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10 - homework 11 JACOBS AMBER Due 1:00 am Question 1 chap...

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homework 11 – JACOBS, AMBER – Due: Apr 22 2008, 1:00 am 1 Question 1, chap 15, sect 6. part 1 of 3 10 points A pendulum that moves through its equilib- rium position once every 1.000 s is sometimes called a “seconds pendulum.” a) What is the period of any seconds pen- dulum? 1. 4.000 s 2. None of these 3. 2.000 s correct 4. 0.250 s 5. 1.000 s 6. 0.500 s Explanation: Basic Concept: A pendulum passes through its equilibrium position twice each cycle. Solution: T = 2(1 . 000 s) = 2 . 000 s Question 2, chap 15, sect 6. part 2 of 3 10 points In Cambridge, England, a seconds pendu- lum is 0.9942 m long. b) What is the free-fall acceleration in Cam- bridge? 1. None of these 2. 9 . 81236 m / s 2 correct 3. 9 . 81341 m / s 2 4. 9 . 79552 m / s 2 5. 9 . 81 m / s 2 6. 9 . 79657 m / s 2 Explanation: Basic Concept: T = 2 π radicalBigg L g Solution: parenleftbigg T 2 π parenrightbigg 2 = L g g 1 = L 1 parenleftbigg 2 π T parenrightbigg 2 = (0 . 9942 m) parenleftbigg 2 π 2 s parenrightbigg 2 = 9 . 81236 m / s 2 Question 3, chap 15, sect 6. part 3 of 3 10 points In Tokyo, Japan, a seconds pendulum is 0.9926 m long. c) What is the free-fall acceleration in Tokyo? 1. None of these 2. 9 . 81 m / s 2 3. 9 . 79552 m / s 2 4. 9 . 79657 m / s 2 correct 5. 9 . 81341 m / s 2 6. 9 . 81236 m / s 2 Explanation: Solution: g 2 = L 2 parenleftbigg 2 π T parenrightbigg 2 = (0 . 9926 m) parenleftbigg 2 π 2 s parenrightbigg 2 = 9 . 79657 m / s 2 Question 4, chap 15, sect 7. part 1 of 3 10 points
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homework 11 – JACOBS, AMBER – Due: Apr 22 2008, 1:00 am 2 A particle rotates counterclockwise in a cir- cle of radius 6 . 6 m with a constant angular speed of 18 rad / s . At t = 0, the particle has an x coordinate of 5 . 3 m and y > 0 . x y (5 . 3 m , y ) Figure: Not drawn to scale. radius 6 . 6 m 18 rad / s Determine the x coordinate of the particle at t = 1 . 71 s . Correct answer: 6 . 59998 m (tolerance ± 1 %). Explanation: LET : x 0 = 5 . 3 m , ω = 18 rad / s , t 0 = 0 s , and R = 6 . 6 m , Since the amplitude of the particle’s mo- tion equals the radius of the circle and ω = 18 rad / s , we have x = A cos( ω t + φ ) = (6 . 6 m) cos bracketleftBig (18 rad / s) t + φ bracketrightBig . We can find φ using the initial condition that x 0 = 5 . 3 m at t = 0 (5 . 3 m) = (6 . 6 m) cos(0 + φ ) , which implies φ = arccos x 0 R = arccos (5 . 3 m) (6 . 6 m) = 36 . 5796 = 0 . 638433 rad . Therefore, at time t = 1 . 71 s , the x coordinate of the particle is x = R cos bracketleftBig ω t + φ bracketrightBig = (6 . 6 m) cos bracketleftBig (18 rad / s) (1 . 71 s) + (0 . 638433 rad) bracketrightBig = 6 . 59998 m . Note: The angles in the cosine are in radians. Question 5, chap 15, sect 7. part 2 of 3 10 points Find the x component of the particle’s ve- locity at t = 1 . 71 s. Correct answer: - 0 . 297822 m / s (tolerance ± 1 %). 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 . 71 s , the argument of the sine is φ 2 ω t + φ = (18 rad / s) (1 . 71 s) + (0 . 638433 rad) = 31 . 4184 rad , and the x component of the velocity of the particle is v x = - ω R sin( φ 2 ) = - (18 rad / s) (6 . 6 m) sin(31 . 4184 rad) = - 0 . 297822 m / s .
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