homework-10-oscillations-and-waves-solutions

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Homework 10 oscillations and waves solutions Laboratory For Phy 302K (University of Texas at Austin) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Homework 10 oscillations and waves solutions Laboratory For Phy 302K (University of Texas at Austin) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A common technique used to measure the force constant k of a spring is the following: Hang the spring vertically, then allow a mass m to stretch it a distance d from the equilibrium position under the action of the “load” m g . 91 m 39 N Find the spring constant k if the spring is stretched a distance 91 m by a suspended weight of 39 N. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 428571 N / m. Explanation: Let : W = 39 N and d = 91 m . The force exerted on the spring is supplied by the weight of the load, so k d = W k = W d = 39 N 91 m = 0 . 428571 N / m . 002 10.0points A load of 62 N attached to a spring hanging vertically stretches the spring 8 . 2 cm. The spring is now placed horizontally on a table and stretched 15 cm. What force is required to stretch it by this amount? Correct answer: 113 . 415 N. Explanation: Let : F = 62 N , x 1 = 8 . 2 cm , and x 2 = 15 cm . The force exerted by the spring is F = k x ∝ x , so F 2 F 1 = x 2 x 1 F 2 = F 1 x 2 x 1 = (62 N) (15 cm) 8 . 2 cm = 113 . 415 N . 003 10.0points The displacement in simple harmonic motion is maximum when the 1. velocity is a maximum. 2. velocity is zero. correct 3. acceleration is zero. 4. kinetic energy is a maximum. 5. linear momentum is a maximum. Explanation: The maximum displacement occurs at the turning points, which are the points where the velocity is zero. 004 10.0points When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the follow- ing is true of the values of its speed and the magnitude of the restoring force? This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 2 Magnitudeof Speed RestoringForce 1. Maximum Zero 2. Zero Maximum correct 3. 1 2 maximum 1 2 maximum 4. Zero Zero 5. Maximum 1 2 maximum Explanation: The maximum displacement occurs at the turning points (where the velocity or speed is zero). The magnitude of restoring force is given by Hooke’s law F = - k x , where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a different perspective, the displace- ment from the equilibrium position can be written as y = A sin » , where » is the phase of the oscillation. When the object is at its maximum displacement sin » = 1 » = Ã 2 so its speed is v = Ë A cos » = Ë A cos Ã 2 = 0 and the restoring force is F = m A Ë 2 sin » = m A Ë 2 Ã 2 = m A Ë 2 , at its maximum value. 005 10.0points A particle executes simple harmonic motion with an amplitude of 2 . 09 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? Correct answer: 1 . 80999 cm. Explanation: The potential energy of a simple harmonic oscillator at displacement x from the equilib- rium point is U osc = 1 2 k x 2 = 1 2 m Ë 2 x 2 , since k = m Ë 2 . When the particle is at maximum displacement A , the energy is all potential: U = 1 2 m Ë 2 A 2 . At other points x , the energy might be both kinetic (speed v ) and potential K + U = 1 2 m v 2 + 1 2 m Ë 2 x 2 . Conservation of energy gives 1 2 m v 2 + 1 2 m Ë 2 x 2 = 1 2 m Ë 2 A 2 , or v 2 + Ë 2 x 2 = Ë 2 A 2 . The speed v is v = - A Ë sin( Ë t ) , and the sine is never more than 1, meaning v max = Ë A . We are asked for the displacement at half this speed, v = ËA 2 , so conservation of energy is now parenleftbigg Ë A 2 parenrightbigg 2 + Ë 2 x 2 = Ë 2 A 2 , or 1 4 A 2 + x 2 = A 2 , This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

