12 - obaidi (aao476) OH-12 Bradley (191) 1 This print-out...

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Unformatted text preview: obaidi (aao476) OH-12 Bradley (191) 1 This print-out should have 17 questions. Multiple-choice 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) OH-12 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 x-axis. 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.

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12 - obaidi (aao476) OH-12 Bradley (191) 1 This print-out...

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