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fl - bello(rtb473 hw12 Demkov(59910 This print-out should...

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bello (rtb473) – hw12 – Demkov – (59910) 1 This print-out should have 44 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 particle rotates counterclockwise in a circle of radius 5 . 9 m with a constant angular speed of 7 rad / s . At t = 0, the particle has an x coordinate of 4 . 5 m and y > 0 . x y (4 . 5 m , y ) Figure: Not drawn to scale. radius 5 . 9 m 7 rad / s Determine the x coordinate of the particle at t = 1 . 14 s . Correct answer: - 4 . 35108 m. Explanation: Let : x 0 = 4 . 5 m , ω = 7 rad / s , t 0 = 0 s , and R = 5 . 9 m , Since the amplitude of the particle’s mo- tion equals the radius of the circle and ω = 7 rad / s , we have x = A cos( ω t + φ ) = (5 . 9 m) cos bracketleftBig (7 rad / s) t + φ bracketrightBig . We can find φ using the initial condition that x 0 = 4 . 5 m at t = 0 (4 . 5 m) = (5 . 9 m) cos(0 + φ ) , which implies φ = arccos x 0 R = arccos (4 . 5 m) (5 . 9 m) = 40 . 2962 = 0 . 7033 rad . Therefore, at time t = 1 . 14 s , the x coordinate of the particle is x = R cos bracketleftBig ω t + φ bracketrightBig = (5 . 9 m) cos bracketleftBig (7 rad / s) (1 . 14 s) + (0 . 7033 rad) bracketrightBig = - 4 . 35108 m . Note: The angles in the cosine are in radians. 002 (part 2 of 3) 10.0 points Find the x component of the particle’s veloc- ity at t = 1 . 14 s. Correct answer: - 27 . 8931 m / s. Explanation: Differentiating the function x ( t ) with re- spect to t , we find the x component of the particle’s velocity at any time t v x = d x dt = - ω A sin( ω t + φ ) , so at t = 1 . 14 s , the argument of the sine is φ 2 ω t + φ = (7 rad / s) (1 . 14 s) + (0 . 7033 rad) = 8 . 6833 rad , and the x component of the velocity of the particle is v x = - ω R sin( φ 2 ) = - (7 rad / s) (5 . 9 m) sin(8 . 6833 rad) = - 27 . 8931 m / s . 003 (part 3 of 3) 10.0 points Find the x component of the particle’s accel- eration at t = 1 . 14 s. Correct answer: 213 . 203 m / s 2 . Explanation: Differentiating the x component of the par- ticle’s velocity with respect to t , we find the
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bello (rtb473) – hw12 – Demkov – (59910) 2 x component of the particle’s acceleration at any time t a x = d v x dt = - ω 2 A cos( ω t + φ ) , so at t = 1 . 14 s , the x component of the particle’s acceleration is a x = - ω 2 R cos φ 2 = - (7 rad / s) 2 (5 . 9 m) cos(8 . 6833 rad) = 213 . 203 m / s 2 . 004 10.0 points When a body executes simple harmonic mo- tion, its period is 1. proportional to its acceleration. 2. independent of its amplitude. correct 3. inversely proportional to its accelera- tion. 4. the reciprocal of its speed. 5. proportional to its amplitude. Explanation: The period of simple harmonic motion is the same regardless of the amplitude. 005 (part 1 of 3) 10.0 points A simple harmonic oscillator takes 17 . 1 s to undergo five complete vibrations. Find the period of its motion. Correct answer: 3 . 42 s. Explanation: Let : t = 17 . 1 s and n = 5 . The period is, by definition, how long it takes the oscillator to complete one full cycle; i.e. , to return to where it started. Since we are told how long it takes to undergo five complete vibrations, the period is T = t n = 17 . 1 s 5 = 3 . 42 s .
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