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Unformatted text preview: homework 11 GAUTHIER, JOSEPH Due: Nov 16 2007, 1:00 am 1 Question 1, chap 15, sect 2. part 1 of 1 10 points The equation of motion of a simple har monic 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 = 3 2 3. T = 6 4. T = 9 2 5. T = 2 9 Explanation: For a simple harmonic oscillator, the equa tion of motion can be written as d 2 x dt 2 = 2 x , where is the angular frequency, so the period of oscillation is T = 2 = 2 9 = 2 3 . Question 2, chap 15, sect 2. part 1 of 1 10 points A body oscillates with simple harmonic mo tion along the xaxis. Its displacement varies with time according to the equation A = A sin( t + / 3) , where = radians per second, t is in sec onds, and A = 1 . 1 m. What is the phase of the motion at t = 2 . 2 s? Correct answer: 7 . 9587 rad (tolerance 1 %). Explanation: Basic Concepts: 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 = bracketleftBig ( rad / s) (2 . 2 s) + 3 bracketrightBig = 7 . 9587 rad . Question 3, chap 15, sect 1. part 1 of 3 10 points A 424 g mass is connected to a light spring of force constant 3 N / m and it is free to os cillate on a horizontal, frictionless track. The mass is displaced 5 cm from the equilibrium point and released form rest. 3 N / m 424 g 5 cm x = 0 x Find the period of the motion. Correct answer: 2 . 36212 s (tolerance 1 %). Explanation: This situation corresponds to the special case x ( t ) = A cos t, with A = 5 cm = 0 . 05 m . Therefore, the frequency is = radicalbigg k m = radicalBigg 3 N / m . 424 kg = 2 . 65998 s 1 . For the period we find T = 2 = 2 . 36212 s . homework 11 GAUTHIER, JOSEPH Due: Nov 16 2007, 1:00 am 2 Question 4, chap 15, sect 1. part 2 of 3 10 points What is the maximum speed of the mass? Correct answer: 0 . 132999 m / s (tolerance 1 %). Explanation: The velocity as a function of time is given by v ( t ) = A sin( t ) , so the maximum speed of the mass is equal to v max = A = (2 . 65998 s 1 ) (0 . 05 m) = 0 . 132999 m / s . Question 5, chap 15, sect 1. part 3 of 3 10 points What is the maximum acceleration of the mass? Correct answer: 0 . 353774 m / s 2 (tolerance 1 %). Explanation: The acceleration as a function of time is given by a ( t ) = 2 A cos( t ) , so the maximum acceleration of the mass is equal to a max = 2 A = (2 . 65998 s 1 ) 2 (0 . 05 m) = 0 . 353774 m / s 2 . Question 6, chap 15, sect 1. part 1 of 1 10 points A large block with mass 25 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency 1 . 51 Hz . Block smaller block with mass 7 kg rests on it, as shown in the figure, and the coefficient of static friction between the two is s = 0 . 646 ....
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This note was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

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