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

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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.

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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.
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