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Unformatted text preview: quiz 04 GAUTHIER, JOSEPH Due: Dec 5 2007, 10:00 pm 1 Question 1, chap 15, sect 2. part 1 of 1 10 points A simple harmonic oscillator has amplitude . 68 m and period 3 . 2 sec. What is the maxi mum acceleration? 1. . 417243 m / s 2 2. . 0664062 m / s 2 3. 8 . 38916 m / s 2 4. . 2125 m / s 2 5. 1 . 31081 m / s 2 6. 2 . 62161 m / s 2 correct Explanation: For a simple harmonic oscillator, the dis placement is x = A cos( 2 T t + ) So the acceleration is a = d 2 x dt 2 = A ( 2 T ) 2 cos( 2 T t + ) The maximum acceleration is 4 2 A T 2 . Question 2, chap 15, sect 1. part 1 of 1 10 points The body of a 1272 kg car is supported on a frame by four springs. The spring con stant of a single spring is 2 . 47 10 4 N / m. Four people riding in the car have a combined mass of 266 kg. When driven over a pothole in the road, the frame vibrates and for the first few seconds the vibration approximates simple harmonic motion. What is the period of vibration of the car? 1. . 783934 s correct 2. . 808518 s 3. . 833894 s 4. . 859809 s 5. . 886821 s 6. . 914267 s 7. . 942793 s 8. . 972216 s 9. 1 . 00289 s 10. 1 . 03441 s Explanation: Basic Concept: T = 2 radicalbigg m k Given: m c = 1272 kg k = 2 . 47 10 4 N / m m p = 266 kg Solution: m = m p + m c 4 = 266 kg + 1272 kg 4 = 384 . 5 kg T = 2 radicalBigg 384 . 5 kg 24700 N / m = 0 . 783934 s Question 3, chap 18, sect 5. part 1 of 1 10 points A small plastic ball (similar to a pingpong ball) with mass 2 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is totally sub merged, the tension in the thread is 0 . 028 N . The density of water is 1000 kg / m 3 and the acceleration of gravity is 9 . 81 m / s 2 . 2 g quiz 04 GAUTHIER, JOSEPH Due: Dec 5 2007, 10:00 pm 2 Determine the diameter of the ball. 1. 1 . 8924 cm 2. 1 . 96227 cm 3. 2 . 02749 cm 4. 2 . 10075 cm correct 5. 2 . 17053 cm 6. 2 . 249 cm 7. 2 . 3212 cm 8. 2 . 40155 cm 9. 2 . 47686 cm 10. 2 . 55686 cm Explanation: V sphere = 4 3 r 3 = 4 3 parenleftbigg d 2 parenrightbigg 3 = 1 6 d 3 . Let : w = 1000 kg / m 3 , g = 9 . 81 m / s 2 , T = 0 . 028 N , and m = 2 g = 0 . 002 kg . Consider the forces acting on the PingPong ball B mg T Applying Archimedes principle, we have B = w f = m f g = w V ball g = 1 6 w g d 3 . Applying summationdisplay F y = 0 to the ball, B mg T = 0 1 6 w g d 3 mg T = 0 . Then we have d = 3 radicalBigg 6 ( T + mg ) w g = 3 radicalBigg 6 [0 . 028 N + (0 . 002 kg) (9 . 81 m / s 2 )] (1000 kg / m 3 ) (9 . 81 m / s 2 ) 100 cm m = 2 . 10075 cm . Question 4, chap 16, sect 2. part 1 of 1 10 points Consider the sinusoidal wave pictured in the figure....
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This homework help 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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