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yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 3 from which we see x = ± √ 3 2 A = ± √ 3 2 (2 . 09 cm) = ± 1 . 80999 cm . 006 10.0points Two waves have the same speed. The first has twice the frequency of the second. Compare the wavelengths of the two waves. 1. The have the same wavelength. 2. The second has half the wavelength of the first. 3. The first has twice the wavelength of the second. 4. The second has square root two times the wavelength of the first. 5. The first has half the wavelength of the second. correct Explanation: 007 10.0points The wavelength of light visible to the human eye is on the order of 7 × 10 2 7 m. Find the frequency of the lightwave if the speed of light in air is 2 . 99792 × 10 8 m / s. Correct answer: 4 . 28275 × 10 14 s 2 1 . Explanation: Let : » = 7 × 10 2 7 m and c = 2 . 99792 × 10 8 m / s . c = f » f = c » = 2 . 99792 × 10 8 m / s 7 × 10 2 7 m = 4 . 28275 × 10 14 s 2 1 . 008 10.0points The figure shows two wave pulses that are approaching each other. P Q Which of the following best shows the shape of the resultant pulse when the centers of the pulses, points P and Q , coincide? 1. correct 2. 3. 4. 5. Explanation: Notice that the two pulses have the same width and amplitude. Choosing the the point P (the same as point Q when the two pulses coincide) as the origin, the two pulses can be described as: P : y 1 = A , - d ≤ x ≤ d Q : y 2 = braceleftBigg A , - d ≤ x < 0 - A , 0 < x < d This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 4 Using the principle of superposition, the re- sultant pulse is y = y 1 + y 2 = braceleftbigg 2 A , - d ≤ x < 0 0 , 0 < x < d P Q P + Q 009 10.0points You are given f 1 ( x ), a transverse wave that moves on a string that ends and is FIXED in place at x = 5 m. As the problem begins, the wave is moving to the right at v = 1 m/s. v 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Amplitude (centimeter) Distance (meter) Consider the image of the wave reflected about the FIXED point x = 5 m in the fol- lowing diagram. The image will be moving to the left at v 2 = - 1 m/s (in the opposite direction from the real wave). v v 2 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Amplitude (centimeter) Distance (meter) What is the shape of the wave on the string after 3 s? 1. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 2. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 3. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 5 4. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 5. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 6. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) correct 7. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 8. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 9. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) 10. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Distance (meter) Explanation: The initial wave (real) on the string is rep- resented with a dashed line and its reflected wave (imaginary) is represented with a dotted line. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Amplitude (centimeter) Distance (meter) Initialtime, t = 0 s After 3 s the positions of the two waves have This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

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yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 6 both moved 3 meters in opposite directions. The resultant sum of the two waves is the light gray line. 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Amplitude (centimeter) Distance (meter) Superpositionat t = 3 s 0 1 2 3 4 5 6 7 8 9 10 - 3 - 2 - 1 0 1 2 3 Amplitude (centimeter) Distance (meter) Resultantat t = 3 s keywords: 010 10.0points A dog can hear sounds in the range from 15 to 50,000 Hz. What wavelength corresponds to the lower cut-off point of the sounds at 20 ç C where the sound speed is 344 m / s? Correct answer: 22 . 933 m. Explanation: Let : v = 344 m / s and f = 15 Hz . The wavelength is » = v f = 344 m / s 15 Hz = 22 . 933 m . 011 10.0points A sound wave has a frequency of 694 Hz in air and a wavelength of 0 . 55 m. What is the temperature of the air? As- sume the velocity of sound at 0 ç C is 331 m / s. Correct answer: 90 . 0371 ç C. Explanation: Let : f = 694 Hz , » = 0 . 55 m , v 0 = 331 m / s , and T 0 = 0 ç C + 273 = 273 K . The relationship between the speed of sound in air and Kelvin temperature is v = v 0 radicalbigg T 273 K = f » T = (273 K) parenleftbigg f » v 0 parenrightbigg 2 - 273 = (273 K) bracketleftbigg (694 Hz) (0 . 55 m) 331 m / s bracketrightbigg 2 - 273 = 90 . 0371 ç C . 012 10.0points A rock group is playing in a bar. Sound emerging from the door spreads uniformly in all directions. The intensity level of the music is 126 dB at a distance of 7 . 79 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing? Disregard absorption. Correct answer: 1 . 55431 × 10 7 m. Explanation: Let : ³ 1 = 126 dB , r 1 = 7 . 79 m , and I 0 = I t = 1 . 0 × 10 2 12 W / m 2 . I 1 = I 0 10 β 1 / 10 . This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 7 The intensity at a distance r from the source is I = P 4 Ã r 2 ∝ 1 r 2 , so I 1 I t = r 2 t r 2 1 r t = r 1 radicalbigg I 1 I t = r 1 radicalbig 10 β 1 / 10 = (7 . 79 m) √ 10 12 . 6 = 1 . 55431 × 10 7 m . 013 10.0points The sound level produced by one singer is 82 . 9 dB. What would be the sound level produced by a chorus of 41 such singers (all singing at the same intensity at approximately the same distance as the original singer)? Correct answer: 99 . 0278 dB. Explanation: Let : ³ = 82 . 9 dB and n = 41 . The total sound intensity is the sum of the sound intensities produced by each individual singer: I n = n I 1 , so ³ n = 10 log parenleftbigg I n I 0 parenrightbigg = 10 log parenleftbigg n I 1 I 0 parenrightbigg = 10 log parenleftbigg I 1 I 0 parenrightbigg + 10 log n = ³ 1 + 10 log n = 82 . 9 dB + 10 log 41 = 99 . 0278 dB . 014 10.0points A phone cord is 4 . 1 m long. The cord has a mass of 0 . 24 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0 . 882 s. What is the tension in the cord? Correct answer: 80 . 9539 N. Explanation: Let : 3 = 4 . 1 m , m = 0 . 24 kg , and Δ t = 0 . 882 s . One complete trip is a distance of 2 3 , so the speed is v = d total Δ t = 4 (2 3 ) Δ t = 8 3 Δ t . The linear density is μ = m 3 , so v = radicalBigg F μ F = μ v 2 = m 3 parenleftbigg 8 3 Δ t parenrightbigg 2 = 64 m 3 (Δ t ) 2 = 64 (0 . 24 kg) (4 . 1 m) (0 . 882 s) 2 = 80 . 9539 N . keywords: 015(part1of2)10.0points A cello string vibrates in its fundamental mode with a frequency of 110 1 / s. The vi- brating segment is 105 . 6 cm long and has a mass of 0 . 48 g. Find the tension in the string. Correct answer: 24 . 533 N. Explanation: 3 Let : f 1 = 110 1 / s , 3 = 105 . 6 cm = 1 . 056 m , and m = 0 . 48 g = 0 . 00048 kg . For the fundamental mode, 3 = » 2 , so » = 2 3 . This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 8 With linear density μ = m 3 , v = radicalBigg F μ F = μ v 2 = m 3 ( f » ) 2 = m f 2 3 (2 3 ) 2 = 4 m 3 f 2 = 4 (0 . 00048 kg) (1 . 056 m) (110 1 / s) 2 = 24 . 533 N . 016(part2of2)10.0points 105 . 6 cm Find the frequency of the string when it vibrates in two segments. Correct answer: 220 1 / s. Explanation: The frequency of the second harmonic is f 3 = 2 f 1 = 2 (110 1 / s) = 220 1 / s . 017 10.0points A string is stretched to a length of 153 cm and both ends are fixed. If the density of the string is 0 . 012 g / cm, and its tension is 1340 N, what is the funda- mental frequency? Correct answer: 345 . 335 Hz. Explanation: Let : L = 153 cm = 1 . 53 m , μ = 0 . 012 g / cm , and T = 345 . 335 Hz . Both ends of the string are fixed (immo- bile), so any standing waves on the string must have nodes (stationary points) at both ends, so n » 2 = L , n = 1 , 2 , 3 , · · · . The resulting possible frequencies are f n = n 2 L v = n 2 L radicalBigg F μ , n = 1 , 2 , 3 , · · · . The fundamental frequency is the lowest possible frequency and corresponds to the case n = 1. f n = 1 2 L radicalBigg F μ = 1 2 (1 . 53 m) radicalBigg 1340 N 1 . 2 × 10 2 7 kg / m = 345 . 335 Hz . 018 10.0points The human ear canal is about 2 . 7 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fundamental frequency around which we would expect hearing to be most sensitive? Assume the speed of sound in air to be 339 m / s. Correct answer: 3 . 13889 kHz. Explanation: Let : L = 2 . 7 cm = 0 . 027 m , and v = 339 m / s . Hearing would be best at the fundamental resonance of » = 4 L , so f 1 = v » = v 4 L = 339 m / s 4 (0 . 027 m) parenleftbigg 1 kHz 1000 Hz parenrightbigg = 3 . 13889 kHz . 019(part1of3)10.0points An open vertical tube has water in it. A This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

