EXAM_2023sol

pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

157

Subject

Physics

Date

Jan 9, 2024

Type

pdf

Pages

19

Uploaded by UltraGoldfish3684

Report
Physics 157 EXAM The University of British Columbia 1 December 12, 2023 Working Period Time: 150 minutes Upload Period Time: 10 minutes YOUR NAME: Last Name First Name SIGNATURE: STUDENT NUMBER: INSTRUCTOR’S NAME: (circle one) 101 – Jambuge 102 – Litinskaya 103 – Dierker TUTORIAL SECTION: PLEASE PLACE YOUR UBC STUDENT CARD ON YOUR DESK. THIS EXAMINATION CONSISTS OF 12 MULTIPLE CHOICE QUESTIONS (WORTH 2 POINTS EACH) AND 4 LONG ANSWER QUESTIONS (WORTH 10 POINTS EACH). THE EXAM HAS A TOTAL OF 19 PAGES. YOU MAY TEAR OFF ONLY THE LAST PAGE CONTAINING THE FORMULA SHEET . TO OBTAIN FULL CREDIT FOR LONG ANSWER QUESTIONS, YOU MUST SHOW THE STEPS IN YOUR WORK AND EXPLAIN YOUR ANSWER – SIMPLY WRITING THE FINAL ANSWER IS NOT SUFFICIENT. YOU DO NOT HAVE TO EXPLAIN YOUR MULTIPLE CHOICE ANSWERS – THE ANSWER ALONE IS SUFFICIENT. Exam Rules : Closed Book: You are provided with the Formula Sheet on the last page. Other than that, the exam is closed book. You are not allowed to access any other resources (including your laptop/tablet/phone) during the working period (first 150 minutes). Pen/Pencil/Pages: You may use pen or pencil for writing the exam. You may only use the exam pages provided. Please DO NOT tear off pages of the exam. Keep all pages stapled together so that no pages are lost. Calculators: You may use any calculator with no wireless capabilities, as long as the memory is cleared. Questions: You are not permitted to ask questions during the exam, except in cases of supposed errors or ambiguities in examination questions. Attendance : No student will be permitted to enter the examination room after the first half hour. No student will be permitted to leave the examination room before the end of the upload period. Washroom: Students should be sure to use the washroom before the exam starts. No student will be allowed to visit the washroom during the first half hour. If students need to use the washroom after the first half hour, they should raise their hand and an invigilator will accompany them to the door of the washroom. Academic Integrity : You may not communicate with anyone during the exam except invigilators. You may not share the questions during or after the exam with any other person, website, etc. Uploading: You are not allowed to continue working during the upload period (last 10 minutes). During the upload period, you may ONLY use your laptop/tablet/phone to enter your multiple choice answers into the “Exam Multiple Choice” quiz on Canvas and to scan, convert to pdf, and upload your entire exam (all pages) to the “Exam Written Problems” assignment on Canvas as a single pdf file . Your answers will be graded solely based on what you submit to Canvas. We will collect your exam at the end of the upload period. If you have any technical issues with uploading and are unable to upload by the end of the upload period, please check this box, and we will upload your exam.
Physics 157 EXAM The University of British Columbia 2 Question 1: The wave speed on a uniform string with a length of 1.5 m and a mass of 500 g is 12 m/s. If you use one-fourth of the same string under the same tension and utilize the same source to generate waves on the string, what will be the new wave speed on the string? o 24 m/s o 32 m/s o 3 m/s o 12 m/s o None of the above Question 2: When brick A is placed on top of cylinder A, the bottom of the brick is a distance L above the bottom of cylinder A. When brick B is placed on top of cylinder B, the bottom of the brick is the same distance L above the bottom of cylinder B. The areas of the top faces of the cylinders are related by 𝐴 ? = 2𝐴 ? and the Young’s moduli of the cylinders are related by 𝑌 ? = 2𝑌 ? . The two cylinders had identical lengths before the bricks were placed on them. What is the ratio of the weight of brick A, ? ? , over the weight of brick B, ? ? (i.e., what is ? ? /? ? )? o 0.25 o 0.5 o 1.0 o 2.0 o 4.0 o none of the above The force exerted by cylinder A on brick A is ? ? = 𝑌 ? 𝐴 ? ∆? ? / ? ? , where ∆? ? is the amount that cylinder A is compressed when brick A is placed on it. Similarly, the force exerted by cylinder B on brick B is ? ? = 𝑌 ? 𝐴 ? ∆? ? /? ? . The bricks and cylinders are in equilibrium, so ? ? = ? ? and ? ? = ? ? , where ? ?,? is the weight of brick A and B, respectively. The cylinders start with the same length and the bottom of each brick is the same distance above the bottom of its respective cylinder, so ? ? = ? ? and ∆? ? = ∆? ? . Therefore, ? ? ? ? = 𝑌 ? ? ? Δ? ? /? ? 𝑌 ? ? ? Δ? ? / ? ? = 𝑌 ? ? ? 𝑌 ? ? ? = 4 The speed of a wave on a string is 𝑣 = ඥ? ? , where the mass per unit length, ? = ? ? . Shortening the string does not change ? and the tension is the same, so the speed does not change.
