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

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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.
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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 ,
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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 .
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