HW 31 - Morby Grant Homework 31 Due noon Inst Drummond This...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Morby, Grant – Homework 31 – Due: Apr 21 2006, noon – Inst: Drummond 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider the oscillation of a mass-spring system, where x = A cos( ω t + φ ) . At the time t = 0, the mass m is at x = 0 (the equilibrium point) and it is moving with a positive velocity v 0 . k m v 0 x = 0 x Find the phase angle φ . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. φ = 2 π 2. φ = 1 2 π 3. φ = 5 4 π 4. φ = 3 4 π 5. φ = π 6. φ = 0 7. φ = 7 4 π 8. φ = 3 2 π correct 9. φ = 1 4 π Explanation: Basic Concepts: x = A cos ω ( t + φ ) ω = r m k A B C D E F G H φ ω x The SHM can be represented by the x - projection of a uniform circular motion: x = A cos ω ( t + φ ) . At t = 0 , x = 0. From inspection, it should be either C or G . At C , v < 0; while at G , v > 0. So G is the correct choice, or φ = 3 2 π . 002 (part 2 of 3) 10 points Let the mass be m = 6 . 55 kg, spring constant k = 964 N / m and the initial velocity v 0 = 1 . 02 m / s. Find the amplitude A . Correct answer: 0 . 084078 m. Explanation: v = d x dt = - ω A sin( ω t + φ ) So the velocity amplitude or the maximum speed is v max = ωA ; i.e. , v 0 = ωA , so A = v 0 ω = v 0 r m k = (1 . 02 m / s) s (6 . 55 kg) (964 N / m) = 0 . 084078 m . 003 (part 3 of 3) 10 points Find the total energy of oscillation at t = T 8 ; i.e. , at one-eighth of the period.
Image of page 1

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

View Full Document Right Arrow Icon
Morby, Grant – Homework 31 – Due: Apr 21 2006, noon – Inst: Drummond 2 (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 1 4 m v 2 0 2. E = 5 2 m v 2 0 3. E = m v 2 0 4. E = 3 4 m v 2 0 5. E = 1 2 m v 2 0 correct 6. E = 3 2 m v 2 0 7. E = 2 m v 2 0 8. E = 1 2 2 m v 2 0 Explanation: Since the spring force is a conservative force the total energy is conserved. One may equate the energy at t = 0 where x = 0. So E = K + U = K max = 1 2 m v 2 0 . 004 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . When an unknown weight W was sus- pended from a spring with an unknown force constant k it reached its equilibrium position and the spring was stretched by 12 . 7 cm be- cause of the weight W . Then the weight W was pulled further down to a position 15 cm (2 . 3 cm below its equilib- rium position) and released, which caused an oscillation in the spring.
Image of page 2
Image of page 3
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