Physics 157 EXAM The University of British Columbia 3 Question 3: A mass oscillates on a spring along x-axis, and its initial phase constant is 𝜙 = − 𝜋 6 . If you use a cosine to represent the displacement, then at ? = 0 : o the mass has a displacement in the positive x direction, and moves in the positive x direction o the mass has a displacement in the positive x direction, and moves in the negative x direction o the mass has a displacement in the negative x direction, and moves in the positive x direction o the mass has a displacement in the negative x direction, and moves in the negative x direction Question 4: At a certain time, a metal sphere at a temperature of 250 o C is radiating 100 W of energy which is being completely absorbed by a very large water bath at a temperature of 10 o C. The sphere and the water are otherwise well insulated from their surroundings. What is the rate at which the entropy of the system (sphere plus water) is changing at this time? Choose the closest answer: o -0.25 W/K o -0.16 W/K o 0 W/K o 0.16 W/K o 0.25 W/K A negative phase shifts to the right. A phase of - /6 would reduce the displacement from +A to 0, so a phase of - /6 means the mass has a displacement in the positive x at t = 0, as shown in the figure. A time increases, the displacement increases from its value at t =0, so the mass is moving in the positive x direction. Heat is leaving the sphere, so the rate at which the entropy of the sphere changes is ∆? 𝑤 ∆? = 𝐻 ? ? = 100 ( 273 . 15+250 ) = −0.19 ?/? . Heat is entering the water, so the rate at which the entropy of the water changes is ∆? 𝑤 ∆? = 𝐻 ? 𝑤 = 100 ( 273 . 15+10 ) = 0.35 ?/? . The total rate of change of the entropy of the system is - 0.19+0.35=0.16 W/K.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 4 Question 5: You need to melt 100 g of a low melting temperature alloy by placing the bottom of a stainless steel pot containing solid pellets of the alloy in contact with a bath of boiling water. The alloy you are using has a melting temperature of 70 o C and a latent heat of fusion of 32 kJ/kg. Assume that all of the heat is transferred through the bottom of the pot, which is 10 cm in diameter and 1 cm thick. The thermal conductivity of stainless steel is ? ?? = 16.2 ?/? ∙ ? . Once the alloy has reached the melting temperature, how long does it have to remain in the bath to melt all 100 g of the alloy? You can assume that the temperature of the boiling water does not change and that a negligible amount of water evaporates during the melting. Choose the closest answer: o 2.1 s o 6.0 s o 8.4 s o 23.5 s o 31.7 s o none of above Question 6: An aluminum rod is 79.3 cm long and a brass rod is 40.1 cm long when both rods are at an initial temperature of 20 o C. The rods are placed in line between two rigid walls with a gap of 0.6 cm between them, as shown in the figure below. The rigid walls are separated by 120.0 cm. The coefficients of thermal expansion of aluminum and brass are ? ? = 2.4𝑥10 −5 ? −1 and ? ? = 2.0𝑥10 −5 ? −1 , respectively. The temperature at which the rods first come into contact is closest to: o 160 o C o 180 o C o 200 o C o 220 o C o 240 o C o 260 o C We have ∆? ? = ? ? ? ? ∆? , ∆? ? = ? ? ? ? ∆? , and ∆? ? + ∆? ? = ∆? ??? = 0.6 ?? . Solving for ∆? gives ∆? = ∆? ??? ( 𝛼 ? ? ? +𝛼 ? ? ? ) = 222 ? . So, the temperature at which the rods first come into contact is 242 o C, or closest to 240 o C. The heat transferred through the bottom of the pot is ? = ?𝐴 ∆? ? . We have 𝐴 = 𝜋? 2 = 3.14 (0.05) 2 = 7.83 𝑥 10 −3 ? 2 , ∆? = 30 ? , and ? = 0.01 ? . So, ? = 16.2 7.83 𝑥 10 −3 30/0.01 = 381.7 ? . The amount of heat required to melt 100 g of the alloy is ? = ?? ? = 0.1 ?? ∙ 32 ?? ?? = 3.2𝑥10 3 ? . This will take an amount of time, ? ???? = ? ? = 3.2?10 3 381 . 7 ? = 8.38 ? . This is closest to 8.4 ? .
