exam4-solutions - Version 156 exam4 Chiu(60180 This...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 156 – exam4 – Chiu – (60180) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The oscillation of a mass-spring system x = x m cos( ω t + φ ) , where x m is a positive number. At time t = 0, the mass is at the equilibrium point x = 0 and it is moving with positive velocity v 0 . What is the phase angle φ ? 1. φ = 7 4 π 2. φ = 2 π 3. φ = 0 4. φ = 3 4 π 5. φ = 1 4 π 6. φ = 5 4 π 7. φ = 1 2 π 8. φ = π 9. φ = 3 2 π correct Explanation: Since the initial position is x = 0, we know that cos( φ ) = 0, so φ = 1 2 π, 3 2 π, 5 2 π, . . . . The velocity can be found as v = d x dt = x m ω sin( ω t + φ ) . At t = 0 we have v 0 = x m ω sin( φ ) > 0, so the solution is φ = 3 2 π . 002 10.0 points A 0 . 414 kg mass is attached to a spring and undergoes simple harmonic motion with a pe- riod of 0 . 1 s. The total energy of the system is 4 . 9 J. Find the amplitude of the motion. 1. 0.469471 2. 0.463776 3. 0.84259 4. 0.0774342 5. 0.14255 6. 0.35291 7. 0.447902 8. 0.303 9. 0.417961 10. 0.592183 Correct answer: 0 . 0774342 m. Explanation: Let : m = 0 . 414 kg , T = 0 . 1 s , and E = 4 . 9 J . The period is T = 2 π radicalbigg m k k = 4 π 2 m T 2 = 4 π 2 (0 . 414 kg) (0 . 1 s) 2 = 1634 . 41 N / m , and the energy is E = 1 2 k A 2 A = radicalbigg 2 E k = radicalBigg 2 (4 . 9 J) 1634 . 41 N / m = 0 . 0774342 m . 003 10.0 points A spring stretches 1 . 9 cm when a 13 g object is hung from it. The object is replaced with a block of mass 15 g which oscillates in simple harmonic motion.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 156 – exam4 – Chiu – (60180) 2 The acceleration of gravity is 9 . 8 m / s 2 . Calculate the period of motion. 1. 0.459882 2. 1.05673 3. 0.297179 4. 0.829975 5. 0.641013 6. 0.553647 7. 0.51743 8. 0.634345 9. 0.435483 10. 0.595399 Correct answer: 0 . 297179 s. Explanation: Let : x = 1 . 9 cm = 0 . 019 m , m 1 = 13 g = 0 . 013 kg , and m 2 = 15 g = 0 . 015 kg . The force on the spring is F = k x k = F x = m 1 g x . When the 15 g is placed into simple harmonic motion, T = 2 π radicalbigg m 2 k = 2 π radicalbigg m 2 x m 1 g = 2 π radicalBigg (0 . 015 kg) (0 . 019 m) (0 . 013 kg) (9 . 8 m / s 2 ) = 0 . 297179 s . 004 10.0 points A particle executes simple harmonic motion with an amplitude of 1 . 67 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? 1. 3.68927 2. 2.19104 3. 1.44626 4. 1.46358 5. 3.89711 6. 3.15233 7. 1.07387 8. 1.79267 9. 1.21244 10. 1.9832 Correct answer: 1 . 44626 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 ω 2 x 2 , since k = m ω 2 . When the particle is at maximum displacement A , the energy is all potential: U = 1 2 m ω 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 ω 2 x 2 . Conservation of energy gives 1 2 m v 2 + 1 2 m ω 2 x 2 = 1 2 m ω 2 A 2 , or v 2 + ω 2 x 2 = ω 2 A 2 . The speed v is v = A ω sin( ω t ) , and the sine is never more than 1, meaning v max = ω A . We are asked for the displacement at half this speed, v = ωA 2 , so conservation of energy is now parenleftbigg ω A 2 parenrightbigg 2 + ω 2 x 2 = ω 2 A 2 ,
Image of page 2
Version 156 – exam4 – Chiu – (60180) 3 or 1 4 A 2 + x 2 = A 2 , from which we see x = ± 3 2 A = ± 3 2 (1 . 67 cm) = ± 1 . 44626 cm .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern