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hw31 - homework 31 Turner(58185 This print-out should have...

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This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A weight suspended from a spring is seen to bob up and down over a distance of 19 cm twice each second. What is its frequency? Correct answer: 2 Hz. Explanation: Let : n = 2 . freq = 2 bobs 1 second = 2 Hz . 002 (part 2 of 3) 10.0 points What is its period? Correct answer: 0 . 5 s. Explanation: period = 1 freq = 1 2 Hz = 0 . 5 s . 003 (part 3 of 3) 10.0 points What is its amplitude? Correct answer: 9 . 5 cm. Explanation: Let : d = 19 cm . The amplitude is the distance from the equilibrium position to maximum displace- ment: 1 2 d = 1 2 (19 cm) = 9 . 5 cm . 004 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 = 5 4 π 2. φ = 7 4 π 3. φ = 1 2 π 4. φ = 3 4 π 5. φ = 0 6. φ = 3 2 π correct 7. φ = 2 π 8. φ = 1 4 π 9. φ = π 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 . 005 (part 1 of 4) 10.0 points A 2 . 3 kg mass is suspended on a 1 × 10 5 N / m spring. The mass oscillates up and down from the equilibrium position y eq = 0 according to y ( t ) = A sin( ωt + φ 0 ) .
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