Physics 157 EXAM The University of British Columbia 5 Question 7: You found a tuning fork with a missing frequency label. When you place the unknown tuning fork next to one with a frequency of 220 Hz and excite both of them, you hear a beat frequency of 5 Hz. Next, you place the unknown tuning fork next to one with a frequency of 218 Hz and excite both. Now you hear a lower beat frequency. What is the frequency of the unknown tuning fork with the missing label? o 225 Hz o 218 Hz o 220 Hz o 215 Hz o None of the above Question 8: What is the mass of liquid water that you can heat from 0 o C to 100 o C if it absorbs all of the thermal radiation emitted by a lightbulb in one hour? Assume that the lightbulb has a temperature of 2000 K, its radius is 5 cm, and its emissivity is equal to one. The specific heat of water is 4190 J/(kg K). Choose the closest answer: o 1 kg o 15 kg o 63 kg o 127 kg o 245 kg o 320 kg The first beat frequency is |f 1 – f 2 |, so the frequency of the unknown tuning fork is either 215 Hz or 225 Hz. Since the beat frequency is lower with the 218 Hz tuning fork, this means the frequency of the tuning fork with the missing label is 215 Hz. The power emitted by the light bulb is ? = 𝐴?𝜎? 4 . We have 𝐴 = 4 𝜋? 2 = 4 𝜋 (0.05) 2 = 3.14 𝑥 10 −2 ? 2 , ? = 1 , and ? = 2000 ? , so ? = 3.14 𝑥 10 −2 1 5.67 𝑥 10 −8 (2000) 4 = 2.85 𝑥 10 4 ? . In one hour, this generates heat ? ???? = 2.85 𝑥 10 4 ( ? ? ) 3600 ? = 1.03 𝑥 10 8 ? . The amount of heat required to raise the temperature of ? kg of water by ∆? is ? ????? = ??∆? , so ? = ? ???? ( ?∆? ) = 1 . 03?10 8 ? ቀ4190 ? 𝑘?∙? ∙100?ቁ = 245 ?? .
Physics 157 EXAM The University of British Columbia 6 Question 9: Incident light is reflected from the upper and lower surfaces of a thin film with thickness ? . If 2? = ?? 2 , where m is an integer, then in which of the following situations can the reflected light produce a constructive interference patter n? Note: Mark ALL answers that are correct. o n 1 = 1, n 2 = 1.3, n 3 = 1.4 o n 1 = 1.2, n 2 = 1, n 3 = 1.6 o n 1 = 1.3, n 2 = 1.2, n 3 = 1.6 o n 1 = 1.4, n 2 = 1.5, n 3 = 1 o n 1 = 1.4, n 2 = 1.3, n 3 = 1.2 Question 10: The fundamental normal mode for a pipe with two open ends occurs at a frequency of 300 Hz. What is the frequency of the fundamental normal mode if you close one end of the pipe? o 75 Hz o 150 Hz o 300 Hz o 450 Hz o 600 Hz The fundamental frequency for a pipe with two open ends is ? 1 , ??? ???? ???? = ? 2? and the fundamental frequency for a pipe with one closed end is ? 1,??? ?????? ??? = ? 4? . So ? 1,??? ?????? ??? = 150 ?𝑧 Constructive interference will occur for 2 ? = ?? 2 if neither or both of the reflected waves have a half- cycle phase shift. A reflected wave incident from a medium with index of refraction n a onto a medium with index of refraction n b will have no phase shift if n a > n b and will have a half cycle phase shift if n a < n b . This means we need either n 1 > n 2 > n 3 OR n 1 < n 2 < n 3 . This is satisfied by n 1 = 1.4, n 2 = 1.3, n 3 = 1.2 and by n 1 = 1, n 2 = 1.3, n 3 = 1.4.