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yoon (hy5647) – Homework 10, oscillations and waves – marder – (54720) 9 tuning fork vibrates over its mouth. As the water level is lowered in the tube, the eighth resonance is heard when the water level is 225 cm below the top of the tube. 225 cm What is the wavelength of the sound wave? The speed of sound in air is 343 m / s. Correct answer: 60 cm. Explanation: Let : 3 = 225 cm and N = 8 . When one end of a pipe is closed and the other end open, the wavelength is » = 4 3 2 N - 1 , N = 1 , 2 , 3 , · · · , since there is node at one end, so » = 4 3 2 N - 1 = 4 3 2 (8) - 1 = 4 (225 cm) 2 (8) - 1 = 60 cm . 020(part2of3)10.0points What is the frequency of the sound wave; i.e. , the tuning fork? Correct answer: 571 . 667 s 2 1 . Explanation: Let : v = 343 m / s . The frequency is f = v » = 343 m / s 60 cm · 100 cm 1 m = 571 . 667 Hz . 021(part3of3)10.0points The water continues to leak out the bottom of the tube. When the tube next resonates with the tun- ing fork, what is the length of the air column? Correct answer: 255 cm. Explanation: The next resonance will occur when the open vertical tube has a length 1 2 » greater than its initial length: 3 2 = 3 + » 2 = 225 cm + 60 cm 2 = 255 cm . This study source was downloaded by 100000830645158 from CourseHero.com on 03-29-2022 22:17:11 GMT -05:00 https://www.coursehero.com/file/40921894/Homework-10-oscillations-and-waves-solutionspdf/ Powered by TCPDF (www.tcpdf.org) Downloaded by ThatGuyTravis (omegaturtle0619@gmail.com) lOMoARcPSD|38944138

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