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Unformatted text preview: obaidi (aao476) OH12 Bradley (191) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This is Online Homework No. 12. It is due 07:00 AM on Thursday 20 November (UAE Time), unless you are otherwise notified. 001 10.0 points An oscillator is described by 1 2 3 4 5 6 7 8 9 10 11 12 1 2 x (t) t (sec) What is the angular frequency ? 1. 3.1 rad/s 2. 2.0 rad/s 3. 6.2 rad/s 4. 1.6 rad/s correct 5. 4.0 rad/s Explanation: From the graph, T = 4 s. = 2 T = 2 4 sec = 2 1 / sec = 1 . 5708 1 / s . 002 10.0 points The equation of motion of a simple harmonic oscillator is d 2 x dt 2 = 9 x, where x is displacement and t is time. What is the period of oscillation? 1. T = 2 3 correct 2. T = 6 3. T = 3 2 4. T = 9 2 5. T = 2 9 Explanation: d 2 x dt 2 = 2 x, where is the angular frequency, so the pe riod of oscillation is T = 2 = 2 9 = 2 3 . 003 (part 1 of 2) 5.0 points A piston in an automobile engine is in simple harmonic motion. Its amplitude of oscillation from the equilibrium (centered) position is 7 . 05 cm and its mass is 1 . 242 kg. Find the maximum velocity of the piston when the auto engine is running at the rate of 5380 rev / min. Correct answer: 39 . 7192 m / s. Explanation: Let : A = 7 . 05 cm , m = 1 . 242 kg , and f = 5380 rev / min . = 2 f = 2 (5380 rev / min) 1 min 60 s = 563 . 392 rad / s . The simple harmonic motion is described by x = A cos t, where is the frequency in rad/s if t is in seconds. The velocity is v = d x dt = A sin( t ) obaidi (aao476) OH12 Bradley (191) 2 meaning the maximum velocity is v max = A = (0 . 0705 m) (563 . 392 rad / s) = 39 . 7192 m / s , since sine has a maximum value of 1. 004 (part 2 of 2) 5.0 points Find the maximum acceleration of the piston when the auto engine is running at this rate. Correct answer: 22 . 3775 km / s 2 . Explanation: The acceleration is a = d 2 x dt 2 = d v dt = A 2 cos( t ) , so the maximum acceleration is a max = A 2 = (0 . 0705 m) (563 . 392 rad / s) 2 = 22 . 3775 km / s 2 , since cosine has a maximum value of 1. 005 10.0 points A body oscillates with simple harmonic mo tion along the xaxis. Its displacement varies with time according to the equation A = A sin parenleftBig t + 3 parenrightBig , where = radians per second, t is in sec onds, and A = 5 . 9 m. What is the phase of the motion at t = 4 . 4 s? Correct answer: 14 . 8702 rad. Explanation: Let : t = 4 . 4 s and = . x = A sin( t + ) The phase is the angle in the argument of the sine function, and from the problem state ment we see it is = t + 3 = ( rad / s) (4 . 4 s) + 3 = 14 . 8702 rad ....
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This note was uploaded on 02/16/2009 for the course PYHS 191 taught by Professor Curtise during the Spring '08 term at Abilene Christian University.
 Spring '08
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