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 7 Question 11: A diatomic gas ( ? ? = 5 2 ? ) expands from 2 L to 6 L at a constant pressure of 100 kPa. How much heat is added to the gas? o 400 J o 600 J o 1000 J o 1400 J o 1800 J Question 12: The two plots below show a snapshot graph and a history graph of a harmonic wave. The snapshot graph shows the displacement versus position at time ? = 0 ? , while the history graph shows the displacement versus time at position 𝑥 = 1 ?? (this point is marked as a dot in both plots). What is the velocity of the wave in the x direction (take the velocity as positive if the wave travels in the positive x direction)? o + 3 cm/s o – 3 cm/s o +0.33 cm/s o -0.33 cm/s o none of the above The velocity is given by 𝑣 = ? ? . From the history graph, we can see that the period, ? = 3 ? . From the snapshot graph, we can see that the wavelength is ? = 1 ?? , so the speed is 𝑣 = 1/3 ?? ? = 0.33 ??/? . At t = 0 s, the displacement is increasing at the position 𝑥 = 1 ?? . From the snapshot graph, this means that the wave is moving to the left, i.e., in the negative x direction, so the velocity is 𝑣 ⃗ = −0.33 ??/? in the positive-x direction. An ideal diatomic gas has a heat capacity ? ? = 5 2 ? so ? 𝑃 = ? ? + ? = 7 2 ? . At constant pressure, ? = ?? 𝑃 Δ? = ?? 𝑃 ( ? 2 − ? 1 ) . We also have ? ? = ???????? , so ? 2 − ? 1 = ? 1 ? 2 ? 1 1 = ? 1 ? 2 ? 1 1 = 2 ? 1 , and ? = ?? 𝑃 2? 1 = ? 7 2 ?2? 1 = ?7?? 1 . From the ideal gas law, ? 1 = 𝑃 1 ? 1 ?? , ? = ?7? 𝑃 1 ? 1 ?? = 7? 1 ? 1 = 7 100 𝑥 10 3 ( ? ? 2 ) 2 𝑥 10 −3 ? 3 = 1400 ?
Physics 157 EXAM The University of British Columbia 8 Question 13: A spherical black asteroid (emissivity equal to 1) of diameter 2 km is orbiting the Sun at a distance from the Sun that is 10 times larger than the distance of the Earth from the Sun. The distance from the Earth to the Sun is 1.5 x 10 11 m. For this problem you can ignore any radiation from the Earth and you can ignore any shadow that the Earth might create. You should take all of the above values as exact and express your final answers with 3 digits. a) What is the temperature of the asteroid? The intensity at the distance of the Earth from the Sun is given by the Solar Constant, ? ?? = 1367 ? ? 2 . The distance of the asteroid from the Sun is ? ?−??? and the distance of the earth from the sun is ? ?−??? , where ? ?−??? = 10 ? ?−??? . The intensity of the radiation from the Sun at the distance of the asteroid is ? ?−??? = ? ?? ( ? ?−??? ? ?−??? ) 2 +1 Evaluating this gives ? ?−??? = 13.67 ? ? 2 +1/2 The diameter of the asteroid is ? ? = 2 ?? and its cross-sectional area is 𝐴 ? = 𝜋 ( ? ? 2 ) 2 +1/2 Evaluating this gives 𝐴 ? = 𝜋 (2000 2 ) 2 = 3.14 𝑥 10 6 ? 2 +1/2 The power absorbed by the asteroid is ? ? , ??? = 𝐴 ? ? ?−??? +1 Evaluating this gives ? ? , ??? = 3.14 𝑥 10 6 ? 2 𝑥 13.67 ? ? 2 = 4.29 𝑥 10 7 ? +1/2 The asteroid will come to equilibrium at a temperature, ? ? , at which it is radiating as much power as it is absorbing, ? ? , ??? = ? ? , ??? +1/2 The amount of power being radiated at temperature ? ? is given by ( ? ? = 1 ): ? ? , ??? = 𝐴 ? ? ? 𝜎? ? 4 = 4 𝜋 ( ? ? 2 ) 2 𝜎? ? 4 = ? ? , ??? +1 Solving this for ? ? gives ? ? = 𝑃 ? , ??? 4𝜋 ( ? ? 2 ) 2 𝜎 1 4 +1 Evaluating this gives ? ? = 88.1 ? +1/2
