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# 12 - obaidi(aao476 OH-12 Bradley(191 This print-out should...

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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 ω ? 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. ω = 2 π T = 2 π 4 sec = π 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 π 3 correct 2. T = 6 π 3. T = 3 2 π 4. T = 9 2 π 5. T = 2 π 9 Explanation: d 2 x dt 2 = - ω 2 x , where ω is the angular frequency, so the pe- riod of oscillation is T = 2 π ω = 2 π 9 = 2 π 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 . ω = 2 π f = 2 π (5380 rev / min) 1 min 60 s = 563 . 392 rad / s . The simple harmonic motion is described by x = A cos ω t , where ω is the frequency in rad/s if t is in seconds. The velocity is v = d x dt = - A ω sin( ω t )

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obaidi (aao476) – OH-12 – Bradley – (191) 2 meaning the maximum velocity is v max = A ω = (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 ω 2 cos( ω t ) , so the maximum acceleration is a max = A ω 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 0 sin parenleftBig ω t + π 3 parenrightBig , where ω = π radians per second, t is in sec- onds, and A 0 = 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 ω = π . x = A 0 sin( ω t + φ ) The phase is the angle in the argument of the sine function, and from the problem state- ment we see it is φ = π t + π 3 = ( π rad / s) (4 . 4 s) + π 3 = 14 . 8702 rad . 006 (part 1 of 4) 3.0 points A 12 . 4 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 ) . Find the angular frequency of the oscillat- ing mass. Correct answer: 89 . 8027 s 1 . Explanation: Let : M = 12 . 4 kg and k = 1 × 10 5 N / m . When the mass moves out of equilibrium, it suffers a net restoring force F net y = F spring - Mg = - k ( y - y eq ) = - ky , and accelerates back towards the equilibrium position at the rate a y = F net y M = - k M y .
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12 - obaidi(aao476 OH-12 Bradley(191 This print-out should...

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