = m ( 400 N )( 0.750 m ) 2.25 × 10 −3 kg = 365 m s When the wire vibrates in its third harmonic, l = 2L = 0.500 m, so the frequency of the vibrating wire and the sound produced by the wire is fS = vT 365 m s = = 730 Hz 0.500 m l Since both the wire and the wall are stationary, the frequency of the wave reflected from the wall matches that of the waves emitted by the wire Thus, as the student approaches the wall at speed vO , he approaches one stationary source and recedes from another stationary source, both emitting frequency fS = 730 Hz The two frequencies that will be observed are (f ) O ⎛ v + vO ⎞ ⎛ v − vO ⎞ and ( fO )2 = fS ⎜ = fS ⎜ ⎟ ⎟ ⎝ v ⎠ ⎝ v ⎠ The beat frequency is so 14.75 ( ⎛ v + vO − v − vO fbeat = ( fO )1 − ( fO )2 = fS ⎜ ⎜⎝ v ) ⎞⎟ = f ⎟⎠ S vO v ⎛ f ⎞ ⎡ 8.30 Hz ⎤ vO = ⎜ beat ⎟ v = ⎢ ⎥ ( 343 m s ) = 1.95 m s ⎝ fS ⎠ ⎣ ( 730 Hz ) ⎦ The speeds of the two types of waves in the rod are vlong = Y = r Y Y ( A⋅ L) = and vtrans = mV m Thus, if vlong = vtrans , we have F = m F = m L F⋅L m Y ( A⋅ L) ⎛ F⋅L⎞ = 64 ⎜ , or the required tension is ⎝ m ⎟⎠ m 10 ⎡ −2 ⎤ Y ⋅ A ( 6.80 × 10 N m ) ⎣p ( 0.200 × 10 m ) ⎦ = 1.34 × 10 N F= = 64 64 56157_14_ch14_p426-456.indd 455 10/12/10 2:52:41 PM www.elsolucionario.org 456 14.76 Chapter 14 (a) For the fundamental mode of a pipe open at both ends, L = l1 or the wavelength of the waves traveling through the air in the pipe is l1 = 2L = 2(0.500 m) = 1.00 m If the frequency of this fundamental mode is f1 = 350 Hz, the speed of sound waves within the pipe must be v = λ1 f1 = (1.00 m)(350 Hz) = 350 m s From v = (331 m s) 1+ TC 273, the Celsius temperature of the air in the pipe is ⎤ ⎡⎛ ⎡⎛ 350 m s ⎞ ⎤ ⎞ v TC = ( 273°C ) ⎢⎜ − = 273°C ( ) ⎥ ⎢⎜ ⎟ − 1⎥ = 32.2°C ⎟ ⎢⎣⎝ 331 m s ⎠ ⎥⎦ ⎢⎣⎝ 331 m s ⎠ ⎥⎦ (b) If the temperature rises to T ′ = T + 20.0°C = 52.2°C, the speed of sound in the air will be v ′ = (331 m s) 1+ TC′ 273 = (331 m s) 1+ 52.2 273, and the new length of the pipe will be L ′ = L0 [1+ a (ΔT )] = (0.500 m)[1 + (19 × 10 −6 °C−1 )(20.0°C)] The new fundamental wavelength is l1′ = L ′, and the new fundamental resonance frequency will be f1′ = 56157_14_ch14_p426-456.indd 456 (331 m s) 1+ 52.2 273 v′ = = 3.6 × 10 Hz l1′ 2(0.500 m)[1 + (19 × 10 −6 °C−1 )(20.0°C)] 10/12/10 2:52:44 PM ... 10/13/10 9:54:06 AM PREFACE This manual is written to accompany College Physics, Ninth Edition, by Raymond A Serway and Chris Vuille For each chapter in that text, the manual includes solutions... 1.12 In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of significant figures For a given rounding, different small adjustments are made,... Chapter – Solids and Fluids 17 49 89 129 169 209 243 288 Part – Thermodynamics Chapter 10 – Thermal Physics Chapter 11 – Energy in Thermal Processes Chapter 12 – The Laws of Thermodynamics 322 344