Physics 157 EXAM The University of British Columbia 9 b) What is the wavelength of radiation that the asteroid emits with maximum intensity? c) What would the intensity of the radiation emitted by the asteroid be for an observer on Earth when the distance from the asteroid to the Earth is 9 times larger than the distance of the Earth from the Sun? From the Wien Displacement Law, ? ? , ??? = (2.90 𝑥 10 −3 ? ∙ ? ) (88.1 ? ) +1 Evaluating this gives ? ? , ??? = 3.29 𝑥 10 −5 ? = 32.9 ?𝑖???? +1/2 The distance from the asteroid to the Earth is ? ?−????ℎ = 9 ? ?−??? The intensity of the radiation from the asteroid at the distance ? ?−????ℎ is ? ?−????ℎ = ? ? , ??? 4 𝜋? ?−????ℎ 2 +1 Evaluating this gives ? ?−????ℎ = 1.88 𝑥 10 −18 ? ? 2 +1/2
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 10 Question 14: A thin flat sheet of rubber (black) separates two steel plates (hatched gray), as shown in the drawing. The steel plates are held between two rigid walls that cannot move. At a temperature of 20 o C, the separation between the steel plates is 0.2 cm, the thickness of the two steel plates are each 0.8 cm, and the rubber is compressed and under a stress of 0.5 kPa. For this problem, you can take Y rubber = 7.0x10 3 N/m 2 , Y steel =2.0 x 10 11 N/m 2 , rubber = 7.7x10 -5 /K, steel = 1.3x10 -5 /K. You can ignore any changes in dimensions of the steel due to stress in the steel since Y steel is so much larger than Y rubber . You should take all of the above values as exact and express your final answers with 3 digits. a) What is the unstressed thickness of the rubber sheet at 20 o C? The change in thickness of the rubber from its unstressed thickness, ? ? ? , due to the intial stress, ? ? 𝑖 = 0.5 ??? is ∆? ? 𝑖 = ? ? 𝑖 − ? ? ? +1/2 We know that ∆? ? 𝑖 is negative because the rubber is compressed by the initial stress. ∆? ? 𝑖 will be given by ∆? ? 𝑖 = − ? ? 𝑖 ? ? ? 𝑌 ? +2 Combing (1) and (2), we can write ? ? 𝑖 = ? ? ? − ? ? 𝑖 ? ? ? 𝑌 ? = ? ? ? ൫1 − ? ? 𝑖 𝑌 ? +1 We are given the initial stressed length of the rubber, ? ? 𝑖 = 2 ?? , so we can find the unstressed thickness of the rubber, ? ? ? , as ? ? ? = ? ? 𝑖 ቀ1−? ? 𝑖 𝑌 ? +1/2 Evaluating this gives ? ? ? = 2?10 −3 ( 1−0 . 5 / 7 ) = 2.15 ?? +1/2
Physics 157 EXAM The University of British Columbia 11 b) What will the stress in the rubber be if the steel plates and rubber are both heated up to a temperature of 100 o C? The initial gap between the steel parts is ? ? 𝑖 = 2.0 ?? . At ? ? = 100 ? ? , the change in gap between the steel parts, ∆? ? , must equal the change in thickness of the rubber from its initial thickness ? ? 𝑖 , so ∆? ? = ∆? ? +1/2 As each steel plate heats up, it will expand, reducing the gap between the plates. The total change in the gap between the steel parts is given by ∆? ? = 2 ? ? 𝑖 ? ? ∆? +2 where ? ? 𝑖 = 8.0 ?? is the initial thickness of each steel plate. The change in thickness of the rubber will be due to both thermal expansion and the change in stress, so the change in thickness is given by (continuing with the sign convention is that positive stress is taken to give compression) ∆? ? = ? ? 𝑖 ? ? ∆? − Δ? ? ? ? 𝑖 𝑌 ? +1 where ∆? ? = ? ? ? − ? ? 𝑖 Combining equations and solving for ∆? ? gives ∆? ? = 𝑌 ? ∆? ൬? ? + 2? ? 𝑖 ? ? 𝑖 ? ? +1 The final stress at ? ? is thus ? ? ? = 𝑌 ? ∆? ൬? ? + 2? ? 𝑖 ? ? 𝑖 ? ? + ? ? 𝑖 +1/2 Evaluating this gives ? ? ? = 7 𝑥 10 3 80 (7.7 𝑥 10 −5 + 8 1.3 𝑥 10 −5 ) + 500 = 601 ?? +1/2
Physics 157 EXAM The University of British Columbia 12 Question 15: A compressed air lift is shown in the sketch. Air can be let in or out of the cylinder to raise or lower the platform sitting on top of a movable piston, then a valve is closed so that there is a constant amount of gas in the cylinder. The combined mass of the piston, platform, and cargo is 2200 kg, and the diameter of the piston, D, is 40 cm. The walls of the cylinder and the piston can be taken to be perfect thermal insulators. A person having mass 75 kg stands on the platform and it is raised to a steady height of 1.4 m. The person then steps (gently, so as to not apply any extra force) off the platform and the platform begins to oscillate. You can take the air in the piston to have heat capacity, ? ? = (5 2 ) ? , the atmospheric pressure outside of the piston to be 10 5 Pa, and the temperature to be 293 K. You should take all the given values as exact and express your final answers with 3 digits. a) What is the initial pressure of the compressed air with the person on the platform? There are three forces acting on the piston (taking up as positive, M = 2200 kg as the mass the platform, and m = 75 kg as the mass of the person): ? ??? = ? ??? 𝐴 +1 /4 ? ??? = −? ??? 𝐴 +1/4 ? ???? = ( ? + ? ) ? +1/4 where the area of the piston, 𝐴 , is given by 𝐴 = 𝜋 ( ? 2 ) 2 = 0.1257 ? 2 +1/2 The net force acting on the piston is: ? ??? = ? ??? + ? ??? + ? ???? +1/4 The initial gas pressure can be found by setting ? ??? = 0 with the person on the platform: ? 𝑖 = ? ??? + ( ?+? ) ? ? +1/2 Evaluating this gives platform: ? 𝑖 = 10 5 + (75 + 2200) 9 . 81 0 . 1257 = 277.60 ??? +1/2
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 13 b) What is the amplitude of the oscillation after the person steps off of the platform? The amplitude of the oscillation will be the difference between the equilibrium height of the platform with the person on it, which is the initial height, 𝑖 = 1.4 ? , and the equilibrium height without the person on the platform. The pressure at the equilibrium position after the person steps off can be found by setting ? ??? = 0 with the person not on the platform: ? ? = ? ??? + ?? ? = 10 5 + 2200 9 . 81 0 . 1257 = 271.75 ??? +1/2 The walls of the cylinder and the piston are perfect thermal insulators, so the compression/expansion of the gas takes place adiabatically. Thus, ?? 𝛾 = ????? +1/2 where ? is given by ? = ( ? ? + ? ) ? ? = 7 5 = 1.4 +1/2 Taking ? 𝑖 as the initial gas pressure at the initial height, 𝑖 = 1.4 ? , the pressure ? at height is given by: ? = ? 𝑖 ( 𝑖 ) 𝛾 +1 From this we can find the height ? at pressure ? ? to be: ? = 𝑖 ( ? 𝑖 ? ? ) 1 𝛾 +1/2 Evaluating this gives ? = 1.4215 ? +1/2 So, the amplitude of the oscillation, 𝐴 ? , is 𝐴 ? = ? − ℎ 𝑖 = 2.15 ?? +1/2
Physics 157 EXAM The University of British Columbia 14 c) What is the period of the oscillation? d) What is the magnitude of the maximum acceleration of the platform? The effective spring constant of the piston (after the person stops off) is given by ? = ( ?? ??? ?ℎ )| ? +1/4 Evaluating the derivative gives: ? = ( ? ( 𝐴? 𝑖 ( 𝑖 ) 𝛾 −? ??? 𝐴 − ?? ) ?ℎ )| ? = ?𝐴? 𝑖 𝑖 𝛾 ? −𝛾−1 +1 Evaluating this expression gives: ? = 33632 ? / ? +1/2 The oscillation angular frequency, 𝜔 , is given by 𝜔 = ඥ? ? +1/4 which evaluates to: 𝜔 = 33632 2200 = 3.91 ??? / ??? +1/4 The period, ? , is given by ? = 2 𝜋 𝜔 = 1.61 ? +1/4 The magnitude of the maximum acceleration is given by ? ??? = 𝐴 ? 𝜔 2 +1/2 Evaluating this gives ? ??? = 2.15 𝑥 10 −2 (3.91) 2 = 0.328 ? / ? 2 +1/2
Physics 157 EXAM The University of British Columbia 15 Question 16: A superhero with spider-enhanced powers leaps off the Lions Gate bridge to test a new spider silk. He ties one end of the spider silk to the bridge, grabs the other end, and jumps. He falls completely vertically with no initial velocity a distance of 63 m from the bridge to the water below. The spider silk material is quite elastic and he oscillates at the end of the fiber. He has spun just exactly the right amount of fiber to just touch the water when he reaches the lowest point in the motion. He is wearing a low friction suit such that damping effects are small. The mass of our superhero is 75 kg and he finds that he has a natural oscillation frequency corresponding to a period of 1.0 s as he bounces up and down at the end of the fiber. You can ignore damping for parts (a)-(f) and only consider damping in part (g). You should take all of the above values as exact and express your final answers with 3 digits. a) What is the spring constant of the spider silk fiber? b) How much does the spider silk stretch to reach the water? The period, T, is related to the angular frequency, 𝜔 , by ? = 2 𝜋 / 𝜔 +1/4 And the angular frequency is related to the spring constant and the mass by 𝜔 = ඥ? ? +1/4 Solving for the spring constant in terms of the period gives ? = ? (2 𝜋 / ? ) 2 +1/2 Evaluating this gives ? = 75(2 𝜋 /1) 2 = 2.96 𝑥 10 3 ? / ? +1/2 The change in gravitational potential energy of our superhero when he falls a distance = 63 ? , will equal the change in potential energy of the spiker-silk when it is fully stretched from its unstretched length by a distance ? : ??ℎ = k ? 2 /2 +1 Solving for ? gives D = 2 ??ℎ / ? +1/2 Evaluating this gives D = 5.60 ? +1/2
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 16 c) What is the unstretched length of the spider silk? d) What is the amplitude of our super hero’s oscillation? e) Write down a mathematical expression for the displacement of our super hero as a function of time, starting from ? = 0 ? when he first reaches the water (take the positive displacement direction as downward). Give values for all parameters in your expression. f) What is the speed of our super hero when he first reaches the equilibrium position? The unstretched length, ? ? , of the spider silk will be ? ? = h D +1/4 This evaluates to ? ? = 57.4 m +1/4 At the equilibrium position, the gravitation force downward will equal the restoring force upward, so ?? = k ∆? +1/2 where ∆? is the amount the fiber is stretched from its free length. Solving for ∆? gives ∆? = ?? /k +1/2 Evaluating this gives ∆? = 0.248 ? +1/2 The distance of the equilibrium position from the bridge, ? , is ? = ? ? + ∆? = 57.7 ? +1/2 The amplitude is thus 𝐴 = ℎ − ? = 63.0 ? − 57.7 ? = 5.3 ? +1/2 𝑥 ( ? ) = 𝐴 cos( 𝜔? + 𝜙 ) +1/4 𝐴 = 5.30 ? +1/4 𝜔 = 6.28 ? −1 +1/4 𝜙 = 0 +1/4 At the equilibrium position, the speed is a maximum, given by 𝑣 ??? = 𝐴𝜔 +1/2 Evaluating this gives: 𝑣 ??? = 5.30 6.28 = 33.3 ? / ? +1/2
Physics 157 EXAM The University of British Columbia 17 g) Our super hero finds that at ? = 10 ? , he is a distance, ? 10 = 61.5 ? below the bridge. What is the damping constant for his low friction suit? The amplitude as a function of time is given by 𝐴 ( ? ) = 𝐴 ? ? −? / ? ? +1/4 A distance below the bridge of ? 10 corresponds to an amplitude at ? = 10 ? of 𝐴 ( ? = 10) = ? 10 − ? = 61.5 57.7 = 4.1 ? +1/4 Where ? ? is the time constant. Solving for ? ? gives ? ? = −? /ln( 𝐴 ( ? )/ 𝐴 ? ) = 10 lnቀ 4 . 1 5 . 3 = 39.0 ? +1/2 The damping constant, ? , is given by b = 2? ? ? = 3.85 ?? / ? +1/2
Physics 157 EXAM The University of British Columbia 18 (This page is extra space for writing)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Physics 157 EXAM The University of British Columbia 19 Possibly useful formulae: ? ? = ? ? + 273.15 ∆? ?ℎ = ??∆? ∆? = ??∆? ∆? 𝐴 = 𝑌 ∆? ?? ? ? = ??∆? ? = ± ?? ? = ?𝐴 ( ? 𝐻 − ? ? ) ? ? = ? ? 𝐴 = 𝜋? 2 𝐴 = 4 𝜋? 2 ? = (4 3 ) 𝜋? 3 ? ??? = ? ? ? = 2.90 𝑥 10 −3 ? ∙ ? ? = 𝐴?𝜎? 4 𝜎 = 5.67 𝑥 10 −8 ? /( ? 2 ? 4 ) ? = ? (4 𝜋? 2 ) ? ?? = 1367 ? ? 2 ? 𝑃 = ? ? + ? ? = ? 𝑃 ? ? [ ? ? = 3 ? 2 ( ???????𝑖? 𝑖???? ??? )] ?? = ??? ? = 8.31 ? /( ??? ∙ ? ) ?? 𝛾 = ????? ?? 𝛾−1 = ????? Δ? = ? − ? Δ? = ?? ? Δ? ? = ?Δ? ? = ??? ln ൫? ? ? 𝑖 ? = ?? 𝑃 Δ? ? = ? ??? ? 𝑖? ? = ? ? ? 𝑖? ? ?????? = 1 − ? ? ? 𝐻 ? ?????? = ? ? ( ? 𝐻 − ? ? ) ?? = ?? ? ?? ?→? = ?? ? ? ? ? ? ∆? ?→? ( 𝑖???ℎ??𝑖? ) = ?? ? ln ? ? ? ? ∆? ?→? ( 𝑖?????𝑖? ) = ?? 𝑃 ln ? ? ? ? ? ( ? ) = 𝐴 cos( 𝜔? + 𝜙 ) 𝜔 = 2 𝜋? = 2 𝜋 ? 𝜔 = ඥ? ? 𝜙 = tan −1 −? ? 𝜔? ? 𝐴 = ඥ𝑥 ? 2 + 𝑣 ? 2 𝜔 2 ? ( ? ) = 𝐴 ? ? −? ? ? cos( 𝜔? + 𝜙 ) 𝜔 = ඥ? ? − ? 2 (4 ? 2 ) [ ? ? = 2 ? ? for ? = −?𝑥 − ?𝑣 ] ? = −?Δ𝑥 ? = − ?? ??? ?𝑥 ൫𝑥 ?? ?𝑥 ? ?𝑥 = ? 𝑥 ?−1 ? = ? ( Δ𝑥 ) 2 2 ? = ?𝑣 2 2 ? ( 𝑥 , ? ) = 𝐴 cos( ?𝑥 ± 𝜔? ) ? = 2 𝜋 ? 𝑣 = ?? = ? ? = 𝜔 ? 𝑣 = ඥ? ? 𝑣 ????? 340 ? ? 𝑣 ?𝑖?ℎ? 3 𝑥 10 8 ? ? Interference: Constructive: ? 𝛥? 12 + 𝛥𝜙 12 = ? 2 𝜋 Destructive: ? 𝛥? 12 + 𝛥𝜙 12 = ? ??? 𝜋 Thin film case: Constructive: 2 ? = ቀ? + | ∆𝜙 12 | 2𝜋 ቁ ? Destructive: 2 ? = ቀ? + 1 2 | ∆𝜙 12 | 2𝜋 ቁ ? ? = ? 𝑣 ? ? sin 𝜃 ? = ? ? sin 𝜃 ? ? = 9.81 ? ? 2